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1 . One can show that it converges ( but not absolutely ) for all z with l z l 1 except for the point z 1 . ( For fixed z =f 1 with z + z 2 + . + zk lzl 1 , the ( moduli of the ) partial sums ak k 1 + (z - z ) / ( 1 - z ) are bounded ( by 2/ ( 1 1 - z l ) ) . An application of Dirichlet's Test ( applied separately to the real and imaginary parts ) gives the stated convergence. ) =
=
=
=
·
=
·
The value of R introduced above is called the radius of convergence of the power series I:: ::=:' o a n (z - zo ) n ; where we say that R = 0 if the series converges absolutely only for z zo , i.e., only for w 0, and that the series has an infinite radius of convergence if it converges absolutely for all values of z - z0 , i.e. , for all values of z . The disc D ( z0 , R) is called the disc of convergence of the power series.
=
4. 7
=
The Derived Series
The ( complex ) derivative of the typical term a n (z - z0 ) n in the power series is na n (z - zo) n - 1 • This leads to a new power series, called the derived series. We wish to show that a power series is differentiable everywhere inside its disc of convergence, and, moreover, that its derivative is got by simply differentiating term by term. To avoid notational complications, we shall consider the case zo 0 and then apply the chain rule to recover the general situation. First, however, we must establish convergence of the derived series. =
Proposition 4.7 Suppose the power series f(z) I:: ::=:' o a n z n converges absolutely for l z l < R. Then the derived series g(z) = I::::=:' 1 n a n z n - l also converges absolutely for i z l < R. =
Let z with lzl < R be given and let r satis fy lzl < r < R. Then L:: := o I a n i rn is convergent and so certainly there is some constant K > 0 such that I an i r n � K for all n ( the terms of a convergent series actually converge to zero, so the sequence of terms is bounded ) . Hence Proof.
1 na n z n - 1 1
But if we set t
=
=
n i a n i rn - 1
1 � r-
1
�
n
� 1 � r- . 1
l z/ rl, then t < 1 and so the series 2:: ::=:' 1 n t n - 1 converges
75
Analytic Functions
(to 1/ ( 1 - t ) 2 ) . It follows, by the Comparison Test, that the derived series 0 g converges absolutely, as claimed.
Now let us tackle the question of the differentiability of f . Theorem 4.4 Suppose the power series f (z) = L::= o an z n converges ahsolutely for l z l < R. Then f is differentiable at any z with l z l < R, and its derivative, f' (z) , is given by the absolutely convergent power series I::= 1 n an z n - 1 . In other words, the derivative of a power series is its derived series (inside its disc of convergence).
We have seen that the power series g( z ) = I::= 1 n a n z n - 1 con verges absolutely. We must show that (f( w ) - f( z ) ) / ( w - z) - g(z) converges to zero, as w -+ z . To do this, we use the fact that convergent series can be manipulated termwise and so this expression can be written also as a series. Next, we· split this into two parts and estimate each one separately. Our first observation, then, is that for w =/:. z Proof.
t
w n - zn f (w) - f ( z ) an - na n Z n - 1 - g(z) = w-z w z n= 1 00
=
=
L an (w n - 1 + w n - 2 z + wn - 3 z 2 + . . .
n= 1
. . . + wz n - 2 + z n - 1
_
nz n - 1 )
¢ 1 ( w ) + ¢2 ( w )
where ¢1 (w)
N
=
L an (w n - 1 + wn - 2 z + w n - 3 z 2 + . . . + wz n - 2 + z n - 1 - nz n - 1 )
n= 1
and 00
¢2 ( w )
=
L
n= N+1
a n (w n - 1 + w n - 2 z + w n - 3 z 2 + . . .
We will say something about N shortly. Let l z l < R and
c >
0 be given
76
Lecture Notes on Complex Analysis
and let r satisfy l z l < r < R. Then, for any w with l w l < r, we have
l an ( w n - 1 + w n - 2 z + . . . + wz n - 2 + z n - l - nz n - 1 ) I _.:::: l an l ( l w n - 1 1 + l w n - 2 z l + . . . . . . + l wz n - 2 1 + l z n - 1 1 + n l z n - 1 1 ) 1
_.:::: l a n l 2 nrn - .
The series I::= 1 n I an i r n - l converges (the series for g converges absolutely for I z I = r ) and so we may choose N sufficiently large that I ¢ 2 ( w ) I , the modulus of the second term on the right hand side of ( *) (the tail) , is less than c/ 2 (it is bounded by the tail of a convergent series) . Fix N so that this is so . Next, we consider ¢ 1 ( w ) , the first term on the right hand side of ( * ) · This i s a sum o f N terms, each of which tends t o zero a s w --+ z . I t follows that there is �' > 0 such that l ¢ 1 ( w ) l < c/2, whenever lw· - z l < �'. Now we piece these two arguments together. Let � = min{�', r - l z l } . Then, if 0 < l w - z l < �. we have that 0 < l w - z l < �' and also that l w l _.:::: l w - zl + l z l < ( r - l z l ) + l z l r . Therefore =
l
2 = �(z ) - g (z ) l
f(w
=
l ¢ 1 ( w ) + ¢2 ( w ) l
_.:::: l ¢ t ( w ) l + l ¢ 2 ( w ) l
< 21 c + 21 c
=
c, 0
whenever 0 < l w - z l < �' and the proof is complete.
L::=o an ( z - zo ) n Corollary 4.3 Suppose that the power series f ( z ) converges absolutely for l z - zo l < R. Then f is differentiable at each z E D ( zo , R) with derivative f ' ( z ) = I::= 1 n a n ( z - z0 ) n - 1 . =
Let h ( w ) L::= o a n w n . By hypothesis, the power series for h ( w ) converges absolutely for all l w l < R. In particular, h is differentiable with derivative h ' ( w ) I::= 1 nan w n - 1 , for l w l < R. Let ,P ( z ) = z - zo . Then '¢ is differentiable and ,P ' ( z ) 1 for all z . By the chain rule, h o 'ljJ h ( '¢( z )) is differentiable for z with l '¢( z ) l < R and, for such z, its derivative i s given 0 by ( h o ,P) ' ( z ) h'( ,P ( z ) ) ,P '( z ) , as required . Proof.
=
=
=
=
=
L::= o a n ( z - zo ) n Corollary 4.4 Suppose that the power series f ( z ) converges absolutely for z E D ( zo , R) . Then for any k E N, f is k-times =
77
Analytic Functions
differentiable with k th _derivative given by f ( k ) (z)
00
=
L an n ( n - 1 ) . . . ( n - ( k - 1) ) ( z - zo) n - k ,
n= k
where this last series converges absolutely for z E D ( z0 , R) . In particular, f ( k ) (zo)
=
k! a k .
The proof is by induction on k . We know, by corollary 4.3, that the result is true for k 1 . Suppose it is true for k = m. Write
Proof.
=
g(z)
=
f (m) ( z )
00
=
L an n ( n - 1 ) . . . (n - (m - 1 ) ) (z - zo t - m
n= m 00
=
L am +j ( m + j ) (m + j - 1 ) . . . (j + 1 ) ( z - zo)i .
j= O
By corollary 4 . 3, g is differentiable at z E D ( z0 , R) with derivative given by the absolutely convergent power series
g' ( z )
00
=
L am +J ( m + j) ( m + j - 1 ) . . . (j + 1 )j ( z - z0 )i - 1 •
j= l
Relabelling, the result follows for k = m + 1 , and, by induction, the proof of the formula fo r f ( k ) (z) is complete. Setting z z0 completes the proof since only the first term in the series for f ( k ) (z0) survives. 0 =
What ' s going on? These results tell us that power series behave very much like polynomials--as long as we stay inside their discs of convergence. The behaviour on the boundary of these discs varies from power series to power series and can be very complicated.
4.8
Identity Theorem for Power Series
We will need the following result later on. Theorem 4 . 5 (Identity Theorem for Power Series) Suppose that n the power series f ( z ) = I:;�=O a n ( z - zo) and g( z ) = I:�=O bn ( z - zo) n both converge absolutely for all z E D ( zo , R) . Suppose, further, that there is some sequence ( (k ) in D ( zo , R) , with (k =/:. zo for all k, such that (k --+ zo as k --+ oo, and such that f ( (k ) g( (k ) for all k. Then a n bn for all n , that is, f g . =
=
=
78
Lecture Notes on Complex Analysis
a n - bn · Then Write h (z) L;�= O en (z - zo) n , where Cn h ((k ) 0 for all k. Suppose that Cm is the first non-zero coefficient of h, i.e., Cn 0 for n < m and Cm =/:. 0. In this case, we can write h as h(z ) = ( z - zo ) m ( Cm + Cm+ l ( z - zu ) + ). cp( z ) Proof.
=
=
=
=
·
· ·
0, for 0 but (k =/:. zo and so we must have that
1 , we see that the term in brackets on the right hand side above is bounded above by L::'=l 1 /2m = 1 and so k 1
L� j=O J .
for any k E
N.
<e
0 .
Suppose that 0 < x :-::; 1 . Then
x 2 x4 x6 cos x = 1 - I + I - 6 ,. + . . . 4. 2. x 2 x4 xB > 1 - 2! - 4! - 6! - . . . 1 1 1 >1- -... 2 23 25 ' x4 1 1 x2 < 1 1 < < 3 , . . . etc. , since - 2 ' 4 . - 4. 3 .2 2 2! --
=1-
1
2
(1 -
�)
=
1
3 .
Hence sin' x = cos x > i on [0, 1] and so (by the Mean Value Theorem) , sin x is strictly increasing on [0, 1] . But sin O = 0 and therefore sin x > 0 for 0 < x :-::; 1 . It follows that sin x = 2 sin � cos � > 0 whenever 0 < x :-::; 2 , since, in this case, sin � > 0 and cos � > i ·
D
The Complex Exponential and Trigonometric Functions
Corollary 5 . 5
87
The function cos x is strictly decreasing on [0, 2] .
We have cos ' x = - sin x which is strictly negative on (0, 2 ) , by the theorem. By the Mean Value Theorem, it follows that the function D cos x is strictly decreasing on [0, 2] .
Proof.
Theorem 5 . 3 Proof.
sin 4 < 0.
From the definition of sin x, 43 45 47 49 sin 4 = 4 - - + - - - + 9! 5! 7! 3! 2 11 13
0 for x E (0, 2 ) and so, in particular, sin 1 > 0. We have also shown that sin 4 < 0. Now, the map x �----> sin x is continuous on IR, and so, by the Intermediate Value Theorem, there is some real number, which we will denote by 1r, with 1 < 1r < 4 and such that sin 1r = 0. We must now show that 1r is the only value in (0, 4 ) obeying sin 1r = 0. To this end , suppose that sin a = 0 with 0 < a < 4. Then
0
=
sin a = 2 sin � cos � .
Hence either sin � = 0 or cos � = 0 ( they cannot both vanish because the sum of their squares is equal to 1 ) . But if 0 < a < 4, then 0 < � < 2 and we know that sin x > 0 on (0, 2), so sin � cannot be zero. Hence, we must have that cos � = 0 . In particular, cos � = 0 . However, cos x is strictly decreasing on [0, 2] and so there can be at most one solution to cos x = 0 in this interval. Hence � = � so that a = 1r , and the uniqueness is established . Next, sin2 z + cos2 z = 1 implies that sin 2 � = 1 , since cos � = 0. Hence sin � = 1 because sin � > 0 ( since 0 < � < 2 ) . Finally, we have cos 1r = 2 cos2 � - l = - 1 . D
88
Lecture No tes on Complex Analysis
What 's going on? blue
as
it
were, by
We have defined the trigonometric functions, out of the means of power series. This approach avoids any app eal to
geometry and right-angled triangles. However, having chosen this route, stick with it. In particu lar , it is necessary to get to
1r
we
must
via these definitions rather
than by drawing trian gles or circles . As we have seen , t h is is all perfectly possible. Our view here is that the trigonometric fu nctions are A ny results must be deduced
as
as
we have defined them.
consequences of these power series definitions .
Remark 5 .6 Numerical investigation yields 1r 3 .14159 . . . . It is known that 1r is irrational ( in fact, transcendental ) . It is something of a sport ( involving some fascinating numerical analysis ) to calculate the value of 1r to a large number of decimal places and this has been done to over a million decimal places. Such programs have been used to test the computational integrity of supercomputers by checking to see whether they get these digits correct or not . =
Theorem 5 . 5
For any z E C and n E Z , sin ( z + n1r ) cos ( z + n1r )
Proof.
=
=
( - 1 ) n sin z ( - 1 ) n cos z .
We use the trigonometric formulae; sin ( z + 1r )
= =
sin z cos 1r + cos z sin 1r - sin z + 0,
and cos ( z + 1r )
cos z cos 1r - sin z sin 1r = - cos z - 0.
=
For n > 0, the result now follows by induction. Substituting z the result then follows for n < 0.
=
w - n1r,
D
Remark 5 . 7 Particular cases o f the above formulae deserve mention. If we set z = 0, then we see that sin( n1r ) = 0 and cos ( n1r ) = ( - 1 ) n , for any n E z . The functions sin and cos are periodic ( with period 27r ) . In particular, for any x E lR and any n E Z,
sin ( x + 2n7r )
cos ( x + 2n7r )
=
=
sin x cos x.
and
The Complex Exponential and Trigonometric Functions
89
If we piece together all the information gained above, we recover the familiar picture of sin x and cos x, for x E IR, as periodic 'wavy' functions. In fact, sin x is an odd function, so it is determined by its values on x � 0. It is periodic, so it is determined by its values on [0 , 21r] . But, sin(x + 1r) sin x cos 1r + cos x sin 1r = - sin x and so sin x is determined by its values on the interval [0 , 1r] . Now, =
(
sin �
±
x
)
=
sin � cos x ± cos � sin x
=
cos x,
so we see that sin x is symmetric about x = l It follows that sin x is completely determined by its values on [0 , � ] . Furthermore, sin (x + �) = cos x and so the graph of cos x i s got by translating the graph of sin x by � to the left. 5.7
Inverse Trigonometric Functions
From the analysis above, we see that sin x is strictly increasing on the interval [- � , �] , cos x is strictly decreasing on the interval [0, 1r] and is strictly increasing on the interval [ -1r, 0] . In particular , this means that sin is a one-one map of [- � , � ] onto [- 1 , 1 ] , cos is a one-one map of [0, 1r] onto [- 1 , 1 ] and also of [-1r, 0] onto [-1 , 1 ] . For t E [- 1 , 1] , let ¢( t ) be the unique element of [- � , � ] such that sin ¢(t) t, let 'lj; (t) be the unique element of [0 , 1r] such that cos 'lj; (t) = t and let p(t) be the unique element of [-1r, OJ with cos p(t) = t. Thus, ¢ is the inverse of sin : [- � , � ] --+ [- 1 , 1 ] , 'lj; is the inverse of cos : [0, 1r] --+ [- 1 , 1] and p is the inverse of cos : [-1r, OJ --+ [ - 1 , 1 ] . The standard inverse trigonometric functions sin- 1 and cos- 1 are given by sin- 1 (t) = ¢(t) and cos- 1 (t) = '!f; (t) for t E [- 1 , 1] so that sin - 1 takes values in [- � , �] whilst cos- 1 takes values in [0 , 1r] . =
Suppose that f : [a, b] --+ [c, d] is a strictly increasing {or decreasing) continuous map from [a , b] onto [c, d] . Then the inverse map f - 1 : [c, d] --+ [a, b] is continuous. Theorem 5 .6
Proof. The idea of the proof is straightforward , but it is a nuisance having to consider the end-points c and d of [c , d] and the interior (c , d) separately. To avoid this, we shall first consider the case of f : IR --+ R Suppose, then, that f : IR --+ IR is, say, strictly increasing and maps IR onto R Let Yo E IR and £ > 0 be given. Let x0 E IR be the unique point such that f ( xo ) = Yo , thus, xo = f - 1 (Yo ) . Set Y 1 = f ( xo - £ ) and Y2 = f ( xo + c) ,
90
Lecture Notes on Complex Analysis
X Fig. 5 . 1
Continuity of the inverse function.
as shown in Fig. 5.1 . Then Y 1 < Yo < Y2 · Furthermore, if Y 1 < y < Y2 , then xo - € < f- 1 (y) < Xo + €. Putting o = min{yo - Y 1 , Y2 - Yo } we see that I Y - Yo i < 8 implies that j ! - 1 (y) - f- 1 (yo) j < €. That is, f- 1 is continuous at Yo and therefore on all of JR. The case where f is strictly decreasing is similarly proved. Alternatively, one can note that f- 1 (y) = ( - f) - 1 (-y) and that -f is increasing if f is decreasing. Returning now to the case where f is strictly increasing and maps [a , b] onto [c, d] , we simply extend f to the whole of JR. Define F : lR --+ lR by
F(x)
=
{
x + f(a) - a , x < a a�x�b f(x) , X + f(b) - b, b < X.
Then F maps lR onto JR, is strictly increasing, and is equal to f on [a, b] . As above, we see that F - 1 is continuous. In particular, F- 1 = f - 1 is 0 continuous on [c , d]. Theorem 5 .7 Proof.
Each of the functions ¢,
1/J ,
and p is continuous.
This follows immediately from the preceding theorem.
0
The Complex Exponential and Trigonometric Functions
5.8
91
More o n exp z an d the Zeros o f sin z an d cos z
Having established the familiar properties of the real trigonometric func tions , we return to a discussion of the complex versions. We attack these via the exponential function, which determines them all.
The equality exp z = exp w, for z, w E C, holds if and only if there is some k E Z such that z = w + 27rki .
Proposition 5 . 3
Multiplying both sides by exp( -w) , we see that exp z = exp w if and only if exp(z - w) = 1 . Now, exp(27rki) = cos 27rk + i sin 27rk = 1 and therefore exp z exp w if z - w 27rki. Conversely, suppose that exp(z - w) = 1 , and write z - w = a + i{3, with a, {3 E JR. Then
Proof.
=
=
1 = exp(z - w) = exp(a + if3) = exp a exp if3 = exp a (cos f3 + i sin {J) . It follows that exp a = 1 (taking the modulus of both sides) , and so cos {3 = 1 and sin {3 = 0 (equating real and imaginary parts) . This implies that a = 0 and {3 is of the form {3 27rk, for some k E z . It follows that D z - w = 21rki, for some k E Z, as required. =
We have extended the definition of the functions sin x and cos x from the real variable x to the complex variable z . It is of interest to note that these complex trigonometric functions have no new zeros, as we show next. Proposition 5 .4 For z E C, sin z = 0 if and only if z 1rk for some k E Z, and cos z = 0 if and only if z = (2k + 1 H , for some k E Z. =
Proof.
We have sin z = 0 - -
- -
-
as claimed .
exp( iz) - exp( -iz) =0 2i exp( iz ) = exp( - iz) exp(2iz) = exp O 2iz = 0 + 21rki, for some k E Z, z = 1rk, for some k E Z,
92
Lecture Notes on Complex Analysis
A similar argument is used for cos z. Indeed, cos z = 0
exp(iz) + exp(-iz) =0 2 exp( iz) = - exp( -iz) exp(2iz) = - 1 exp(2iz) = exp(irr) , since exp(irr) = - 1 , 2iz = irr + 2rrki, for some k E Z, z = (2k + 1 H , for some k E Z,
and the proof is complete. 5.9
D
The Argument Revisited
We know, informally, that any non-zero complex number can be written as r( cos 9 + i sin 9) = rei8 , where r is its modulus and 9 is some choice of argument (angle with the positive real axis) . We shall establish this formally and also consider the argument mapping in more detail. Suppose, then, that z =/: 0. Write z = a + ib, with a, b E JR. Then r = j z j .../a 2 + b2 =/: 0 and z r(a + i,B) where a = ajr, ,8 bjr. Clearly, a 2 + ,8 2 1 and so ja j � 1 and j,Bj � 1 . We would like to show that it is possible to find 9 such that a = cos 9 and ,8 = sin 9. Since j,Bj � 1 , there is a unique 9 E [- � . �] such that sin O = ,B. Now , cos 2 9 = 1 - sin 2 9 = 1 - ,8 2 = a 2 • It follows that cos 9 = ±a. If cos 9 a, we have z rei8 and we are done. (Note that cos 9 � 0 since - � � 9 � � . ) If not , we must have cos 9 = -a. Let 9' = 71' - 9. Then, using the trigonometric formulae above, we find that sin 9' sin 71' cos 9 - cos 71' sin 9 = sin 9 and cos 9' = cos 71' cos 9 + sin rr sin 9 = - cos 9 = a. Therefore, in this case, we ' can write z as z = re;o . Consider the family of complex numbers given by z (t) e 2 11'it cos 2rrt + i sin 2rrt, for t � 0. Clearly, j z (t) j = 1 and z(O) 1 . As we continuously increase t, the complex number z (t) moves continuously anti clockwise around the circle centred on the origin and with radius one. For -1 , t = t . we find z ( t ) = ( 1 + i) / ,;2. Also , we have z ( -! ) i , z ( ! ) z ( � ) = - i and z(1) = 1 . We have seen that exp z exp w if and only if there is some integer k E Z such that w = z + 2rrki. It follows that we can always write z = rei8 ==
=
==
==
=
=
=
=
=
=
=
=
=
93
The Complex Exponential and Trigonometric Functions
Fig. 5.2
The point z(t)
=
e 2" it moves anticlockwise around the unit circle.
where the argument () E IR is only defined up to additional integer multiples of 21r, i.e . , z rei0 = rei>. if and only if .X = () + 21rk for some k E Z. In particular, we can always find a unique value for () in the range ( -1r, 1r] . (The difference between successive possible values for () is 271", so there must be exactly one such value in any open-closed interval of length 21r.) This value of the argument is given a special name, as already discussed (though somewhat informally) . =
Definition 5 . 3 For any z # 0, the principal value of the argument of z is t li.e unique real number Arg z satisfying -7r < Arg z � 1r and such that z = l z l ei Arg z . (Arg z is not defined for z = 0.)
Notice that if z is close to the negative real axis, then its imaginary part is small and its real part is negative. If its imaginary part is positive, then the principal value of its argument is close to 1r, whereas if its imaginary part is negative then the principal value of its argument is close to -1r. In the example above, the principal value of the argument of the complex number z (t) increases from 0, when t 0, through � ' when t -!, to 1r, when t � and z = - 1 . However , as z(t) crosses the negative real axis, from above to below, the principal value of its argument j umps from 1r to "nearly" -7r (it never assumes the value -1r ) . (The limit from above is 1r , whereas the limit from below is -1r.) It continues to increase, as t increases , until it has the value 0 , when t = 1 and z = 1 . The negative real axis is a line of discontinuity for Arg z . (Recall that Arg z is not defined if z 0.) =
=
=
=
94
Lecture Notes on Complex Analysis
Arg z
=
95
100 rr
'
Fig. 5.3
5.10
'
'
.... - - - - "'
/
/
I
I I I
Arg z is discontinuous across the negative real axis.
Arg z is Continuous i n the Cut-Plane
We shall show next that Arg z is continuous everywhere apart from on the negative real-axis. Theorem 5 . 8 The map z f---> Arg z is a continuous mapping from the cut-plane C \ { z : z + l z l 0 } onto the interval ( -rr, rr ) . =
Proof. Let z E C \ { z : z + l z l 0 } , i.e. , z is any point of C not on the negative real axis (including 0 ) . Write z x+ i y = r(cos Arg z + i sin Arg z ) , where r l z l . Consider first the region where Im z y = r sin Arg z > 0 , that is, the upper half-plane. Then sin Arg z > 0 and s o 0 < Arg z < rr. But r cos Arg z = x and so Arg z = ,P(xjr) where 'ljJ is the inverse to cos : [0 , rr] -+ [ - 1 , 1] . But this map is continuous, by theorem 5 .7, and so Arg z is continuous on the upper half-plane { z : Im z > 0 } Next consider the region x = Re z > 0 (right half-plane) . For such z , we have cos Arg z > 0, so that - � < Arg z < � . However, r sin Arg z y and therefore Arg z ¢(yjr) , where ¢ is the inverse to sin : [- � , � ] -+ [- 1 , 1] . Again, by theorem 5. 7, we conclude that Arg z is continuous on this region, namely the right half-plane { z : Re z > 0 } . Finally, consider the region with y = I m z < 0 . Here, sin Arg z < 0 , so that -rr < Arg z < 0. Since r cos Arg z = x, it follows that Arg z p(x/r) , where p is the inverse to cos : [ -rr, O] -+ [-1 , 1] . Again, by theorem 5.7, we see that Arg z is continuous on { z : Im z < 0 } , the lower half-plane. We conclude that z f---> Arg z is continuous on C \ { z : z + l z l = 0 } · =
=
=
=
=
=
=
95
The Complex Exponential and Trigonometric Functions
If z -:/- 0 does not lie on the negative real-axis, then Arg z -:/- rr and so Arg z E ( - rr , rr ) . On the other hand, for any - rr < () < rr, we have Arg( cos () + i sin B) = (). It follows that z �----> Arg z maps C \ { z : z + l z l = 0 } D onto the interval ( - rr , rr ) . What's going on? plane C
\ {z : z
+
The claimed continuity of the function Arg
lzl
=
0}
is clear from a diagram.
z
on the cut
However, our starting
point has been the power series definitions of the trigonometric functions and the function Arg
z
is formally constructed in terms of these
( via
"obvious"
)
behaviour.
) ( but
suit able inverses .
This has meant that we have had to do some work to wring out the required
Chapt er 6
The Complex Logarithm
6.1
Int roduction
The logarithm is an inverse for the exponential function: if x = e t , then In x t, where t E JR. Note that the natural logarithm In x is very often also written as log x. C an we mimic this construction to get a logarithm for complex variables? If z E C , we want log z w , whatever it turns out to be, to satisfy the relation e w = z . To see how we might proceed, write z as z = x + iy = r e i 0 , with r = Jx 2 + y 2 and note that (} is not uniquely determined-we can always add 27rk, for any k E z. Nevertheless, suppose that we have made a choice for 0. Then we want to construct log z log ( rew ) . We try the formula =
=
=
log z
=
log ( r e ; 0 ) = log r .......,..,
usual log
=
ln r
+ +
log e i0 ...__.,
undefined as yet
iO
as an apparently reasonable attempt. Then we find that whatever our choice for (}
since r > 0 and so
e 1n r
= r. In fact , if w = In l z l + iO + i27rk then ew
= =
=
=
e 'n l z l + i0 +i27r k e 'n l z l e iO e i 27r k l z l eiO e i27rk z,
97
=
z e i27r
k
98
Lecture Notes on Complex Analysis
for any k E Z. In other words, for given z =f. 0, the equation
has infinitely-many solutions, namely,
w
=
ln l z l + i arg z ,
where arg z is any real number such that z l z l ei arg z . For z = 0 , the equation becomes ew = 0, which has no solution. (We know that the complex exponential function is never zero.) These heuristics suggest that setting up a theory of complex logarithms might be a little more involved than for the real case. =
6.2
The Complex Logarithm and its Properties
Definition 6 . 1 For z E C \ { 0 } , we say that a logarithm of z is any particular solution w to ew = z .
I f W I and w2 are solutions to ew = z , then ew1 = z = ew2 and so we see that ew1 -w2 = 1 . Putting WI - w 2 = a + ib, with a, b E JR , this becomes e a e1b = 1 and so a = 0 and b = 21rk, for some k E Z. In other words , the difference between any two possible choices for the logarithm is always an integer multiple of 27ri. The arbitrariness of the complex logarithm reflects the ambiguity in the choice of the argument of a complex number. When there is no chance of confusion, one often j ust writes log r to mean the usual real logarithm of any positive real number r. Definition 6 . 2 The principal value of the logarithm of z =f. 0 is that obtained via the principal value of the argument and is denoted Log z; thus
Log z = ln I z I + i Arg z . We see that -1r < Im Log z � 1r. Moreover, since possible choices o f the argument must differ by some integer multiple of 27r, it follows that any choice of logarithm of z , log z , can be written as log z = Log z + i27rk, for some k E z. Of course, k may depend on z . Various properties o f any such choice , log z , o f the logarithm o f z are considered next.
The Complex Logarithm
99
Suppose that for each z E C \ { 0} a value for log z has been chosen. Then the following hold.
Theorem 6 . 1
(i) elog z = z . (ii) log(e z ) = z + 2 7r ki, for some k E Z . (iii) log(z1 z2) = log Z 1 + log z2 + 21rki, for some k E Z . (iv) log
(;.) = - log z + 27r ki, for some k E
z.
Proof. By definition, any choice w log z o f the logarithm o f z satisfies ew = z , which is (i) . To prove (ii) , let w = log( e z ) be any choice of the logarithm of e z . Then ew = e z . It follows that w - z = 21rki, for some k E Z, as required. Let w 1 = log z1 and w2 = log z2 be any choices of the logarithms of z1 and z2 , respectively, and let w 3 = log(z1 z2) be some choice for the logarithm of the product z 1 z2. By the definition, =
It follows that there is some k E Z such that w 3 = w 1 + w2 + 21rki , which proves part (iii). Finally, suppose that w is a choice of log � and let ( be any choice of log z . Then ew = 1 /z and e( = z. It follows that z = 1 /ew and therefore
We deduce that ( = - w + 27rki , for suitable k E Z, which proves (iv) .
0
Examples 6 . 1
( 1) Possible choices o f the logarithm o f 1 are log 1 = 0, o r 27ri, o r 47ri , or . . . , or -2 7ri, or -47r i , or . . . . ? (2) log 1 = log 1 2 � log 1 + log 1 . This is only true if we make the choice log 1 0. No choice log 1 = 21rki , with k E Z, k -:/- 0, will work. ( 3) Consider the principal value of the logarithm of the product i ( - 1 + i ) . =
We have
Log i( - 1 + i)
ln l i ( - 1 + i) l + i Arg i ( - 1 + i) = ln v'2 - i 3_; . =
100
Lecture Notes on Complex Analysis
Now, Log i = ln I ii + i Arg i = 0 + i � and Log( - 1
+
i) is given by
Log( - 1 + i) = ln l -1 + il + i Arg( - 1 + i) = ln v'2 + i 3,r . We see that Log i + Log( - 1 + i)
=
i � + ln v'2 + i 3,r
# Log i ( - 1 + i) .
=
ln v'2 + i 5;
( 4) For which values of z is sin z ei z , so 3 ? To solve this, put w 1 that sin z (w - w- ) /2i. Then sin z = 3 becomes the quadratic equation w 2 - 6iw - 1 0 with solutions w = (3 ± 2 v'2)i. But ei z = w means that iz is a choice of log w , that is , iz must be of the form iz = Log w + 2k1ri for some k E Z. Hence iz ln 1 3 ± 2 v'2 1 + i � + 2kni and so z = -i ln 1 3 ± 2 J2 1 + � + 2k1r for k E Z . =
=
=
=
=
Remark 6. 1
If z is real and positive, z = x + iy , x > 0 and y = 0, then Log x
=
=
Log z = ln I z I + i Arg z = ln x + iO ln x .
In other words , on the positive real axis, { z : z real and strictly positive } , the principal value o f the logarithm agrees with the usual real logarithm. 6.3
Complex Powers
For real numbers , a and b, with a > 0 , the definition of the power ab is given by ab = exp(b log a) . Now that we have a meaning of the logarithm for complex numbers we can try to similarly define complex powers. Definition 6 . 3
For given z, ( E C, with z # 0 , w e define the power z'
=
exp(( log z ) .
Evidently, z ( depends o n the choice o f the logarithm log z . That is, we must first make a choice of log z before we can define z( . Put another way, different choices of log z will lead to different values for z( . The principal value of zC , for z # 0, is defined to be exp( ( Log z ) .
101
The Complex Logarithm
Examples 6 . 2 (1)
1
What are the possible values of 8 3 ? We have
8t
=
=
e � log S = e � (ln 8+ 2 7rki)
�
e 'n 2+ 1rki
=
2e
� 1rki
for k E Z. Taking k 0, 1 , 2 gives all the possibilities ( further choices of k merely give repetitions of these three values ) . (2) The possible values of i i are =
ii
=
=
=
e i log i
=
ei.,./2
e i ( L og i+2k7ri ) ,
for k E
e i( i f + 2 k7ri ) e - � - 2 k11" ,
which are all real. (3) Taking principal values, e � L og( - 1 )
=
and
for k E Z ,
( -i ) 1 12
i 1 12
=
z,
=
e � L og i
e � L og ( - i ) =
ei.,./ 4
'
=
e - i.,./ 4 , ( - 1 ) 1 /2
so that
This provides an example of complex numbers
w
and ( for which
where the principal value of the square root is taken. Remark 6.2 There are one or two consistency issues to worry about. We seem to have two possible meanings of z m when m E Z , namely, as the product of z with itself m times ( or the inverse of this if m is negative ) or as the quantity exp ( m log z ) , for some choice of log z . In fact, there is no need to worry. Let log z be any fixed choice of the logarithm of z ( where z -:/- 0) . Suppose that m E N. Then
exp ( m log z )
=
exp ( log z ) . . . exp ( log z ) m factors
=
�,
m factors
Lecture Notes o n Complex Analysis
102
which shows that exp(m log z) reduces to the usual "product of m terms" meaning of z m . For negative m, set k -m. Then, as above, =
exp(m log z)
=
exp( - k log z)
1
=
=
=
1 exp (k 1 og z )
k terms z - k , usual meaning z m , usual meaning.
For m 0, we have exp( m log z) exp 0 = 1 = z0 which is the usual meaning of a number to the zeroth power . Another concern is with our agreed notation ez for exp z . Does this conflict with the meaning of the real number e raised to the complex power of z? From the definition, we see that ez = exp(z log e) with some choice of the logarithm being made. Now , any such choice has the form log e = Log e + 2rrki , for some suitable k E z. This means that we can always write exp(z + 2zrrki) = exp z exp(2zrrki) , since ez as exp(z(Log e + 2rrki) ) Log e 1 . For this to equal exp z , we must insist that exp(2zrrki) = 1 , that is, 2zrrki 2rrmi for some m E Z . I n general, this is only possible when k 0 , in which case m = 0. We come to the conclusion that the complex power ez agrees with exp z provided that we always take the principal value of the power . We shall adopt this convention if there is any doubt. In fact , ez is really only usually used as a notational shorthand for exp z. So we can choose not to use it,· and always use exp z instead, or to use it and remember exactly what we are doing, or use it and always interpret it as the principal value of the power. There is unlikely ever to be any confusion. =
=
=
=
=
=
Proposition 6. 1 Suppose that z -1- 0 and for m E N let ( ( m = z, that is, ( is an mth root of z . Proof.
=
z 1 / m . Then
We have that ( = exp( ;k log z) for some choice of log z . But then (m
as required.
=
1 1 exp( - log z) . . . exp( - log z) = exp log z m m m terms
=
z, 0
103
The Complex Logarithm
Example 6.3 Let a be any choice of yCI (i.e . , of ( - 1) ! ) and let b be any choice of y'l (i.e . , of d ) . Then
/-1 VT
{1 Y�
=
=
yCI
= a.
But
yCI y'l
For equality, i.e., for
� and y'l b yCI
=
=
�. a
R = R = Jl f£ r,
y1
11 =
y -1
- , we require that
-1
This, in turn, requires that a 2 b2 o r - 1 = 1 , which evidently can never hold (no matter what choices a and b are made) . We conclude that whilst =
since both are j ust some choice of yCI, nevertheless
This means that the "equalities"
simply cannot ever both be simultaneously true. 6.4
Branches of the Logarithm
We have seen that log z depends on a choice for arg z. A natural question to ask is whether or not there is some consistent choice of arg z so as to make z �---> log z continuous. The answer depends on where exactly we want to define log z , i .e. , its domain of definition. For some regions, for example an annulus around the origin such as { z : 1 < l z l < 2 } , this cannot be done, as we will show below .
104
Lecture Notes on Complex Analysis
Proposition 6 . 2 The choice z 1--+ Log z of the logarithm on the cut plane C \ { z : z + l z l = 0 } .
is
continuous
Proof. For z E C \ { z : z + I z I = 0 } , Log z = ln I z I + i Arg z . Now, we have seen that Arg z is continuous at each such z and so is z �-----+ l z l and 0 hence also z 1--+ ln lz l . The result follows.
Example 6.4 There is no choice of logarithm making the map z 1--+ log z { z : l z l = r } . To see this, continuous everywhere on the circle C suppose that z 1--+ f(z) is such a choice of logarithm, for z E C. We know that z 1--+ Log z is continuous for z not on the negative real axis , and so the map z 1--+ f(z) - Log z is continuous for z E C \ { -r } . e f(z) , by definition of a logarithm. But z = e Log z Now, we have z e L og z for z E C. It follows that / (z) Log z + 21rik(z) and so e f(z) for some k(z) E Z, depending possibly on z E C. However, both f(z) and Log z are continuous on C \ { - r } and so , therefore, is their difference f(z) - Log z = 27rik(z) . r (cos t + i sin t) = reit . Then the map For -1r < t < 1r, set z(t) t �---+ k(z(t) ) is a continuous map from ( - 1r , 1r ) -+ Z and so must be constant (by the Intermediate Value Theorem) . We conclude that there is some fixed k E Z such that =
=
=
=
=
f(z) = Log z + 27rik = ln r + i Arg z + 21rik for all z E C \ { -r} . However, the left hand side of ( *) is continuous at each z E C, by hypothesis, whereas the right hand side has no limit as z approaches - r . (If z approaches -r from above (i.e . , through positive imaginary parts) then Arg z converges to 1r, but if z approaches -r from below, then Arg z converges to - 1r . ) This contradiction shows that such a continuous choice of logarithm on C cannot be made . Definition 6.4 A branch of the logarithmic function is a pair (D, /) , where D is a domain and f : D -+ C is continuous and satisfies exp f(z) z for all z E D . (Note that D cannot contain 0 since the exponential function never vanishes .) The principal branch is that with D C\{ z : z+ l z l 0 } and f(z) = Log z. =
=
=
By suitably modifying Log z in various regions of the complex plane, we can construct other branches of the logarithm . Example 6.5 Let D C \ { z : z - l z l = 0 } , the complex plane with the positive real axis (and {0}) removed. We define f(z) , for z E D, in terms =
The Complex Logarithm
105
of Log z , as indicated in Fig. 6 . 1 . For given z E D, set
f (z ) =
{
Log z ,
Log z + 27ri ,
Im z � 0 Im z < 0 .
Notice that f ( z ) takes the value Log z for z on the negative real axis . Log z
Log z
Log z 0
-
Log z + 27l"i
Fig. 6 . 1
- - - \: �e�oved Log z + 27l"i
A branch of t h e logarithm o n C \ { z
: z
-
lzl
=
0 }.
It is clear that f, as defined here, is continuous at any z not on the real axis . The function Log z j umps by -27ri on crossing the negative real axis from above to below, so the construction of / , via the addition of an extra 27ri in the lower half-plane, ensures its continuity, even on the negative real axis. Example 6.6 Let S be the set S = { z : z = t + it (t - 1 ) , t � 0 } and let D C \ S. By way of example, we seek a branch (D, f) of the =
Log z + i 671"
f( -2)-ln 2 H 771" Log z + i 8 71" Fig. 6.2
Log z + i 8 71"
A branch of log z on D with f( - 2) = In 2 + i 771".
106
Lecture Notes on Complex Analysis
logarithm on the domain such that f( - 2) ln 2 + i 7-rr . Notice that D is the complement of part of a parabola. (Let x t and y = t (t - 1) so that y x ( x - 1) for x � 0 . ) The branch (D, f) is as indicated in the Fig. 6 .2. Note that f(z) i s equal t o Log z + i 61r on the negative real axis (dashed) . =
=
=
Remark 6.3 If (D, f) is a branch of the logarithm, then, for any fixed k E Z, so is (D, g) , where g(z) f(z) + 21rki , for z E D. Indeed, g exp f(z) exp 27rki is continuous, and exp g (z) exp f (z) z . The converse is true as we shall show next. =
=
=
=
Suppose (D, f) and (D, g) are branches of the logarithm on the same domain D . Then there is k E Z such that
Theorem 6.2
g(z)
=
f(z) + 21rik
for all z E D (-the same k works for all z E D) . Proof.
Set h(z) = g(z) - f(z) , for z E D . Then exp h (z)
exp (g (z) - f( z ) ) exp g (z) exp ( - f( z ) ) z z --1 - exp f (z) z =
=
-
·
Hence, for each z E D, there is some k(z) E Z such that h(z) = 27rik(z) . Let z 1 , z2 E D be given. Since D is connected, we know that there is some path 'Y : [a, b] --+ C in D joining z 1 to z2 . Now, h, and therefore k is continuous on D. Hence the map t �---> k('Y(t) ) from [a, b] into Z is continuous. It is therefore constant, by the Intermediate Value Theorem. Hence k("!(a) ) k ("!(b)) , that is, k(z 1 ) = k ( z 2 ) and we deduce that k is D constant on D. =
The next result tells us that any continuous choice of the logarithm is automatically differentiable and that its derivative is exactly what we would guess it to be, namely, 1 / z . Theorem 6 . 3 Let (D, f) be a branch of the logarithm.. Then f E H(D) and f' (z) = 1 / z , for all z E D .
z , for z E D , means that 0 rJ_ D . Proof. First we recall that exp f ( z) Also, i f z -=/- w, then f ( z ) -=/- f( w) (since otherwise, z = ef (z) = ef ( w ) w) . =
=
1 07
The Complex Logarithm
Let z , w E D, with z =I w. Then
f(w) - f(z) w-z
- ef(w) -- ef(z) 1 ( ef(w) ef(z) ) = _
f(z)
f(w)
_
f(w) - f(z)
Now, the continuity of f implies that if w --+ z then f(w) --+ f(z ) . Further more, for any ( E C ,
e� - e< < ----+ exp' ( = e �-(
as � --+ (, with � =I (. Hence, as w --+ z
(*)
--+
( e f(z)) - 1
=
1
ef(z)
=
1 . that 1s, f E H(D) and f'(z) = - , for any z E D.
1 ;' 0
z
What's going on?
The notion of a complex logarithm is straightforward , but
complicated by the fact that there is an infinite number of ways in which it c an be
defined . This is simply a consequence of the ambiguity in the choice of the polar angle, the argument of a complex number . To talk sensibly about a logarithm requires specifying some particular choice. continuity considerations.
This done, one then inquires about
This leads to the notion of branch of the logarithm
where the region of defi nition of the logarithm is highlighted . It is not possible to make a continuous choice of logarithm in some domains - for example, in
an
annulus centred on the origin . The basic definition of a logarithm together with continuity is enough to imply its differentiability. might expect .
Its derivative is
1/ z,
as one
O n c e o n e has some notion o f logarithm, it c a n then be used to construct complex powers. As a corollary, we obtain an alternative proof of the above result on the uniqueness, up to an additive constant multiple of 27l'i , of the branch of the logarithm on a given domain .
Corollary 6 . 1
Suppose ( D, f) and ( D, g) are branches of the logarithm on the same domain D. Then there exists some integer k E Z such that g(z) = f(z) + 27l'ki , for all z E D . Since f and g are both logarithms , it follows that for each z E D there is an integer k(z) E Z such that h(z) = g(z) - f(z) = 27rk(z)i. By the theorem, both f and g are differentiable o n D and so, therefore, is h Proof.
Lecture Notes on Complex Analysis
108
and hence so is k . However , k(z) E JR, for all z E D and so k(z) is constant on the domain D . Note that one could also argue that since f and g both have the same derivative on the domain D, namely 1/ z, then their difference h satisfies h' (z) = 0 on D. This means that h is constant on D and therefore of the D form 21rki for some fixed k E z .
Example 6 . 7 Let S be the set S = { z : z = t e it , where t E JR , t � 0 } . We construct the branch (D, f ) of the logarithm on the domain D = C \ S with f(l) = Log 1 . The idea is to build on the values of Log z by compensating for the j ump in its value as z passes through the negative real axis . the curved regions enclosed between the Denote by R_ 1 , Ro , R 1 , spiral S and the negative real axis, as shown in the Fig. 6 .3, so that D = U�- 1 Rk . The section ( -1r, 0) of the negative real axis is included in R_ 1 , the section ( - 3 7f , 7f) is included in Ro and so on. •
•
•
-
Fig. 6.3
The branch of logarithm on D with f ( l )
= 0.
The Complex Logarithm
109
The function f is defined on the domain D by setting
f(z) = Log z + 2k7ri ,
for z E Rk , k = - 1 , 0, 1 , 2, . . . .
Defined in this way, f is continuous on D and determines a branch of the logarithm. Furthermore, / ( 1 ) Log 1 , since 1 E Ro. This uniquely specifies the branch (D, f). =
Chapter 7
Complex Integration
7. 1
Paths and Contours
In this chapter, we consider integration along a contour in the complex plane. Recall that a path in C is a continuous function 'Y : [a, b] --+ C , for some a ::::; b in JR. The set of points { z : z 'Y(t) , a ::::; t ::::; b } is the trace of 'Y which we denote by tr "(. =
Example 7. 1
Let "f 1 (t) tq1
Now suppose that 'Y2 (t) tq2
=
=
=
=
e 2""i t for 0 ::::; t ::::; 1 . Then we find that
{ z : lzl
=
1}
=
unit circle.
e _4,. i t for 0 ::::; t ::::; { z : lzl
=
1}
=
1.
We sec that again
unit circle.
The paths 'YI and 'Y2 have the same trace but are different paths: the path "( 1 goes round the circle anticlockwise once, whereas 'Y2 goes round twice in a clockwise sense.
Remark 7 . 1 The convention is to take the anticlockwise sense as being positive. This is consistent with the convention that the positive direction is that with increasing polar angle. Definition 7. 1 The path 'Y : [a, b] --+ C is said to be closed if "f(a) 'Y(b) . The path 'Y is said to be simple if it does not cross itself, that is, 'Y(s) =f. 'Y(t) whenever s =f. t with s , t in (a, b) . ( The possibility that 'Y(a) 'Y(b) is allowed . ) =
=
Example 7.2 The path "( 1 in the example above is a simple closed path , whereas 'Y2 is closed but not simple. The path 'Y(t) e -S ,-it ; 0 ::::; t ::::; 1 , has the unit circle as its trace, but 'Y is neither closed nor simple. =
111
112
Lecture Notes o n Complex Analysis
Definition 7.2 Let "( : [a, b] ---+ C be a path. The reverse path ::Y is given by ::Y : [a, b] ---+ C with ::Y(t) = "f ( a + b - t) .
Evidently, ::Y is "'Y in the opposite direction" . Note that the parametric range for ::Y is the same as that for "(, namely [a, b] . It is also clear that tr ::Y tq. =
Definition 7.3 The path "( : [a, b] ---+ C is smooth if the derivative 'Y'(t) exists and is continuous on [a, b] (with right and left derivatives at a and b, respectively) . In other words, if 'Y(t) x(t) + iy(t) , then 'Y is smooth if and only if both x and y are differentiable (real functions of a real variable) and such that the derivatives x' (t) and y' (t) are continuous functions of the parameter t E [a, b] . A contour is a piecewise smooth path, that is, 'Y : [a, b] ---+ C is a contour if and only if there is a finite collection a = ao < a1 < < an b (for some n E N) such that each subpath 'Y : [ai - l ! ai] ---+ C is smooth, 1 :::; i :::; n . + "fn , where 'Yi is the restriction of "( to the We write 'Y 'Yl + "(2 + subinterval [ai- l , ai] · =
·
=
· ·
· ·
=
·
Example 7.3 According to our (reasonable) definition, the path given by t r---> 'Y(t) cos3 (27rt) + i sin3 (27rt) , for 0 :0:::: t :0:::: 1 , is smooth. Its trace is illustrated in the figure, Fig. 7.1 . =
-1
Fig. 7. 1
A
technically smooth path.
1 13
Complex Integmtion
7.2
The Length of a Contour
Next, we wish to develop a means of formulating the length of a path. By way of motivation, suppose that "'( : [a, b] ---+ 0, as n -> oo. Furthermore, each Tn is closed, so we can apply Cantor's Theorem, theorem 3 .3, to conclude that there is some zo such that {zo} = nn fn . Next, we note that z0 E D and so f is differentiable at zo because f is analytic in D. Define the function T on D by the formula
{
T (z ) =
o) :... f (:... ...c z )_ ..:. -....:.!--'-.. ( z--'- ! ' ( z0 ) , z - zo 0,
£
or z
for z
-L
-r=
z0 zo.
Then T is continuous on D and, in particular, T ( z ) -> 0 as z -> zo. We can express f in terms of T as follows,
f ( z ) = f( zo) + ( z - zo ) f ' ( zo ) + ( z - zo ) T ( z ) , for z E D. Hence
r�
f larn
18Tn� { J( zo) + (z - zo ) f'(zo) + (z - zo ) T ( z ) } dz = 0 + 0 + r � ( z - zo ) T ( z ) dz, larn =
since fafn dz fafn ( z - zo ) dz = 0 because the integrands have primitives ( on the whole of C ) . We must estimate the ( modulus o f the ) third integral i n the equation above. ( For large n, it is an integral around a very small triangle, located around the point zo, of a function which is close to zero in this region. We therefore expect this integral to have small modulus. However, the earlier estimate for I far ! I in terms of I fafn !I involves the factor 4n , so we must pay attention to the details here. ) Let c: > 0 be given. Since T ( z ) -> 0 as z -> z0, we know that there is some o > 0 such that I T ( z ) l < c: whenever l z - zol < o . Furthermore, we =
130
Lecture Notes on Complex A nalysis
know that diam Tn = 2 - n diam T --> 0 as n --> oo , and so we may choose n so that diam Tn < 8. Then any for z E 8Tn , the boundary of the triangle Tn , it follows that lz - zo l diam Tn < 8, since zo E Tn · Hence IT(z ) l < e and therefore
�
l ( z - zo ) T ( z ) l
for any z E 8Tn - It follows that
� diam Tn
e
I JraTn ( z - zo ) T (z ) dz I � diam Tn
e L ( 8 Tn )
n4 L(8T)
�
e diam T
and therefore
Hence, finally, we have
I !aT J I � 4n l faTn J I < - 4n diam4Tn L(8T) diam T L ( 8T ) . e
=
e
This holds for any e > 0 and so we conclude that
far f 0. =
D
Let D be a star-domain. Then every function analytic in D has a primitive, i. e. , if f E H(D) , then there is F E H(D ) such that F' = f on D.
Theorem 8.2
Proof.
Let zo be a star-centre for D. Define F on D by
F(z)
=
{ f J[zo,z]
for z E D. Note that F is well-defined since [z0, z] c D. We claim that F is differentiable and that F' = f everywhere in D. To show this, let z E D be given. Since D is open, there is r > 0 such that D(z, r) � D. We wish to show that
F (z + () - F (z ) (
_
f(z)
__,
0
131
Cauchy 's Theorem
as ( - 0 . We are assured that F(z + ( ) is defined for all ( with 1 ( 1 < r, since D(z , r ) � D, as noted above. Suppose that 1 ( 1 < r from now on. We will appeal to Cauchy's Theorem to rewrite F(z + () - F (z ) in another form. Indeed, F(z + () - F(z ) =
1
1
f-
[zo ,z +(]
f.
[zo , z]
Let T denote the triangle with vertices z0 , z + ( and z . Since zo is a star centre for D, and since [z + (, z] , the line segment from z + ( to z, lies in D( z , r ) � D, it follows that T C D. Indeed, any point w on the line segment [z + (, z] lies in D and so, therefore, does the line segment [zo , w] . By varying w , we exhaust the triangle T. Now, by Cauchy's Theorem, far f = 0 . However, the contour integral around a triangle is equal ( by definition ) to the sum of the integrals along its sides. Hence, we have
1
f+
1
[z +(,z]
[zo ,z +(]
f+
1
[z , zo]
f
=
0.
Reversing the direction of the contour is equivalent to a change in sign of the integral, and therefore we may rewrite the above equation as
1
f-
[zo ,z +(]
1
[zo ,z]
f=-
In terms of F, this becomes F(z + () - F(z)
=
1
1
!=
1
! ( � ) d� .
[z + ( , z]
[z , z +(]
f
[z , z+(]
Hence F (z + () - F(z )
(
1 1 =
- f (z) = � (
[z, z+(]
[z , z+(]
! ( � ) d� - f (z)
! ( � ) - f (z)
(
�
since f (z ) ft z , z+(] d� = f( z) ( . It remains to estimate this last integral. To do this, we use the continuity of f at z . Let t: > 0 b e given. Then there i s 8 > 0 such that I f ( � ) - f (z) l < t: whenever I � - z l < 8. It follows that if 1( 1 < min { 8, r } , then I � - z l < 8 for
132
Lecture Notes on Complex Analysis
every � E [z, z + (] so that I f ( � ) - f( z) l < c for such �- Therefore
I F(z + (� - F(z) - f(z) I = 1 ; 1 1 1[ , +(] ( /(�) - f (z) ) d� I '>
::::;
zz
j(f c L( [z , z + (J ) 1
= RT1 c 1 (1
= €
0
and the result follows.
What 's going on? The method is constructive in that an explicit formula for a primitive is given, rather than it merely being argued to exist. To achieve this, use is made of the existence of a star-centre for D. The fact that zo is a star-centre means that every line segment (zo , z] lies in D whenever z does. This, together with the continuity of f ensures that F is well-defined. The fact that F actually is a primitive for f is a consequence of the continuity of I and the fact that the integral of f around any triangle is zero. Of course, these two facts are consequences of the (assumed) analyticity of I in D, but the proof that F is a primitive for I carries through if the hypothesis "I is analytic in D" is directly replaced by "I is continuous in D and its integral around any triangle in D is zero" . Stating the theorem this way might seem rather artificial or contrived, but we will see later (see Morera's Theorem, theorem 8.8) that it turns out to be quite relevant.
As a consequence of this last result, we show that if f is does not vanish on a star-domain, then f has an analytic logarithm and also an analytic nth_root. Theorem 8.3 Suppose that f is analytic in the star-domain D and that f (z ) -=1- 0 for any z E D . Then there is g E H(D ) such that f (z) exp g (z) for all z E D . In particular, for any n E N there is h E H(D ) such that hn f in D.
=
=
Proof. Since f does not vanish in D, the function f ' If is analytic in D. But D is star-like and so, by Theorem 8.2, f' If has a primitive there, say F E H(D ) . Then F' f ' If in D. (Experience from calculus suggests that F might be a good candidate for a logarithm of f . This is almost correct. ) Now let 'f/; ( z) f (z ) e - F( z ) for z E D. Then we calculate the derivative =
=
'1/J ' (z)
=
f ' (z ) e- F( z) - F' (z ) f( z ) e - F( z)
=
0
133
Cauchy 's Theorem
because F'(z ) f (z ) = f '(z ) . Since D is connected, it follows that 1/J is constant on D. Fix zo E D. Then
f ( z ) e - F (z)
=
1/J(z )
=
=
1/J(zo)
for any z E D. Rearranging and using f (zo) f (z )
Let g(z )
=
=
f (zo) e - F (z o )
=
e Log / (za ) , we get e F (z) - F ( zo ) + Log f(zo ) .
F(z ) - F(zo) + Log f (zo) . Then g
eg (z)
=
E
H(D) and obeys
f( z ) ,
as
required. It is now easy to construct an analytic nth_root of f in D. Indeed, for any n E N , set h(z ) = exp ( � g(z ) ) . Then hn (z)
=
exp(g (z) )
=
f (z ) D
and the proof is complete. 8.2
Cauchy's Theorem for Star-Domains
Theorem 8.4 (Cauchy's Theorem for Star-Domains) Let f be analytic in a star-domain D. Then J"Y f = 0 for any closed contour 'Y with tr -y C D . Moreover, if cp and 1/J are contours in D with the same initial point zo and the same final point z 1 , then J f = f,p f . "'
Proof. By theorem 8.2, any f E H(D ) has a primitive on D, that is, there is F E H(D) with F' = f on D. Let 'Y : [a, b] -+ C be any contour in D. Then
1 1 l f
=
F'
=
F('Y(b ) ) - F ('Y( a ) ) = 0 ,
if 'Y is closed. For the last part, as above, we have f
as required.
=
F(z 1 ) - F(zo)
=
Lf,
D
1 34
Lecture Notes on Complex Analysis
This central result can be generalized to more general domains, but troubles may arise when D has holes. Example 8.1 Suppose that f E H ( D) and that D, as indicated in Fig. 8 . 2 .
Fig.
8. 2
J""t f
"'f
is a contour lying in
may still vanish i n certain non star-like domains.
We see that J""t f 0 even though D is clearly not star-like. The reason is that we can put in cross-cuts and use the fact that each sub-contour is in a star-domain in which f is analytic. The contributions to the integral from the extra cross-cuts cancel out because they are eventually traversed in both directions. In other words, we may sometimes be able to piece together overlapping star-domains to get a non star-like domain for which the theorem nevertheless remains valid. This kind of argument is often done "by inspection" , that is to say, the precise way the contour is cut up will generally depend on the particular case in hand. =
Example 8.2 It is important to realize that this trick cannot be used if D has holes and the contour "'f goes around such a hole. For example, suppose that the domain D is D = B -> C -> D is the same as that from A to D along the circular arc -y(t) = J2e i t , -311"/4 � t � 37!"/4, namely
1"Y dzz !-3,./ 4 J2J2iee itit dt =
3 ,. / 4
= 37ri
2
.
Next, we note that { z : Re z < 0 } is star-like and so the integral of 1 / z along the side D -> A of the square is equal to that from D to A along the arc (dotted in Fig. 8.3) cp ( t ) = J2e i t , 37r/4 � t � 57r/4, namely
1'P dzz
Adding, we find
=
{ 5,. / 4 J2 ie i t dt = }3 ,. / 4 J2e i t
7ri . 2
z 2 + -1 dz 1r -1r -z1 dz + = 1"Y -dzz + 1'P -dzz 2 1ri . z3 =
0
=
Lecture Notes on Complex A nalysis
136
8.3
Deformation Lemma
Before proceeding, it is convenient to introduce some terminology. In order to avoid some gratuitous circumlocution, let us agree to call a contour 'Y a simple circular curve, or j ust a circle, if it has the form 'Y(t) ( + p e 2 1rit , 0 � t � 1 , for some ( E C and p > 0 . Here, tr "f is a circle with radius p and with centre at the point ( in C. The path is traced out in the usual positive (i.e., counter-clockwise) sense and makes j ust one single circuit. 'Y is a simple closed smooth path. If w is any point whose distance from the centre ( is smaller than p , then we shall say that "f encircles w (or goes around w ) or that w lies inside the circle 'Y· Note, by the way, that we could just as well have parameterized this circle as 'Y(t ) ( + p e it with 0 � t � 27r. =
=
Lemma 8. 1 (Deformation Lemma) Suppose that ( belongs to the disc D(a, R) and that g is analytic in the punctured set D(a, R) \ {(} . Let 'Y and r be simple circular curves encircling the point ( and lying inside D( a, R) , as indicated in Fig. 8.4 . Then
In particular, if f is analytic in D(a, R) , then
1 wf(w) dw "�
-(
=
r wf (w)( dw .
lr
-
(I + P i e 2 1ri t and r (t) = (2 + The hypotheses are that 'Y(t ) 2 p 2 e 1ri t , for 0 � t � 1 , for some (I , (2 E C and for some PI > 0 and p2 > 0. Furthermore, tq C D(a, R) , tr r C D(a, R) , 1 ( - (I I < PI and I ( - (2 1 < P2 - Without loss of generality, we may assume that the circle "f has a small radius and so is inside r, as illustrated in Fig. 8.4. (If not, we j ust show that each integral is equal to that around one such small circle and so are equal to each other.) The idea is to put in cross-cuts and apply C auchy's Theorem in suitable resultant star-like regions. With the notation of Fig. 8.4, we see that Proof.
=
r
jA+B+C+D+A'+B'+C'+D'
g=
r
lr +;y
g
since the contributions to the contour integral along the parts B and B' cancel out, as do those along D and D'.
137
Cauchy 's Theorem
D ( a, R) _
.
..
· · ·
· · . . · · · .
A
r
· · · · · · · ·
Fig.
8.4
Put in cross-cuts.
A + B + C + D is a closed curve which is contained in a star-domain D 1 � D( a, R ) \ { ( } , as shown in Fig. 8.5 , and g is analytic in D 1 • (We can take D1 to be the star-domain D ( a, R) n (C \ L t ) , where L1 is any straight line (ray) from ( which does not cut A + B + C + D, as shown in Fig. 8.5.)
Now,
Hence, by Cauchy's Theorem, theorem 8.4,
r
jA+B+C+D
g
= 0.
.
Fig.
8.5
..
There is an enclosing star-domai n.
138
Lecture Notes on Complex A nalysis
Similarly, (replacing L1 by another suitable straight line cut (this time up rather than down ) )
r
jA' + B ' +C'+D'
and we conclude that
g=o
Hence as required. For the last ( important ) part, take g (w )
=
f (w ) j (w - () .
0
Remark 8.1 This says that the contour 'Y can be moved or "deformed" into the contour r without changing the value of the integral, provided the change is through a region of analyticity of g. As can be seen from the proof of the Deformation Lemma, the disc D ( a, R) could be replaced by a more general shaped domain, and the closed contours 'Y and r need not be circular. However, the case with circles is of particular interest as we will see later. 8.4
Cauchy's Integral Formula
We now use the Deformation Lemma to obtain another of the basic results of complex analysis. Theorem 8.5 (Cauchy's Integral Formula) Suppose f is analytic in the open disc D ( a , R) and that zo E D ( a, R) . Let r be any simple circular contour around z0 and lying in D ( a , R) . Then
f (zo) =
f (w) � 21r2 lrr w - zo dw.
Let 'Y ( t ) zo + p e 2 1r it , 0 :S: t :S: 1 , where p > 0 is chosen small enough that tr -y C D ( a , R) . Using the Deformation Lemma, we have
Proof.
=
r
f (w) dw w - zo r J
=
=
1 wf-(w)zo dw 1 f (w)w -- f(zo zo ) dw + 1 wf (z-ozo) dw . "�
-y
-y
1 39
Cauchy 's Theorem
Now,
1 wf(zo) dw - zo
1
{ f(zo) 21ri e 2 Tr i t dt Jo p e 2 Tr z t p
=
"�
= 21ri f(zo) · f( w) f(zo) . . 'T' ..LO estimate dw, we use t he contmmty o f f at zo . w zo 'Y Let c. > 0 be given. Then there is 8 > 0 such that lf(z) - f(zo) l < c. whenever l z - zo l < 8. Choose p so that 0 < p < 8. Then for any w E tr ]', we have that lw - zo l = p < 8 so that l f(w) - f(zo) l < c. and therefore f(w) - f(zo) l f(w) - f(zo) l :. I = < . t follows that w - z0 p p •
I
1 I
11 f(w)w -- f(zo) dw l � :_ L(J') p z0 "�
= =
E.
p 27rp
-
21rc. .
Using this, we see that
f(w) - f(zo) I }rr wf(w) dw I dw - 27rif(zo) I 1 1 w - zo - zo "� =
�
21rc..
This holds for any c. > 0. Hence, the left hand side (which, incidentally, D does not depend on p) must be zero. 8.5
Taylor Series Expansion
The next theorem is yet another result of fundamental importance.
Suppose that f is analytic in the disc D(z0 , R) . Then, for any z E D(z0, R) , the function f(z) has the power series expansion
Theorem 8.6 (Taylor Series Expansion)
f (z) = where a n =
r f (w � � 27rt Jr ( w - zo n
+1
00
L a n (z - zo) n
n =O
dw for any circle r encircling zo and lying
in D(zo , R) . The series converges absolutely for any z E D(z0 , R) .
140
Lecture Notes on Complex Analysis
z E D(zo, R) be given. Let r > 0 satisfy lz - zol < r < R and put -y(t) zo + re 2 1rit , for 0 ::=; t :::; 1 . Notice, firstly, that lz - zol / r < 1 and, secondly, that z lies inside the circle -y . Proof.
Let
=
The idea of the proof is to apply Cauchy's Integral Formula
f (z) = � 21Tt
j"' wf (w)- z dw
1 w --z in npowers of ( z - zo) . The formula 1 - a ( 1 - a)(1 + a + a 2 + · · · + an - ) can be rewritten
and expand
=
to give
1 1 -a
--
=
1
1 1 + a + a2 + · · · + an - 1 + an --. 1-a
Hence, we may write
1 w-z
1 (w - zo) - (z - zo) =
=
{
(
1 (w - zo)
) ( ) ( ) ( )
1 z - zo + -z - zo 2 + . . . 1 + -w - zo w - zo (w - zo) n-1 1 Z - Zo n . . . + wZ -- Zo + w - zo 1 - ( �) zo w - zo n 1 (z - zo) n 1 + (z - zo) + + (z - zo) - + (w - zo) (w - zo) 2 · · · (w - zo) n (w - zo) n (w - z) " _
}
Therefore, using Cauchy's Integral Formula, we have
f (z)
=
=
1 (wf (w)- z) dw � 27rt 1
-
27ri
"'
1 (wf-(w)z0) dw + 27ri1 1 (wf-(w)zo) 2 (z - zo) dw + . . . "' "' 1 f . . . + 21Tt. 1 ( (w)zo ) n ( z - zo) n - 1 dw + . . . "' 1 1 f (w) (z - zo) n dw _ . . . + 27ri _ (w - zo) n (w - z) -
-
W -
"'
141
Cauchy 's Theorem
where ak =
1 (w -f (wzo� k+I � 21ft "I
dw , for k = 0, 1, 2 , . . . The value of ak is
independent of 'Y as long as it encircles the point zo . This follows from the f (w ) . . . DeformatiOn Lemma, lemma 8.1, w1 th ( = zo and g (w) = (w - z0 ) k +l We wish to show that R, --> 0 as n --> oo. To see this, we note that the continuity of f and the compactness of tr 'Y together imply that there is some M > 0 such that l f(w) l � M for any w E tr -y ( proposition 4.3) . Also, z ¢:. tr 'Y and so there is o > 0 such that lw - z l � o for any w E tr 'Y· (In fact, o could be any positive real number less than r - l z - zo l - ) This means that 1/ lw - z l � 1/8, whenever w E tr -y. Therefore
M I (w - zof )(w)n (w - z) I .x ) . Then l f(x) l :., but f'(O) = >., which can be chosen as large as we wish.
147
Cauchy 's Theorem
Theorem 8 . 1 0 (Liouville's Theorem) If f is entire and bounded, then
f is constant. In other words, no entire function can be bounded unless it is constant. Suppose that f is entire and that l f(z) l .S M for every z E C. Then f is analytic i n C and s o has a Taylor series expansion about zo 0 which is valid for z in any disc D(O, R) , i .e., for all z , Proof.
=
f(z)
00
=
L an z n ,
n=U
.
With a n
=
J( n ) (O) n.
--1- .
Applying Cauchy's Inequality, theorem 8.9, to f in the disc D(O, R) , we find that
=
0, 1 , 2 , . . . . This holds for any R > 0 . Fixing n 2: 1 and letting we conclude that f(n ) (O) = 0 . It follows that f (z) = a0, that is, the function f is constant. 0
for any n
R --+
oo,
Example 8.5 Suppose that f is entire and satisfies l f (z ) l � 1 + lz l m for all z E C. It follows that lf (z) l � 1 + R m for all z E D(O, R) and so Cauchy's inequality implies that
for any n > m . So if f(z ) = l:::'= o a n z n is the Taylor series expansion of f about zo 0, then all the coefficients an with n > m vanish. In other words, f(z) is a polynomial of degree at most m .
as R --+
oo,
=
We can use Liouville's Theorem to give a fairly painless proof of the Fundamental Theorem of Algebra.
Every non constant complex polynomial p has a zero, that is, there is zo E C such that p(zo) = 0.
Theorem 8 . 1 1 (Fundamental Theorem of Algebra)
Proof. We may write p as p(z) = a n z n + an - I Z n - l + · · · + a 1 z + ao, where n 2: 1 and a n # 0. To show that p has a zero, we suppose the contrary and
obtain a contradiction. Thus p is entire and, assuming it is never zero, 1/p
148
Lecture Notes on Complex A nalysis
is also entire. However 1
p(z)
=
1
a n z n + a n- l z n - l + 1
(
·
·
·
+ a 1 z + ao
)
a a ao z n a n + n- 1 + + n-1 1 + -n z z z a1 1 a n- I ao As l z l � oo , a n + + + -+ -n � a n and so � 0. In nl p(z) z z z < 1 , whenever lz l > R. particular, i t follows that there is R such that p z) On the other hand, if p is never zero, then 1 /p is analytic and so is certainly continuous . In particular, 1/p is bounded on the closed (and so M compact) disc D (O, R) , that is, there is M > 0 such that p z) � , whenever l z l � R.
(
--
--
·
·
·
· ·
·
) ItI --
Combining these remarks , we may say that
t
p z)
--
ltl
i s entire and obeys
I M + 1 for any z E C. Hence by Liouville's Theorem, 1 /p is lconstant, p tz ) say 1/p = a . But then p = 1 / is also constant, a contradiction. �
a
We conclude that p does, indeed, possess a zero, that is, p(zo) = 0 for some
zo E C.
0
Let p(z) = an z n + + a1z + ao , with a n -1- 0, be a polynomial of degree n . Then there exist (t , . . . , (n E IC and a E IC su c h that Corollary 8.2
·
p(z )
=
·
·
a(z - (I ) . . . (z - (n )
for all z E C. {N.B. The (j s need not be all different.) In other words, any polynomial of degree n has exactly n zeros-counted according to their multiplicity. Proof. We shall prove this by induction. For each n E N, let Q (n) be the statement that any polynomial of degree n can be written in the stated form. Any polynomial of degree 1 has the form p(z) = az + b with a -1- 0. Clearly, such p can alternatively be written as p(z) = a(z - c) , where c -b/a. Hence Q ( 1 ) is true. Next we show that the truth of Q (n) implies that of Q (n + 1 ) . So suppose that Q ( n) is true. Let p be any polynomial of degree n + 1 , =
1 49
Cauchy 's Theorem
By the theorem, p has a zero, zo , say. Then p(zo)
p(z)
=
0 and so
p(z) - p(zo) an+ I Z n + I - an+ I Zon+ I + a n z n - an z0n + · + a i Z - ai zo + ao - ao a n + I (z n + I - z� + I ) + an (z n - z�) + · · · + a I (z - zo) (z - zo) {a n + I (z n + z n - I zo + · · · + z�) + a n (z n - I + z n - 2 zo + · · · + z� - I ) + · · + a 2 (z + zo) + a ! } = ( z - zo)q(z) =
=
·
·
=
=
·
where q(z) is a polynomial of degree n. By induction hypothesis, namely, that Q (n) is true, we can write q(z) as
q(z)
=
,B(z - (I ) . . . (z - (n )
for some ,B and (I , . . . , (n E C. It follows that Q (n + 1 ) is true. Hence, by D induction, Q (n) is true for all n E N. A further direct consequence of Liouville's Theorem is the interesting observation that the range of any non-constant entire function permeates the complex plane, C.
Suppose that f is entire and not constant. Then for any w E C and any c. > 0 there is some ( E C such that f (() E D(w, c.) . In other words, f assumes values arbitrarily close to any complex number.
Proposition 8 . 1
Suppose that w E C and c. > 0 are given. To say that there is no ( with f(() in the disc D(w, c.) is to say that l f(z) - wl 2": c. for all z E C. In particular, f(z) - w -I- 0 and so g 1 / (f - w) is entire. But then g obeys lg (z) l : 0 such that D(wo , r) � D. Now, h(z) is analytic in the disc D(wo , r) and so has a Taylor series expansion about wo valid for all z E D(wo, r) . But wo E A, so by the Identity Theorem for power series, we conclude that h vanishes throughout D(wo , r) and therefore every point of D(w0 , r) is a limit point of Z. Now, cp is continuous at to, and therefore there is some b > 0 such that cp(t) E D(wo , r) whenever l t - to l < b and t E [a, b] . That is, cp(t) E A and so 9(t) 0 for all such t. It follows that 9 is continuous at to. (9 is constant in a neighbourhood of to.) =
=
Case ( ii ) : wo ¢ A. This means that wo is not a limit point of zeros of h ( but this does not preclude the possibility that h(wo) 0) . It follows that there is some p > 0 such that h is never zero in the punctured disc D' (wo , p) ( otherwise wo would be a limit point of zeros of h ) . Furthermore, by construction, 9(to) 1 . Once again the continuity of cp at to means that there is b' > 0 such that cp ( s ) E D( w0 , p) whenever =
=
152
Lecture Notes on Complex A nalysis
I s - t o l < fJ' and s E (a , b] . However, for any such s, either rp(s ) wo, in which case g (s) = 1 (because rp(s) wo � A) or rp(s) E D' (wo , p ) and so h(rp(s)) -I- 0 and therefore g (s) = 1 because rp(s) is certainly not a limit point of zeros of h. It follows that g is continuous at to. We see then, that g is continuous at each t0 E (a, b] . Now, g only assumes at most two values, either 0 or 1 . By the Intermediate Value Theorem, it follows that g is constant. But g (a) 0 and so g (t) = 0 for all t E (a , b] Therefore g (b) 0 and so w = rp(b) E A and, in particular, h(w) = 0. This =
=
=
.
=
0
completes the proof.
Remark 8.4 In practice, the set S is usually a line segment, or a disc or part of a disc or some similarly simple geometric region. This result expresses a remarkable rigidity property enjoyed by analytic functions. For example, suppose that f is entire. Suppose that g is also entire and that g ( z ) = f ( z ) for z in, say, the very small and very remote line segment S = (10 66 , 10 66 + 10-999] . Nonetheless, by the theorem, it follows that g ( z ) = f (z ) throughout the whole complex plane.
The complex functions exp z , cos z and sin z are the only entire functions which agree with their real counterparts exp x, cos x and sin x, respectively, on the real axis. That is, if f is entire and f ( x) exp x for all x E JR, then f (z) = exp z for all z E C, and similarly for cos z and Corollary 8 . 3
=
sin z .
Proof.
Take S =
lR
and apply the theorem.
0
Remark 8 . 5 Suppose that D is a domain and suppose that f E H(D) is not identically zero on D. Let S denote the (possibly empty) set of zeros of f ;
S = { zeros of f } = { z E D : f (z ) = 0 }.
Then S can have no limit points in D--otherwise, by the Identity Theorem, f would be zero throughout D. In other words, the zeros of f are isolated: if zo E D is such that f (zo) = 0 then there is r > 0 such that f has no other zeros in the disc D(zo, r) . Definition 8.2 Suppose that f is analytic in a domain D. The point zo E D is said to be a zero of f of order m (with m � 1 ) if f (zo) = 0 and the k Taylor series expansion of f about zo has the form f (z ) = L�= m ak ( z - zo ) where a m -I- 0.
Cauchy 's Theorem
153
This is equivalent to demanding that the derivatives f(zo ) , f'(zo ) , . . . , f ( m - Il (zo ) all vanish, but f ( m l (zo ) # 0. Note also, that according to the discussion above, the Identity Theorem implies that either f vanishes throughout D or every zero of f is isolated and so has some (finite) order. (In the latter case, the Taylor series cannot be the zero series and so must have at least one non-vanishing coefficient. ) A s the next example shows, the set o f zeros may well have a limit point not belonging to the domain. Example 8.6 Let D = C \ { 0 } , the punctured plane and let f(z) sin(1 /z) for z E D. Now, sin( 1 /z) = 0 whenever 1 / z = k7r for some k E Z, i . e. , when z = 1 / (br) , k E Z \ { 0 } . Let S = { z : z 1 / (krr) , k E Z \ {0} } . We see that S � D , f vanishes on S and that S has (the single) limit point 0. However, f does not vanish on the whole of D. This does not contradict the Identity Theorem because the limit point 0 does not belong to the domain D. =
Recall that a set is said to be countable if it has a finite number of elements (or is empty) or if its elements can be listed as a sequence (that is, can be labelled by N) . For example, the sets N, Z and Q are countable, but one can show that the sets (0, 1 ) and lR are not.
Suppose that D is a domain and that f E H(D). Then either f vanishes throughout D or the set of zeros of f is countable.
Theorem 8 . 1 3
Suppose that f is not identically zero on D and let Z = { z E D : f (z) = 0 } denote the set of zeros of f in D. Let K � D be compact. We show that K n Z is either empty or has only a finite number of elements. Indeed, suppose that K n z is an infinite set. Let W I E K n z . Now let w 2 E (K n Z) \ { WI } and let w3 E (K n Z) \ { WI , w 2 } . Continuing in this way, we construct a sequence (wn ) in K n Z such that Wn # Wm for any n # m. (This construction works because (K n Z) \ { WI , w 2 , . . . , Wn } is not empty.) Now, K is compact and so (wn ) has a convergent subsequence, Wn k ---+ (, say, with ( E K . But each Wn k is a zero of f and so ( is a limit point of zeros in D. By the Identity Theorem, this is impossible and so we conclude that K n Z cannot contain infinitely-many elements. To complete the proof, we observe that by proposition 3.6, the set D has a compact exhaustion, D U:'= I Kn where each Kn is compact. B ut then Z = U := I (Kn n Z) which is the union of a sequence of finite (or empty) sets and so is countable. 0 Proof.
=
Lecture Notes on Complex Analysis
154
8.10
Preservation of Angles
Consider a path 1 and a point zo on the path. Let us say that 'Y makes an angle 0 with respect to the positive real direction at the point zo = r (t0) if r( t) =I- zo for sufficiently small t - to > 0 and if
r (t) --zo--, ----> e ·& ir (t) - zo l
.,...--'-: --
as
t l to .
•
The angle between two paths 12 and 11 , each passing through is then 82 - 81. where o1 is the angle 11 makes with the positive real direction at zo, for j 1 , Now suppose that f is analytic in some disc D(zo, r ) and is not constant there. Then we know that f has a Taylor series expansion
zo
=
2.
f(z)
00
=
L an (z - zo t . n=O
By hypothesis, f is not constant and so there must be some non-zero an with n � 1 . Suppose that a m is the first such non-zero coefficient so that the Taylor series for f has the form 00
00
f(z) = L an (z - zo) n ao + L an (z - zo t f(zo) + (z - zo) m g(z) n=m n =O where m � 1 and g(zo) = a m =/=- 0. Let r1 (t) = J ( r1 (t)) for j = 1 , Since f is not constant, the Identity Theorem implies that fj (t) =I- f(zo) for all sufficiently small t - to > 0 (where zo = r (to)). Hence f( rj (t)) - f(zo) rj(t) - f(zo) i fJ (t) - f(zo) l lf h1 (t)) - f(zo)l ( /j (t) - zo ) m g( rj (t)) - lrJ (t) - zo l lg(rJ (t)) l g(zo) as t l to, ( ei0i ) m lg(zo) I' = eim oi ei Arg g(zo) =
=
2.
---->
and so r1 makes an angle mOj + Arg g(zo) with respect to the positive real direction at f(z0). But then this means that the angle between f 2 and f1 at f(zo) i s m( 02 - 01) .
Cauchy 's Theorem
155
In other words, if the derivatives j(r) ( zo } = 0 for all r = 1, . . . , m - 1 , but f( m ) ( zo } =/:- 0, then the angle between paths intersecting at zo is multiplied by m under f. In particular, if f' ( zo) =/:- 0, then f preserves the angle between intersecting paths. ( Such maps are said to be conformal. )
Chapter 9
The Laurent Expansion
9.1
Laurent Expansion
In this chapter, we discuss a generalization of the Taylor series expansion. Rather than considering analyticity in some disc, we assume only analyticity in an annulus. This leads to a representation in terms of both positive and, possibly, negative powers of z, the Laurent expansion. The starting point is Cauchy's Integral Formula for the given function f. Here the circle of integration can be deformed, as in the Deformation Lemma, to produce , in fact, two contour integrals. Each of these is the integral of f(w)/(w - z) around a certain circle. The idea is then to expand 1 / ( w - z ) in powers of z - zo , in one case , and in powers of 1 / (z - zo ) in the other. Doing the integrals gives the coefficients and all that remains is to worry a bit about the convergence of the two series thus obtained. Now for the details. 9 . 1 Suppose that f is analytic in the annulus A A(zo ; R1 , R2 ) = { z : R1 < l z - zo l < R2 } . Let z E A, and let R1 < r1 < l z - zo l < r2 < R2 . Then
Lemma
27rif(z)
=
r
f(w)
r
f(w)
dw + }02 w - z lc, w - z
dw,
where c1 is the circle zo + r1 e 2 rrit and c2 is the circle zo + r2e 2 rr it ' 0 :::; t :::; 1 . We argue j ust as in the proof of the Deformation Lemma, lemma 8. 1 . Insert line segments between the two circular contours, as shown in Fig. 9. 1 . The idea is to put in sufficiently many cross-cuts so that each of the contours /j is so narrow that it is contained in the star-domain DJ which is itself inside the annulus A. Clearly, if r2 - R1 is small, there will need to be many such cross-cuts. We number the /j S so that 11 goes around the point z , as indicated. Proof.
1 57
158
Lecture Notes on Complex Analysis
Fi g . 9. 1 Construct cross-cuts to g ive many narrow simple closed contours ')'1 , /'2 , . . . , "Yn . The poi nt z is encircled by /'1 ·
Next, let r be a circle around z with sufficiently small radius that it is encircled by 11 , as shown in Fig. 9.2.
Fig . 9 . 2
Star- like domains D i contai nin g "Yi .
Using Cauchy's Integral Formula and arguing as in the proof of the Deformation Lemma, lemma 8 . 1 , we have
.
2m f (z) =
1 f (w ) dw j"' wf (w- )z dw. w-z r
--
=
--
159
The Laurent Expansion
(w )
, Dn are star-like and wf - z is analytic in Furthermore, since D2 , D3 , each of these domains, it follows, by Cauchy's Theorem for a star-domain, theorem 8.4, that •
1
1'2
•
•
f (w) dw = . . . -Z
W
=
1
f (w) dw = 0. -Z
"'n W
Piecing all these together, and using the fact that the contour integrals along the cross-cuts cancel out, we get
1
1
1
f (w ) d + . . . + f (w) dw + f (w ) dw 1'2 - Z w -Z -Z ) dw + r f (w) dw . = r wf (w lc2 - z }0, w - z
l'l W
W
"'n W
We conclude that
27ri f (z)
=
r 2 wf (-w )z dw - Jrc, wf (-w )z dw,
}0
D
and the proof is complete. As mentioned earlier, the next step is to expand powers. We use the formula
1/ ( w - z ) into suitable
1 = 1 + o: + o:2 + + o: k -1 + -Q: k -· · · 1 - o: 1 - o: valid for any complex number and w =/:- zo, z =/:- zo , we have
1 w-z
o: =1- 1 and any k E N.
Indeed, for any n E N,
1 1 w 1 - zo)/(w - zo)) zo) w zo (z zo) ( (z ( ( ) n 1 z - zo ( z - zo ) - 1 + ( z - zo ) n - (w - zo ) w - zo + . . . + w - zo (w - zo) n - 1 (w - z ) n 1 1 + (z - zo ) + . . . + ( z - zo ) + ( z - zo) n ..,..----.,. .. (w - zo ) (w - zo ) 2 ( w - zo) n (w - zo ) n (w - z ) Sn (zo , z , w ) , say.
(
-:----'----.----c-'----,.,..
=
)
160
Lecture Notes on Complex Analysis
Similarly, for any m E N and w =/:- zo and z =/:- zo , we have 1
1
-1
- --
w-z
(w - zo ) - ( z - zo ) ( z - zo) - ( w - zo) 1 (w - zo) . . . (w - zo)m- l (w - zo)m + + + -:--'-----:-m--7-,----...,.- + - ---:(z - zo) (z - zo) 2 (z - zo) m (z - zo) (z - w ) Sm (zo, w, z). =
=
------,----,-
Notice that the first expression involves positive powers o f z - zo, whilst the second involves negative powers of z - zo . Why should we be interested in both cases? The point is that lw - zo l < lz - zo l whenever w belongs to the inner circle C1 , whereas the reverse inequality holds when w belongs to the outer circle c2 . This means that wz- zoo < 1 in the first case -z so that Sn (zo, z , w) converges as n ---> oo . On the other hand, if w is on the outer circle c2 , then �=:z < 1 and so the second expansion, Sm (z0 , w , z) , the one with negative powers of (z - zo) , will converge as m ---> oo . Applying these considerations, together with lemma 9. 1 , leads to the Laurent expansion, as follows.
1
I
1
I
Theorem 9 . 1 (Laurent Expansion)
Suppose that f is analytic in the annulus A = A(zo; R1 , R2) . Then, for any z E A, f(z)
=
00
00
n=O
n== l
L an (z - zo t + L bn (z - zo) - n
with dw, an = -12ni- Jcr (w -f(w) zo) n + l and bn
=
1
27rZ .
for n = 0, 1 , 2, . . ,
1 (w - zo) n - 1 f (w) dw C
.
=
1
27rZ .
1
C
f(w) dw, ( W - ZQ) - n + l
for n = 1 , 2, . . . , where C is any circle encircling zo such that tr C � A . Furthermore, both of these series are absolutely convergent in the given annulus A. Let z E A = A(z0; R1 , R2 ) be given and let r1 and r2 satisfy the inequalities R1 < r 1 < l z - zo l < r2 < R2 . Let C1 and C2 be the circles zo + r1 e 2 1l"it and zo + r2 e 2 1l"it, for 0 ::::; t ::::; 1, respectively. Then using lemma Proof.
1 61
The Laurent Expansion
9 . 1 , together with the preliminary discussion above , we see that given by
f (z )
= =
=
f (z)
is
f (w) dw - � r f (w) dw 21rz 1c, w - z c2 w - z � dw + � { f (w) Sm (zo , w, z ) dw 21rz 1c, 2m 1{c2 f (w ) Sn ( zo, z, w ) + + + ao a 1 (z - zo ) · · · an - 1 (z - zo t - 1 + Pn + b 1 + · · · + bm m + Q m (z - zo ) (z - zo )
� r 21rz 1
where the as and bs are as stated in the theorem, and where the remainder terms Pn and Q m are given by
and 1 r (w - zo ) m f (w) Q m - 27ri 1c, (z - zo ) m (z - w ) dw. _
We wish to show that Pn ----> 0 as n ----> oo and that Q m ----> 0 as m ----> oo . The traces of the circles C1 and C2 are compact sets in C and therefore there is some M > 0 such that J f (w) J � M whenever w E tr C1 U tr C2 . Furthermore , since z rJ_ tr c1 u tr c2 , it follows that there is some (j > 0 such that D ( z, fJ) n (tr C1 U tr C2 ) = 0 (because z E C \ (tr C1 U tr C2 ) , an open set) . Hence Jw - zJ � fJ for every w E tr C1 U tr C2 . (One could alternatively apply proposition 3 .7 to reach this conclusion.) We can use this to estimate IPn l and IQ m J , as follows.
JPn J as n ----> oo ,
1
�
27r
---->
0,
Jz - zo J n M 21rr 2 r2n (j
Jz - zo l
since !..._ =-c < 1 . Similarly, T2 1 r m1 JQ m J � _
27r J z - zo J m
---->
0,
M
7 u
21rr 1
162
as
Lecture Notes on Complex Analysis
m � oo ,
then
since
m � oo ,
rt
l z - zo l
< 1
·
Taking the limit, say,
n � oo ,
first, and
it follows that
f(z)
=
00
00
k=O
j= l
L: a k (z - zo ) k + L bi ( z - zo)- i ,
as claimed. By an argument as in the proof of the Deformation Lemma, lemma 8. 1 , we see that the a n and bn integrals are independent of the particular circles Ct and C2 , provided that they lie in A and encircle z0• To show that the two series are absolutely convergent in A, we simply estimate the general term in each case. We have
and
By the Comparison Test, both series converge absolutely.
0
Definition 9 . 1 The (double) series constructed above for the function f, analytic in A, is called the Laurent series expansion of f in the annulus A(zo; Rb R2 ) . The series of negative powers, E:= l b n (z - zo) - n is called the principal part of f.
Our next task is to establish the uniqueness of the Laurent expansion. After all, it is not obvious that it is not possible to change some (or all) of the an s and some (or all) of the bns without affecting f.
The
0
Laurent Expansion
163
z0 + re 2 1rit , for As a preliminary observation, we note that if "Y (t) � t � 1 , is the circular contour centred at zo and with radius r, then =
11
dz
27ri ..., (z - zo) m
=
{0,
1,
m
E Z,
m
= 1.
m
-:/:- 1
This is a consequence o f corollary 7.2, since (z - zo) - m has a primitive in C \ {zo } provided m -:/:- 1 . 9.2
Uniqueness o f the Laurent Expansion
Theorem 9 . 2 (Uniqueness of Laurent Expansion) is analytic in the annulus A = A(zo ; R1 o R2 ) and that
f (z)
00
=
Suppose that f
00
L an (z - zo) n + L f3n (z - zo) - n n=O n= l
where each of the two series converges absolutely in A. Then an an and f3n = bn for all n, where the a n s and bn s are the Laurent series coefficients (as given by the integral formulae in theorem 9. 1). =
0
be given and set Rn (z) = E� n + l ak (z - zo) k and set Tn (z) E� n + l f3k (z - zo)- k , for n E N. The series for Rn converges absolutely in A, i.e . , for z with R1 < lz - zo l < R 2 . It follows that this series converges for all z with lz - zo l < R 2 . In other words, Rn is analytic in the disc D(zo , R2 ) , and, in particular, Rn is analytic in A. However, Proof.
Let E > =
Tn (z)
=
f(z) -
n
n
k=O
k= l
L ak (z - zo) k - Rn (z) - L /3k (z - zo) - k
0
and so Tn is analytic in the annulus A since this is true of the right hand side. In particular, if R 1 < r < R2 and "Y (t) = zo + re 2 1rit , � t � 1 , then both Rn and Tn are continuous on tr "Y· Furthermore , E � o la k l r k and E� 1 1 /3k l r- k both converge and so there is N E N-such that 00
L la k l r k < E
k=n+ l
00
and
L lf3k l r- k < E
k=n+ l
1 64
Lecture Notes o n Complex Analysis
whenever
n
> N. It follows that, for any z E tr "'( (so that lz - zo l
I Rn (z) l �
00
L
k=n+l
=
r) ,
la k l r k < £
and that
I Tn (z) l
�
00
L
k =n +l
whenever n > N . Finally, let m E Z be given. Choose
I .Bk l r- k
lml . We have
n
n k f(z) = L ak (z - zo ) + L .Bk (z - zo)- k + Rn (z) + Tn (z ) . k= l k=O Hence 1 _
27ri
1
-y
}
{
O! m , if m � 0 f(z) 1 dz + (z - zo) m+ l - .B tml • if m < 0 27ri _
But the left hand side is j ust Furthermore ,
1 Rn (z) + Tmn+(z)l dz . "I
(z - zo )
{ b�:;. i[r� � � } , according to their definitions.
1-1- 1 Rn (z) + Tn (z) dz l -< 2._ � 27rr ' 27ri 271" r m + l (z - zo) m+ l "I
for all
n
> N,
2£
=
This is an estimate for the modulus of the difference between a m and O!m or between b t m l and .Bt m l • depending on whether m � 0 or not (and assuming that n is chosen greater than both N and lml ) . Since this holds for any given £ > 0 , we conclude that O! m = a m for m = 0, 1 , 2, . . . and .Bm = b m 0 for m = 1 , 2, . . . , as required. Remark 9 . 1 Suppose that r1 < r < R < R1 and that f is analytic in the annulus A(zo ; r1 , R l ) . Then the uniqueness of the Laurent expansion implies that the coefficients in the Laurent expansion of f in the annulus A(z0; r, R) are precisely the same as those in the Laurent expansion of f in the annulus A(zo ; r1 , R1) .
165
The Laurent Expansion
What ' s
going on? A function f analytic in an annulus can be written as a sum of a series of positive powers and a series of negative powers. Both series converge absolutely in the annulus and the coefficients can be expressed in terms of the function f by means of certain contour integrals. This is the content of the Laurent Expansion Theorem . The absolute convergence of these series is part of the theorem. The uniqueness of the Laurent expansion means that whenever and however one manages to write f as a sum of absolutely convergent powers and inverse powers, then this has to be the Laurent expansion.
Examples
9.1
( 1 ) For any z =/:- 0, the power series definition o f the sine function gives 1 1 1 1 1 + + 5 1 10 - sin 2 = 2 - 7 .1 z 14 3 1. z 6 z z .z
·
·
·
which is the Laurent expansion about z = 0 of the function sin(1/z 2 ) , valid for z in the punctured plane C \ {0} . (2) f (z) = 1 / (z - a) is analytic in the annulus { z : l z - al > 0 } . The Laurent expansion of f about z = a, valid for l z - al > 0, is simply f(z) =
1
. z-a
The function f is also analytic in the annulus { z : l z l > lal 1 1 a a2 1 a f(z) = = - 1 + - + -2 + . . = - + -2 + z ( 1 - a/ z ) z z z z z
(
·
)
} . We find a2
z3
+
·
·
·
is the Laurent expansion of f about z = 0, valid for l z l > l a l . For z in { z : 0 < l z l < l al } , we find that 1 1 f ( z) = --- = - -�---..,.-:z-a a( 1 - z / a ) is the Laurent expansion of f about z = 0, valid for z with 0 < l z l < l a l (where we suppose that a =/:- 0) . This i s also valid for z = 0 . Note that these series converge absolutely for the given ranges o f z . ( 3 ) The function f(z) = z (e 1 f z - 1) i s analytic i n the punctured plane C \ { 0 } . For any z =/:- 0, we have f(z) = z
( (1
1
1
+ - + -- + z 2., z 2
+ ...) - 1 3 ., z 3
1
1 1 1 + -+ =1++, , , 2 4. z3 . . . 2. z 3. z which is the Laurent expansion of f about z = 0.
)
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Lecture Notes on Complex Analysis
Suppose f is analytic and bounded in the punctured disc D'(zo, R) . Then there is a function F analytic in the whole disc D(zo, R) such that f F on D'(zo, R) . In other words, f can be extended so that it is defined at zo in such a way that the resulting function is analytic (in the whole disc).
Proposition 9 . 1 =
The hypotheses imply that f has a Laurent series expansion in the annulus A(zo; r, R) for any 0 < r < R. In particular, we can calculate the coefficients of the principal part of the Laurent expansion as
Proof.
j (z - zo) m - l f (z) dz zo + pe 2 1rit for 0 t 1 and p can be chosen arbitrarily in bm
=
1 -. 2 7!"2
'Y
where 1 (t) � :::::; the range 0 < p < R. By hypothesis, there is M > 0 such that D'(zo, R) . Hence, for any m 2:: 1 , =
lbm l
::=;
1 mP 1M 27rp 27!"
=
lf(z) l :::::; M for all z in
M pm .
This holds for any 0 < p < R and so it follows that b m 0 for all m 2:: 0 . The principal part o f the Laurent expansion of f vanishes and we have =
f(z)
=
00
L an (z - zo)n ,
n =O
for
z E D ' (zo, R) .
This series converges absolutely for all 0 < l z - zo l < R and so certainly for all l z - zo l < R. Hence, we may define F on the disc D(zo , R) by the power series
F(z)
00
=
l:::>n (z - zot .
n =O
The absolute convergence implies that F is analytic in D(zo, R) . Clearly F = f on the punctured disc D'(zo , R) (and F(zo) = ao). 0 What's going on? If f is analytic in a punctured disc in which it is also bounded, then, by suitable adjustment of f at the centre of this punctured disc, one finds that f is really just the restriction of a function analytic in the whole disc to the punctured disc. This is because one can estimate the coefficients occurring in the principal part of Laurent expansion and show that they must vanish. In other words, f is just a sum of positive powers, that is, it is a power series and we know that power series are analytic. We will need this result later.
Chapter 1 0
S i ngularit ies and Meromorphic Functions
10.1
Isolated Singularities
We can think of the principal part of the Laurent expansion of a function f analytic in a punctured disc as encapsulating how badly behaved f is near to the centre. There are essentially only three situations; all coefficients in the principal part vanish, only a finite number of coefficients are non-zero, or an infinite number of coefficients are non-zero. Definition 1 0 . 1 A point zo E C is said to be an isolated singularity of a function f if there is R > 0 such that f is analytic in the punctured disc D'(zo , R) but f is not analytic at zo . Examples 1 0 . 1
0 i s an isolated singularity o f the function f ( z ) � ( 1 ) The point zo This function is not defined at 0 , so is certainly not analytic there. In fact, it is not possible to assign a value at this point so that the resulting (extended) function is analytic. If this were possible, say, with f(O) defined to be a:, then the resulting function would have to be continuous at 0 and, in particular, bounded in some disc around 0. This n is evidently false, as is seen, for example, by observing that f(z) when z 1 /n. (2) Suppose f : C ----> C is defined by f(z) z for z =1- i and f( i ) 3. Then z i is an isolated singularity of f . (f is not even continuous at z i.) 0 whenever z 1 /br with (3) Let f (z) 1 / sin( 1 /z) . Now, sin(1/z) k E Z \ {0} and so f is certainly undefined at such points. However, for any given k E Z with k =1- 0, f is analytic in the punctured disc D' ( 1 jk1r , r ) provided r is sufficiently small (depending on k) . Therefore, for any k E Z with k =1- 0, the point z 1 /br is an isolated singularity =
=
=
=
=
=
=
=
=
=
=
167
=
Lecture Notes on Complex Analysis
168
of f. f is not defined for z 0 and f is not analytic in any punctured disc D'(O, r ) for any r > 0 (because any such disc will contain points of the form 1 /k1r) . Evidently z 0 is a singularity of f (in that f is not analytic there) but according to the definition, z = 0 is not an isolated singularity of f. (4) The principal value logarithm, f(z) = Log z , is analytic in the cut plane C \ { z : z + l z l = 0 } . Each point of the negative real axis { z : z + l z l = 0 } is a point of discontinuity of Log z but none of these points are isolated singularities. Note that Log z is not defined for z = 0 but is defined in the punctured plane C \ {0}. =
=
Let zo be an isolated singularity of the function f. Then, by definition, f is analytic in some punctured disc D' (zo, R) . It follows that f has a Laurent expansion valid for all z in this punctured disc: f (z)
=
00
l:::>
n=O
n n (z - zo) +
00
L )n (z - zo ) -n .
n= l
Definition 10.2
(i) If all bn 0 , then z0 is said to be a removable singularity. (By defining or redefining f at z0 to be equal to ao , we get a function analytic in n the whole disc D(zo , R) , namely, L: �= O an (z - zo ) .) (ii) Suppose that only a finite (but positive) number of the bns are non zero; say, b m -:f. 0 , but bn 0 for all n > m. Then zo is said to be a pole of f of order m. (Sometimes one uses the terms simple pole or double pole for the cases where m = 1 or m = 2, respectively. ) (iii) If an infinite number of the bns are non-zero, then zo is said to be an isolated essential singularity of f. =
=
Examples 1 0 . 2
(1) Consider the function f (z)
=
cos z - 1 z2
=
z22, + z 4, (1 - 4 ·
1
·
z2
-
z2 z4
z66,
·
+ ...) - 1
= -, + , - , + . . . 4. 6. 2.
Evidently z
=
0 is a removable isolated singularity.
169
Singularities and Meromorphic Functions
(2) Consider f ( z)
We see that z
(3) The function
=
=
zsin z = z3
z 3 + 5T zs
31
z3
-
...
1 z2 1 + = 2 3! 5T z
·
· ·
0 is a double pole.
f(z)
=
1
exp - = z
1 1 1 1 + - + 21 2 + 1 +... z .z 3. z -
3
has z = 0 as an essential singularity. (4) Let
f (z) =
(sin z) 4 4 + cos z , z
for z E m ) . Evidently, Proof.
1 70
Lecture Notes on Complex A nalysis
required. Conversely, suppose that lim z -+ zo ( z - zor f (z) = a '# 0. Let F be given by F(z) = (z - zo) m f(z) for z E D' (zo, R) . Since f and therefore F is analytic in this punctured disc, it follows that F has a Laurent expansion as
F(z)
=
00
00
n =O
n= l
L An (z - zo) n + L Bn (z - zo) - n
valid (absolutely convergent) in D'(zo, R). However, by hypothesis, we have that F(z) -+ a as z -+ zo and so F is bounded in the neighbourhood of zo, say, in D'(zo , r) . (IF(z) l is close to lal if z is sufficiently close to zo.) It follows that Bn = 0 for all n E N (as in proposition 9.1). Hence F(z) = E::'= o An (z - zo) n and 0 -:f.
a =
lim
z -�o zo
F(z)
=
Ao .
Therefore
so that
Ao A1 f(z) = ( Z - Zo ) m + ( Z - zo ) m _ 1 + · · · + A m + Am + I (z - zo) + . . . for any z E D'(zo, R) . Since Ao '# 0, we see that zo is a pole of f of
order
= a
m.
0
1 - is undefined at those points z ez - 1 for which e z 1 , i.e. , when z 2k7ri for k E Z . f is analytic everywhere else and so these are isolated singularities. Fix k E Z and set w z - 2k7ri. Then f(z) f(w + 2k7ri) 1/(ew + 2 k1ri - 1) 1/ (ew - 1 ) . Using the power series expansion of ew , we see that (ew - 1)/w -+ 1 as w -+ 0 and so we deduce that (z - 2k7ri) f (z) -+ 1 -:f. 0 as z -+ 2k7ri. It follows that for each k E Z, z 2k7ri is a simple pole of f. Proposition 1 0 . 1 Suppose zo is an isolated singularity of f and suppose that l f(z) l -+ oo as z -+ zo . Then zo is a pole of f {so f is meromorphic at zo). Conversely, if zo is a pole of f , then lf(z) l -+ oo as z -+ zo . Example 1 0 . 3
The function f(z)
=
=
=
=
=
=
=
=
Suppose l f(z) l -+ oo as z -+ zo . Then, for any given M > 0, there is R > 0 such that f E H(D' (zo, R)) and lf(z) l > M for all z E D'(zo, R) . In particular, f(z) -:f. 0 on the punctured disc D'(zo , R) . It follows that g(z) = 1/ f(z) E H(D'(zo, R)) and g satisfies l g (z) l < 1/M on D'(zo , R) .
Proof.
171
Singularities and M eromorphic FUnctions
By proposition 9. 1, we see that zo is a removable singularity of g and that g can be written as (using lim z -+ zo g ( z) 0) =
g(z)
=
A 1 (z - zo) + A2 (z - zo) 2 + . . .
for any z in D'(zo , R) . Furthermore, g is not zero in this punctured disc (because g(z)f(z) = 1) and therefore not all of the Ans are zero. That is, there is m � 1 such that Am # 0 but An 0 for all n < m . It follows that =
t
f z)
=
g(z)
=
(z - zo) m (Am + Am + l (z - zo) + . . . )
and so
1 -:----:----:-:--:- ---+ (z - zo) m f(z) A m ---+
�
z ---+ zo, i.e. , (z - zo) m f (z) # 0. By the theorem, theorem 10.1, A it follows that zo is a pole o f f o f order For the converse, suppose that zo is a pole of f o f order 2:: 1 , say. Then, by theorem 10. 1 , (z - zo) m f (z) ---+ # 0 as z ---+ zo. Hence, for z # zo (and in some neighbourhood of zo so that f(z) is defined)
as
m.
m
a
m 1 1 - 1 (z (z- zo)- zo)m f(z) 1_ I -+ I Q 1 - 0 f(z) a
as z ---+ zo and the result follows.
-
0
Suppose that zo is an isolated essential singularity of f . Then f is neither bounded near zo nor does 1/(z) l diverye t o oo as z ---+ zo . In other words, there are sequences (zn ) and ((n ) such that both Zn ---+ zo and (n ---+ zo and such that lf(zn) l ---+ oo as n ---+ oo but ( f ((n )) is bounded. Corollary 10.1
Proof. If f were bounded near zo , then zo would be a removable isolated singularity. On the other hand, if 1 / (z) l ---+ oo as z ---+ zo , then we have seen that this would imply that zo is a pole of f. By hypothesis, neither of these possibilities hold and so the result follows. 0
Lec ture Notes on Complex Analysis
172
To say that limz-.z0 (z - zo ) m f ( z ) = G of. 0 means that one can think of f ( z) as behaving rather like D oo as z -> zo . However, the converse, although true (as we have seen) , is not quite so obvious . After all , one could be forgiven for imagining that it was possible for f to have an infinite number of terms in the principal part of its Laurent expansion in some punctured disc around zo (and so zo would be an esse ntial singularity) and still be such that l f ( z ) l -> oo as z -> zo . That this cannot happen is perhaps something of a surprise. What ' s going on?
Example 10.4 any z 1- 0,
Define f(z) , for any z 1- 0, by f(z) = cos � · Then, for
f(z) = cos - = 1 - -1-2 + -1-4 - . . z 2. z 4. z and we see that zo = 0 is an essential singularity of f. Taking Zn = i/n and (n = 1 /n , for n E N, we see that Zn ----> 0 and (n ----> 0 as n ----> oo b ut f(zn ) cos (n/i) � (e n + e- n ) ----> oo as n ----> oo, whe re as the sequence (f((n ) ) = (cos n) is bounded.
1
1
1
•
=
10.3
=
Behaviour
as
lzl
- oo
Let f be entire and non-constant. Then f is a polynomial if and only if l f (z) l ----> oo as l z l ----> oo .
Theorem 1 0 . 2
Proof.
Suppose, first, that f i s a (non-constant) polynomial,
f(z)
=
a n z n + · · · + a 1 z + ao
with a n 1- 0. Then
I f tz) I I an zn + � · · + ao I ----> 0 ao an- � . . I zn ( an + -' + · + -n ) 1 z z =
=
as l z l ----> oo . Conversely, suppose that lf (() l ----> has a Taylo r expansion (about z 0)
oo
as 1 (1 ---->
oo .
=
f(z)
00
=
L an zn ,
n=O
for all z .
Since f is entire, it
Singularities and Meromorphic Func tions
For z i=
0,
173
set
g(z)
=
! (�)
00
=
-n '"' � an Z . n =O
Then g E H(C \ { 0 } ) Moreover, this series converges absolutely for lz l > 0 and so is the Laurent expansion of g ( about z 0). Now, as z ----> 0, I � I ----> oo and so , by hypothesis, I f ( � ) I ---> oo , i.e., l g ( z ) l ----> oo as z ----> 0. I t follows that z = 0 i s a pole of g and therefore .
=
for some
m
am a1 g (z) = ao + - + · · · + zm z � 1 , where a m =/= 0. Hence
f(z) = for z i=
f(O)
=
0.
g(� )
=
ao + a1 z +
· · ·
+ am zm
However, from the Taylor expansion of f above , we see that
ao and so
f(z) for all z E
C ( including z
=
=
ao + a 1z + · · · + a m z m
0).
That is,
f is a polynomial.
0
Any entire non-polynomial function f has the property that there is a sequence (zn ) such that lzn l ----> oo and lf(zn ) l ----> oo as n ----> oo and another sequence ((n ) , say, such that l (n l ----> oo as n ----> oo but the sequence (f( (n )) is bounded. Corollary 1 0 . 2
First we note that by Liouville's Theorem, theorem 8.10, since f is entire it cannot be bounded, unless it is a constant . However, it cannot be constant because it is not a polynomial. The existence of a sequence (zn ) , as above , then follows. Furthermore, again because f is not a polynomial, the existence of some sequence ((n ) , as above, follows from the previous theorem. 0 Proof.
What's going on? It is intuitively clear that if f is a polynomial then lf(z) l must diverge to oo as lzl --+ oo. It is far from obvious that the converse is true. After all, one might suspect that a non-polynomial function such as e • + e i z , or something similar, would exhibit this behaviour. Not so. (In this particular example, set z = it - t for t E lR and t > 0. Then e • + e i z e it e - t + e - t e - it -t 2e cos t which stays bounded as t --+ oo . ) =
=
1 74
10.4
Lecture Notes
on
Complex Analysis
Casorati-Weierstrass Theorem
The next result tells us that near an isolated essential singularity a function takes values arbitrarily close to any given complex number.
Theorem 10.3 (Casorati-Weierstrass Theorem) Suppose zo is an isolated essential singularity of the function f. Let r > 0, E > 0 and w E C be given. Then there is some z with l z - zo l < r such that l f (z) - w l < E. Proof. We know that there is some p > 0 such that f is analytic in the punctured disc D' (zo , p) . Let r' min {r, p} . By way of contradiction, suppose that l f (z) - w l 2: E for all z E D' (zo , r ' ) . Then it follows that g (z) = l j ( f (z) - w ) is analytic and bo unded in D' (zo , r') . This means that zo is a removable singularity of g, i.e . , we can extend g to an analytic function in the whole disc D(zo , r' ) . Let us denote this extension by G. Then =
G(z) (f (z) - w )
=
1
for z E D' (zo , r') . If G(zo ) 1- 0, this entails that f be bounded near zo . On the other hand, if G(zo ) 0 then l f (z) l must diverge as z --+ zo . The first case wo uld mean that zo were a removable singularity and the second that it was a pole of f. However, zo is neither of these, it is an essential singularity of f. We conclude that l f (z) - w l 2: E for all z E D' (zo , r') is 0 false , and the result follows. =
Remark 10. 1 In fact, it has been shown that a function takes on all values with at most one exception in any neighbo urhood of an essential singularity, but this is harder to prove (Picard's Theorem) . The example exp ( l /z) shows that there may be an exception. Evidently zo 0 is an isolated essential singularity but the exponential function is never zero . =
Chapter 1 1
Theory o f Residues
11.1
Residues
The function z n has a primitive on 0 , - :i: 7r < e < � 1r } and so it follows that =
J
[ l , i]
dz
-;
=
1 dz '{J
-;
where cp is the quarter-circle cp(t) = eit , 0 � t � � 1r. Applying this idea to the other sides of the square (in the suitably rotated star-domains,
Theory of Residues
179
ein: / 2 D, ein: D and eiJn: / 2 ) and adding the results, we see that
1 dz = 1 dz 'Y
where '1/J is the circle '1/J ( t ) should.
11.3
=
z
"' z
eit , 0 � t � 27r. Hence Ind(-y : O) = 1 , as it
Cauchy's Residue Theorem
The next theorem is yet another fundamental result.
Let f be a mero morphic function with a finite number of poles, (1 , . . . , (m , say, in the star domain D and let ')' be a closed contour with tr ')' � D \ { (1 , . . . , (m } (i. e., the contour does not pass through any of the poles). Then
Theorem 1 1 .3 ( Cauchy's Residue Theorem)
1f 'Y
Proof.
S uppose that
=
2rri
't, Res ( f : (k) Ind( T (k ) ·
k= 1
(k is a pole of order mk.
that
Then there is
Rk > 0 such
00
L a�k l (z - (k t + Pk (z) n =O k for z E D ' ((k , Rk) , where Pk(z) = z=:! 1 b� ) (z - (k)- n . The function Pk k is analytic in 0, 1 � k � n). Suppose further that zf(z) ---) 0 whenever l zl ---) oo with Im z 2: 0 . Then n R limoo f (x) dx 2rri L Res ( ! : Zk ) . Theorem 1 1 .4
1
R-+
-R
=
k= l
Proof. Let c > 0 b e given. B y hypothesis, there i s R so that l zf(z) l < c for all l zl > R with Im z 2: 0. Let E be the semicircular path with centre 0 and radius p and let r [-p, p] + E. Choose p so large that r encloses all the poles z 1 , . . . , Zn. Then Ind ( r : zk ) = 1 for all 1 � k � n and so, by the Residue Theorem, =
1f r
2rri
=
t Res (f : z ) . k
k= l
Now,
hf
and we claim that
hf
....-+
=
0 as
l
f+
- p ,p]
p
---)
hf
To see this, suppose
oo.
p > R.
for any z E tr E, l zf (z) l < c , that is, p if(z) i < c , since lz l = p for z Hence if(z) i < sjp for all z E tr E. By the Basic Estimate, we get
I lr.r f I
and it follows that
hf
....-+
�
0 as
� L ( E)
p
p
---)
1
=
� rrp
p
oo, as
f The result now follows because . [- p,p]
=
=
Then, E
tr E.
c 7r
claimed.
1P f (x) dx. -p
0
Note that in the theorem above, there are no poles on the real axis. Example 1 1 .6
We evaluate
1 00 (x2 + 1 ) (x2dx+ 4) (x2 + 9) . - oo
1 . Evidently, f satisfies the (z 2 + 1 ) (z 2 + 4) (z 2 + 9) hypotheses of the theorem. The poles of f are at ±i, ±2i and ±3i and are
To do this, let f(z)
=
Theory of Residues
all simple poles. By the theorem,
1
00
_ 00
dx = (x 2 + l ) (x 2 + 4 ) (x 2 + 9) p�� = 27ri
1
P
-p
poles
183
dx (x 2 + l ) (x 2 + 4) (x 2 + 9) (
L
in { Im z
>
Res(! : () 0}
= 21ri ( Res( ! : i) + Res(! : 2i) + Res (! : 3i ) ) . All we now need do is to calculate the residues. For example, Res(f : i) = lim (z - i)f (z ) =
.
1
2z X 3 The other two residues can be similarly calculated. z-H
X
8
Chapter 1 2
The Argument Principle
1 2. 1
Zeros and Poles
We shall see that the number of zeros and poles of a meromorphic function f is determined by the behaviour of the quotient f.
Suppose that f is analytic at zo and that zo is a zero of f of order m. Then f is meromorphic at Zo and Res ( r zo ) m. In fact, the point zo is a simple pole of f.
Theorem 1 2 . 1
=
Proof.
By hypothesis, we can write
f (z)
00
=
an (Z - zo) n L n =m
where am =f. 0 and this power series converges absolutely for all z in some disc D(z0 , R) . Since am =f. 0, it follows that f is not identically zero in D(zo , R) and so there is r > 0 such that f(z) =1- 0 for all z in the punctured disc D' (zo , r) ( otherwise f would vanish, by the Identity Theorem) . It follows that 11 f is analytic in D' ( z0 , r) and so, therefore, is f' I f . Hence, z0 is an isolated singularity of f' I f. Write
f(z) = (z - zo ) m (am + am+ I (z - zo) + . . . ) . 'P (Z)
for z E D(z0 , R) . Differentiating, we get
f'(z) = m(z - zo) m - 1 cp(z) + (z - zo) m cp ' (z) . 185
186
Lecture Notes on Complex Analysis
Now, cp ( zo) = a m =f. 0 and so, by continuity, there is p > 0 such that cp ( z) =f. 0 for all z E D ( z0, p ) . Hence, for any z E D' ( z0 , p ) , we may write m
f' ( z )
f( z )
( z - zo )
cp' ( z )
+
cp ( z )
·
Furthermore, cp' I'P is analytic in the disc D ( z0, p) and so has a Taylor series expansion there. Therefore
f' ( z) f ( z)
=
� n � A n ( z - zo ) ( z - zo ) + n=O m
for suitable coefficients (A n ) , where the series converges absolutely in D( zo , p) . This is the Laurent expansion of f ' If at zo and the result follows. 0
There is a corresponding result for poles.
Suppose that zo is a pole of f of order k . Then simple pole at zo with Res ( { zo) = -k .
Theorem 1 2.2
Proof.
f has a
By hypothesis, f has the Laurent expansion
where bk =f. 0, valid in some punctured disc
f ( z)
=
D ' ( z0, R) .
Hence we may write
'1/J( z)
( z - zo) k in D( zo , R))
where '1/J is analytic at zo (in fact , and '1/J ( zo ) = b k =f. 0 . By continuity, it follows that '1/J is non-zero in some neighbourhood of z0 and so there is some p > 0 such that '1/J is analytic and does not vanish in the disc D( z0 , p) . From the equality above, we see that f does not vanish in the punctured disc D' ( z0, p) . Hence
f' ( z) f ( z)
-k
( z - zo )
+
'1/J ' ( z) '1/J(z)
for z E D ' ( z0 , p) . The last term '1/J' N is analytic in this disc (so has a Taylor expansion) and we deduce that z0 is a simple pole of f' If and that Res( ! ' If : zo ) -k, as claimed. 0 =
187
The Argument Principle
12.2
Argument Principle
Applying these results, we obtain the following theorem.
Suppose that D is a star domain and 'Y is a contour in D such that Ind ('Y : ( ) = 0 or 1 for any ( ¢:. tr 'Y. Suppose, further, that f is meromorphic in D, has a finite num ber of poles in D and is such that none of its poles nor zeros lie on tr "(. Then 1 !' = N-y - P-y 21l"i -y f Theorem 1 2 . 3 (Argument Principle)
1
where N-y is the number of zeros of f inside "( (counting multiplicity) and P-y is the number of poles of f inside "( (counting multiplicity). Proof. This is an immediate consequence of the previous two theorems together with the Residue Theorem, theorem 1 1 .3. 0
It is called the Argument Principle for the following rea son. The expression 2 ; i J'Y f ' / f looks as though it should be j ust 2 ;i log f ] 'Y but we have seen that the logarithm must be handled with care. To make some sense of this, suppose that we break the contour 'Y into a number of ( possibly very small ) subcontours, 'Yk · The integral is then the sum of the integrals along these subcontours. Since f does not vanish on tr "f, we can imagine that each subcontour is contained in some disc also on which f does not vanish. (This is a consequence of the continuity of f.) We can also imagine that these discs are so small ( if necessary) that the values f(z) taken by f on each such disc lie inside some disc in C which does not contain 0. This means that there is a branch of the logarithm defined on the values f(z) , as z varies on a given subcontour, 'Yk · Now the integral really is the logarithm Remark 1 2. 1
1 2 1!"�.
1
'Yk
!' f
=
log f(zk ) - log f (zk - 1 )
= log l f (zk ) l + i arg f (zk ) - log lf(zk- d l - i arg f (zk 1 ) = log lf(zk ) l - log lf(zk - d l + i l:!. k arg f , -
where log denotes the branch constructed above, Zk- 1 and Zk are the ends of "fk and l:!. k arg f denotes the variation of the argument of f ( z) as z moves along 'Yk ·
Lecture Notes on Complex Analysis
188
The idea is to do this for each of the subcontours, but being careful to ensure that the choices of branch of logarithm (or argument) match up at the values of f at the ends of the "fk S · Integrating along "Y means adding up these subintegrals and we see that the log lf(z) l terms cancel out leaving j ust the sum
il\ 1 arg f + il\ 2 arg f + . . . , which is the total variation of the argument of f(z) as z moves around the contour "f. (The point is that we have to keep making possibly different choices of the argument as we go along, depending on the winding behaviour of f(z) around zero.) Given "Y : [a, b] --> C, let r be the contour r(t) Then
Remark 1 2 . 2
for a
�
t � b.
=
f( "Y (t)),
b 1 fJ' = 1 f' (J"Y((t))(t)"Y') (t) dt "Y -; b - J r' (t) dt a
-
= =
a
r(t)
l d:
21ri Ind(r : o) .
Hence, by the Argument Principle, we see that the number of zeros minus the number of poles inside "Y is determined by the winding number of the contour r f 0 "( about the origin;
=
N-; - P-;
=
Ind(r : 0) .
In particular, if f E H (D), then the number of zeros of f inside the closed contour "Y is precisely Ind (f o "(= 0) . Now for fixed w0 E C, any solution to f(z) w0 is a zero of the function f(z ) - w0 , and vice versa. By arguing as above (and using (f - wo ) ' !') , =
=
The Argument Principle
we see that
189
1 (! wo ) ' = 1 f' ..., (! - wo) ...,bf -' wo f ('y (t))'y ' (t) = 1 f(-y(t) dt ) Wo -
a
-
=
-
{
dw lr w - wo 2rri Ind (r : wa )
which is to say that the number of solutions to f(z) = w 0 inside 'Y is equal to the winding number Ind ( f o -y: wa ) . 1 2. 3
Rouche's Theorem
This relationship between the number of zeros of a function and certain winding numbers suggests that the comparison of two functions might be attacked by comparing winding numbers. Suppose that -y1 and 'Y2 are two given closed contours, parameterized by [a, b] and neither passing through the origin. If 'Y2 (t) is "close to" 'YI (t) for every t E [a, b] , then we would expect that they have the same winding number around 0. We can think of t as "time" and the pair of points -y 1 (t) and -y2 (t) moving as a composite system around 0 as if j oined by a spring. An alternative picture is an earth moon system. Here the earth and the moon have the same winding number around the sun, even though they have quite different traj ectories. The issue is what is meant by "close to" in this context? To be sure that the two points -y 1 ( t) and 'Y2 ( t) encircle the origin the same number of times, we must ensure that one of them, say 'Y2 (t) , cannot "duck under" the origin whilst the other, -y 1 (t) , goes "over the top" . This is ensured if -y2 (t) always lies in the disc D('y 1 (t) , l"f 1 (t) 1 ) , with its centre at 'YI (t) and with radius l "f 1 (t) l , since in this case -y2 (t) and -y 1 (t) are always "to the same side of 0" . The requirement that -y2 (t) belong to D ('y 1 (t) , 1"f 1 (t) l ) is to demand that l"f 1 (t) - -y2 (t) l < i'YI (t) l . This discussion leads to the following proposition. Proposition 1 2. 1
by [a , b] , such that
Suppose -y 1 and -y2 are closed contours, parameterized
Lecture Notes on Complex Analysis
190
for all a � t � b. Then First we note that neither of -y1 nor -y2 pass through 0 (otherwise the inequality above would fail) . For t E [a, b] , set -y (t ) -y2 (t) h1 (t) . Then -y ' 'Y2 - 'Yi 'Y 'Y2 'Y l and 'Y is a closed contour satisfying Proof.
=
I I - -y (t ) l < 1 for each t E [a , b] . It follows that tr -y indeed, by theorem 8.2) , Ind(-y: O) However, Ind( - y: O)
C
1 = 2m
.
1 jb
D(I , 1) and so, by theorem 8.4 (or,
1 dzz = 0 . '"Y
-
j
dz b -y ' (t) dt I 1 2rri '"Y z 2rri a -y(t) 1 'Y2 (t) 'Y� (t) dt dt - 1 = 2rri a 'Y2 (t) 2rri a 'YI (t) = Ind ('Y2 : 0) - Ind('Y1 : 0) , =
__
and we conclude that Ind('YI : 0)
=
=
__
jb
0
lnd('Y2 : 0) .
Suppose that f and g are ana lytic in the star-domain D and that 'Y is a contour in D, as in Theorem 1 2. 3. Suppose, further, that I f(() - g (() l < lf(() l , for all ( E tr -y . Then f and g have the same number of zeros inside 'Y (counted according to multiplicity).
Theorem 1 2 .4 (Rouche's Theorem)
Let 'YI f o 'Y and 'Y2 g o 'Y so that I'YI - 'Y2 I < I'YI I · Then, by the Argument Principle, theorem 12.3, and proposition 12. 1 , we have
Proof.
=
=
0
as required.
We shall use Rouche's Theorem to show that z 5 + 14z + 2 has precisely 4 zeros inside the annulus { z : � < lz l < 2 } . To show this, set f(z) z 5 and g (z) z 5 + 14z +2 . Then l z l 2 implies that lf(z) l 32 and l f (z) - g (z) l = I I 4z + 2 1 � 1 4 l z l + 2 30. Hence
Example 1 2 . 1
=
=
=
=
=
191
The Argument Principle
lf( z) - g (z) l < lf(z) l on the circle lzl 2.
By Rouche's Theorem, f and g have the same number of zeros inside the circle, that is, in { z : l z l < 2 } . This number is 5. Therefore z 5 + 14z + 2 has 5 zeros in { z : l z l < 2 } . =
Now put
f(z) 14z and leave g as before. For lzl lf(z) l 14 l z l 14 x � 21 =
=
=
=
� ' we find
=
and
l f(z) - g (z) l l z5 + 2 1 � l z l 5 + 2 (�) 5 + 2 < 10. Therefore lf(z) - g (z) l < I J (z) i on i z l = � and so neither f nor g can vanish for such z and 14z and z 5 + 14z + 2 have the same number of zeros in { z : i z l < � } , namely 1. It follows that four of the five zeros of the polynomial z 5 + 14z + 2 lie in the annulus { z : � < lzl < 2 } . =
=
As another example, reconsider the Fundamental Theorem o f Algebra.
For any n E N and any complex numbers ao , a t , . . . , an - 1 , the polynomial g(z) z n + an - 1 Z n - 1 + · · · + a 1 z + ao has precisely n zeros in the complex plane (including multiplicity). zn . Let r > l ao ! + · · · + !an - 1 ! and also r > 1. Then, Proof. Set f(z) for any z with i z l r, we have n- 1 lf(z) - g (z) l = L a k z k
Theorem 1 2 . 5 (Fundamental Theorem of Algebra)
=
=
=
I
I k=O
�
=
n- 1
L: ia k i i z k l k= O n- 1 L ia k i r k k=O
n- 1 �r =
n- 1
L ia k l k =O
,
because r k � rn - 1 , for 0
�
k � n - 1,
l f(z) / .
By Rouche's Theorem, it follows that f and g have the same number of zeros inside the circle 'Y(t) re 2 rrit , 0 � t � 1, namely n. 0 =
192
Lecture Notes on Complex A nalysis
The following result is a corollary to Rouche's Theorem.
Suppose that f is analytic in the star-domain D and 'Y is a contour in D as in Theorem 12. 3. Suppose, further, that f - w0 does not vanish on tr "f. Then there is o > 0 such that the equations w = f ( z) and Wo f ( z) have the same number of roots inside 'Y whenever w E D( Wo ' o) .
Theorem 1 2 . 6
=
Proof. Since f - wo is continuous and never zero on tr 'Y and tr 'Y is compact, it follows that there is some o > 0 such that l f (z) - wo l ;:::: o for all z E tr "(. ( 1 / (/ - wo ) is continuous and bounded.) Let F(z) = f (z) - wo and G(z) f (z) - w and suppose lw - w0 1 < o. Then =
I F(z) - G(z) !
=
!w - wo !
0. Then, by the Maximum Modulus Principle applied to f on the domain D(zo , R) , we find that f is constant on D(zo , R) . In any event, f is constant on the domain D, by the Identity Theorem. 0
An alternative version of this can be given using Rouches Theorem via Theorem 12.8, as follows .
Suppose f is analytic and non-constant in a domain D . Then for any zo E D, there is z E D such that lf(z) l > l f(zo ) l .
Theorem 1 3 . 3
Proof. Since f(zo) E f (D) and f (D) is open (by Theorem 12.8) , there is some p > 0 such that D (f(zo) , p) � f (D) . Suppose that f (z0) Re i9 where R = l f (zo) l and 0 E R Then for any 0 < r < p, f (z0) + re i9 E D(f(zo) , p) � f(D). Hence there is z E D such that f (z) = f(z0) + re i9 . But then =
l f(z) l as required.
=
1 Re i9 + re i9 l
=
R + r = l f (zo ) l + r > lf (zo ) l , 0
This discussion suggests that if I J I is to achieve a maximum, then this should occur on the boundary of a domain-assuming that the function f is defined there and sufficiently well-behaved. A formulation of this is contained in the next theorem. For this, we note that the closure of a set is the union of the set together with its boundary.
Let D be a bounded domain and suppose that f : D ---+ C is continuous and that f is analytic in D . Then either f is constant on D
Theorem 1 3 .4
Maximum Modulus Principle
199
or l fl attains its maximum on the boundary of D but not in D. In fact, for any z E D (and assuming that f is not constant), l f(z) l < sup l f(w) l w ED
= m � lf(() l = max lf(() l . ( E 8D
(ED
First we note that the boundedness of D implies that of D. This set is also closed and therefore is compact. The continuity of f on D implies that of lfl which therefore is bounded on D and achieves its supremum. That is, there is some ( E D such that Proof.
lf (() l = sup lf (z) l = M, say. zED
By the Maximum Modulus Principle, theorem 1 3 . 2 , if f is non-constant then l f(z) l < M for all z E D. It follows that ( E D \ D = aD, the boundary of the domain D. 0 Remark 1 3. 2
A slight rephrasing of this theorem is as follows . Suppose
that D is a domain with D bounded and suppose that f : D ---> C is continuous and that f is analytic in D. If lf(z) l � M for all ( E an , then either l f(z) l < M for all z E D, or else f is constant on D. For unbounded regions the above result may be false. For example , let D be the infinite horizontal strip
Example 1 3 . 1
7r
7r
D = { z : - - < Im z < - } , 2 2 and let f ( z) = exp ( exp z) , z E C. Then f is entire and so is certainly continuous on D. We claim that f is bounded on the boundary of D. To see this , let ( E an , so that ( = X ± i � for some X E R We have
f(() = exp ( exp ( x ± irr/2)) exp ( e"' e±i,. / 2 ) = exp (±ie"' ) = cos e"' ± i sin e"' . =
It follows that lf (() l = 1 for every ( E aD, the boundary of D.
200
Lecture Notes on Complex Analysis
Is l f(z) l < 1 for all z E D? The answer is no. For example, suppose that z = x E lR s;;; D. Then we find that f (z) = f (x) = exp (exp x ) = ee x . Clearly l f (x) l = eex ---> oo as x never mind being less than 1 . 13.3
---> oo
and so 1 / 1 is not even bounded on D
Minimum Modulus Principle
Maxima of I f I correspond to minima of 1 ] 1 , assuming that we do not need to worry about f being zero. This observation leads to the following Minimum Modulus Principle. Theorem 1 3 . 5 (Minimum Modulus Principle) Suppose f is analytic and non-constant on a domain D and that lf(z) l � m > 0, for all z E D . Then l f (z) l > m for all z E D .
If, in addition, D is bounded and f is defined and continuous and non zero on D, then 1/ (z) l > inf 1/(w) l = min 1 /(() 1 wED
(, E 8D
for all z E D. Proof.
Apply the Maximum Modulus Principle to g =
1 y-
0
As motivation for the next theorem, we observe that if w = a + ib, then = ea ei b , so that I e w I = ea. This idea of looking at exponentials leads to maximum and minimum principles for the real and imaginary parts of an analytic function. ew
Let D be a domain and suppose that f = u + iv E H (D) is non-constant. Suppose M is some real number such that u � M on D. Then u < M on D . Similarly, if there is m E lR such that m � u on D, then m < u on D . A similar pair of statements hold for v . Furthermore, if D is bounded and f i s defined and continuous o n D, then the suprema and infima of u and v over D are attained on the boundary of the domain D .
Theorem 1 3 . 6
Proof.
Suppose that u( x , y)
� M
for z = x + iy E D. Set g(z) = exp (f(z)) eu e iv , for z E D . =
201
Maximum Modulus Principle
Then g is non-constant on D ( otherwise f would be ) and lg (z) l = eu ( x, y ) , so that u � M implies that eu � eM. ( The real exponential function is monotonic increasing. ) Hence l g (z) l � eM on D. By the Maximum Modulus Principle, theorem 13.2 , it follows that and therefore u ( x , y) < M for all x + iy E D. Next, suppose that m � u ( x, y) for all z
=
h(z)
=
exp ( - f(z))
=
x
+ iy E D. P ut
e - u e - iv , for z E D.
Then, h is non-constant on D and , for any z E D ,
so
that
on D , by the Maximum Modulus Principle. It follows that m < u ( x , y) for any x + iy E D , as required. The analogous results for v are obtained by considering the functions exp ( ::r= if(z)) . Finally, we note that g and h are defined and continuous on D if f is. Furthermore , if f is non-constant, neither are g nor h. By the Maximum Modulus Principle, the suprema of the moduli of these functions is attained on the boundary of D ( and not on D itself) . This amounts to saying that the maximum and the minimum of u over D are both attained on the boundary of D ( and not in D) . A similar argument applied to the two functions exp ( ::r= if) leads to the similar statements for v rather than u . 0 1 3.4
Functions on the Unit Disc
Theorem 1 3.7 ( Schwarz's Lemma )
open unit disc D (O, 1) and satisfies f(O) Then lf(z) l � M l z l fo r all lz l < 1 .
Proof.
Let
f(z)
=
=
Suppose that f is analytic in the 0 and l f(z) l � M for all lz l < 1 .
ao + a 1 z + a 2 z 2 + . . .
be the Taylor series expansion of f(z) about ao = f(O) = 0, by hypothesis.
z0
=
0, for z E D (O, 1 ) . Then
202
Lecture Notes on Complex Analysis
=
Put g(z ) a 1 + a 2 z + . . . , so that f(z ) z g(z). The function analytic in D(O, 1) and g(z) f(z)/z for 0 < l z l < 1 . Let 0 < r < 1 and suppose that lzl r. Then
=
i g(z) i
=
= f( z) lf(z) I I= r l �
M
r By the Maximum Modulus Principle, it follows that =
Z
g
is
.
M lg(z) l � r
for all lzl � r. Letting r ----> 1 we see that therefore
i f(z) i for all
z E D(O, 1 ) , as required.
lg(z) l
= lzl lg(z) l �
M
� M,
for all
l zl
< 1 , and
izi 0
Remark 1 3. 3 In fact, the inequality lg(z) l � M for lzl < 1 implies that i g(z) i < M for all lzl < 1 or else g is a constant, g(z) o: and lo:l M. =
But then
for all 0
llm z l so large that K e- y� / n < let R be the rectangle
R = { z : 0 < Re z
w = z + a, with a E C. A rotation is any map of the form z �----> w = ei'P z, with cp E JR. A magnification is a map of the form z �----> w = rz with r > 0. Note that if r > 1 , then this is a magnification in the conventional sense, but if r < 1 then it is a contraction rather than a magnification. Illustrations of these three mappings are given in Figs. 14. 1 , 14.2 and 14.3, below.
w=z+a
Fig. 14.1
Translation:
z ,__. w
207
=
z
+ a, a E IC .
208
Lecture Notes on Complex Analysis
Fig. 14.2
Fig. 14.3
Rotat ion: z
>--+ w =
Magnification: z
e i
0.
Notice that under any of these three special transformations, any given region retains its general shape but may be moved around and magnified (or shrunk) . Next , we discuss inversion , which is somewhat less transparent.
209
Mobius Transformations
14.2
Inversion
Inversion is the map z f-> w = � , for z =f. 0. To see what happens under an z inversion , write z r ei 9 . Then =
1 1 -t w = - = -e 9 z r
Thus, points near zero are transformed far away, points in the upper half plane are mapped into the lower half-plane and vice versa. 1 1 z . Indeed , we can wnte z f-> w = - as z f-.-> w = - = -2 .
z
z
lz l
Let us see what happens to circles and straight lines under an inversion. To say that z lies on a circle, we mean that if z x + iy, then x and y satisfy an equation of the form =
for real numbers a, b, c, d, with a =f. 0. Explicit inclusion of the coefficient a means that we can set a 0 to get the equation of a straight line. That is, equation ( * ) determines either a circle (if a =f. 0) or a straight line (if a = 0) in C . This circle/line passes through the origin depending on whether d 0 or not. Actually, if a =f. 0, then by completing the square, we see that ( *) can be written as =
=
This really is a circle (with strictly positive radius) provided b2 + c2 Let us write � = w = u + iv. Then z
1
z = - =
w
1
u+w
--.-
=
u - iv u 2 + v2
=
. x + zy.
It follows that
u u 2 + v2
X = --;:---:::
and y
Now , if x and y satisfy (* ) , then
a (u 2 + v 2 ) bu + 2 2 2 u +v u + v2
=
-
v . u 2 + v2 --::---:::
>
4ad .
Lecture Notes o n Complex Analysis
210
and so
i.e. , ( u , v ) lies on a straight line or a circle (depending on whether d 0 or not) . It follows that the family of circles and straight lines is mapped into itself under the inversion z �--+ w = ! (the points z = 0 and w 0 z excepted) . =
=
(* )
(** )
a ,;:. 0, d ,;:. 0
circle not through (0, 0)
circle not through (0, 0)
a = 0, d 1:- 0
line not through (0, 0)
circle through (0, 0)
a ,;:. 0, d = 0
circle through (0, 0)
a = 0, d = 0
line through ( 0, 0 )
line not through ( 0 , 0 ) line through ( 0, 0 )
So the family of circles and straight lines is mapped into itself under all four of the special maps considered so far, and therefore also under any composition of such maps. 14.3
Mobius Transformations
Definition 14. 1
A Mobius transformation is a map of the form z �--+
T(z ) =
az + b ez + d
-
where a, b, e, d E C satisfy ad - be 1:- 0. Mobius transformations are also called bilinear transformations or frac tional transformations. The condition ad - be 1:- 0 ensures that the map is defined (except, of course, at z = -d/e if e '#- 0 ) and is not a constant. If a 1:- 0 and c 1:- 0, then T(z ) a(acz + bc )/c(acz + ad ) . If be = ad then this reduces to a constant, namely aje. Also, one can check that the condition ad - be 1:- 0 means that T(z ) = T(w ) if and only if z = w. If w = T(z) (az + b)/ (ez + d ) , then "solving for z" , we calculate that =
=
Mobius Transformations
211
z = ( -dw + b)/ (cw - a ) and (-d) ( -a) - be = ad - be =F 0. Hence -dw + b S(w) cw - a =
is also a Mobius transformation and is the inverse of T. Furthermore, if TI and T2 are Mobius transformations, then so is their composition T2 TI (which maps z into T2 (TI (z) ) ) . It follows that the set of Mobius transfor mations forms a group ( under composition ) with the group identity being the identity transformation I ( z ) z. =
I t turns out that any any Mobius transformation T(z)
=
az + b can be cz + d
expressed in terms of the special transformations considered earlier. To see this , suppose first that c =F 0. Then we write
az + b cz + d
(a/c ) (cz + d ) - ( ad/c ) + b cz + d a (ad / c ) - b cz + d c a be - ad = + 2 � c (z + (d/c)) "
be - ad =F 0. Then c2 T(z)
Let A =
=
a A + ( z + (d/ c) ) c ·
Consider the following maps:
d +c 1 WI 1--+ W2 = WI w2 �---+ WJ IAI ei arg >-w2 a WJ �--+ W = WJ + Z
1--+
WI
=
Z
translation inversion magnification / rotation
=
translation.
C
Then
w i.e., z
�---+
=
a a wa + - = AW2 + C
C
=
A a A a +-= +W I C Z + (d/c) C
-
T(z) = w is given by the composition Z
f--->
trans
WI
f---> i
nv
W2
f--->
mag/rot
W3
f--->
trans
W.
212
Lecture Notes o n Complex Analysis
Now consider the case c = 0, which is a little simpler. First note that now ad =f. be demands that ad =f. 0. In particular , d =f. 0. Therefore T(z) = Consider
az + b = � z+ �. d d d magnification/rotation translation
giving T(z) = w via z
�---+
mag/rot
w1
�---+
trans
w.
We have seen that any Mobius transformation can be expressed as a suitable composition of the special mappings, namely, translation, rotation, magnification and inversion. It follows that the family of circles and straight lines is mapped into itself under any Mobius transformation. Let us consider the inversion z 1 / z in more detail . Consider points z E C with Re z constant, that is points of the form z = a + iy, y E IR , for fixed a E R Such zs form a line parallel to the imaginary axis. Suppose that a =f. 0, so that the line does not pass through the origin. 1 1 u - iv Now, if - = w = u + iv, then z = . and so z = a + iy = u 2 + v 2 z u+w giving v u a = --� and y = 2 2 u +v u u That is, w is such that u and v satisfy a = 2 , or (u 2 + v 2 ) - - = 0 u + v2 a or 2 u+ v2 = 1 2 . 4a 2a
......
--
( 2_ )
__
This is the equation of a circle, centred at ( 2� , 0) and with radius � . So w lies on this circle. To find the range of values taken by w, we use 1 1 a - iy . =-= w = u + zv = z a + iy a 2 + y 2
---
to get and
...::.,..
__
213
Mobius Transformations
z varies on the line, i.e., as y varies in JR., we see that u takes all values between 0 and 1/a, including 1/a but not including 0. Thus ( u - � ) takes 2 on all values between ± 2� , not including - 2� and v takes on all values between ± 2� (including both values ± 21 themselves) . It follows that, for any a =f. 0, the image of the line { z : Re z a } under the inversion z f---> is the set { - � I � } \ {0}. This leads to the 21 1 2 following theorem. As
w � =
w : lw
=
w
Inversion z f---> one-one onto the disc { �I
Theorem 14. 1
=
1
- maps the half-plane { z : Re z > 1 } z � } , and vice versa.
w : lw - < We can use the circles, varying a > 1 , above, or prove this directly as follows. Setting z = x + iy, w 1/ z, we have 2-z < l w - � 1 = 1 � - � � < � 1 -2z I 2 21 - z l < l z l 12 - z l 2 < l z l 2 (2 x) 2 + Y 2 < x2 + Y 2 - 4x + x2 + Y 2 < x2 + Y 2 4( 1 - x) < 0 =
Proof.
as
=
1
{==:}
{==:}
{==:} {==:}
_
{==:} 4 {==:}
{==:}
x > l.
Hence inversion maps the half-plane { z : Re z > 1 } into the disc D ( � , �) and it also maps the disc D (� , �) into the half-plane { z : Re z > 1 } . But inversion is its own inverse, so it follows that it maps { z : Re z > 1 } onto D ( � , �) and D ( � , �) onto { z : Re z > 1 } . 0 We can use this fact to help construct Mobius transformations mapping various given half-planes into discs and vice versa by manoouvering via the "standard" half-plane/disc pair, as above.
w : lw -
2 } onto the disc { 21 3 }.
As can be seen from Fig. 1 4 . 4 , the idea is to manipulate the original half-plane into the standard half-plane { w2 : Re w2 > 1 } using rotations and translations (in fact, one of each) . Inversion now transforms this stan dard half-plane into the standard disc. Further rotations, translations and magnifications now bring the image into the required position.
Lecture Notes on Complex Analysis
214
-i1rf2z
=
-iz
0
1
Fig. 14.4 Manreuvring via the standard half-plane/disc pair.
215
Mobius Transformations
The overall transformation is built from a number of simple ones :
z �---+ WI
e - i'Tr / 2 z = -iz , 1 W 2 I-+ W3 = - ' W3 W2 W4 I-+ W 5 = 6 W4 ,
=
I-+
W4
=
W3 - 2I ,
Combining these, we get
w = w5 + 2 = 6w4 + 2 = 6 (w3 - �) + 2 6 6 = -1 6w3 - 1 = - - 1 W2 WI - 1 6 + iz + 1 6 _1= -iz - 1 -iz - 1 iz + 7 -iz - 1 =
14.4
--
Mobius Transformations in the Extended Complex Plane
The Mobius transformation T : z �---+ (az + b)/ (cz + d) is not defined when z = -d/c (when c =f. 0 ) . Now, we can extend the inversion mapping z �---+ 1 / z on C \ {0} to a mapping from Coo to Coo by the assignments 0 1---+ oo and oo ...... 0 . We mimic this for any Mobius transformation as follows. First recall that ad - be =f. 0, so c and d cannot both be zero. If c = 0 ( so that a =f. 0 ) , set
Tz = If c =f. 0, set
Tz
=
{
{ �� + b) jd,
:o: z� C, .
(az + b) / (cz + d) , for z E C and z =f. -d/c, oo , z = -d/c ajc, z = 00 .
In this way, any Mobius transformation T : z ,..... (az + b)j(cz + d) can be extended to a mapping on C00 • Moreover, one checks that the mapping z ...... ( dz - b)/ ( -cz + a) is the inverse to T on Coo , so that the collection of Mobius transformations on Coo also forms a group under composition.
216
Lecture Notes o n Complex Analysis
L et z1 , z2 and Z3 be any three distinct points in Coo . Then there is a Mobius transformation mapping the triple (z1 , z 2 , z3 ) to (0, 1 , oo) .
Proposition 14. 1
Proof.
Suppose first that z1 , z2 and z3 all belong to C. Set
Tz
=
(z - zi ) (z2 - z3) (z2 - z i ) (z - z3)
·
Then we see that T is a Mobius transformation with the required properties . Now, the cases when one of z1 , z2 or z3 is equal to oo are handled as follows. We define T by if Z l if Z2 if Z3 = =
00 ,
=
00 , 00 .
D
Again, one sees that T has the required properties.
Suppose that z1, z2 and z3 are any distinct points in Coo and that S is a Mobius transformation such that S zJ z1 for j 1 , 2 , 3 . Then Sz z for all z E C00 . az + b Proof. Suppose first that (z1 , z 2, z3) (0, 1 , oo) . Writing Sz = cz + d ' we calculate that
Theorem 1 4 . 2
=
=
=
=
O f---4 0
===}
O = b/d
===}
b=O
and 1
f---4
1 and oo f---4 oo
===}
aj(c + d)
=
1
and c = 0
===}
a = d.
It follows that S z = z for all z. Now consider the general case o f any distinct points z1 , z2 and z3 . Let T be a Mobius transformation which maps (z1 , z 2 , z3 ) to (0, 1 , oo). Then the composition T o S o r- 1 is a Mobius transformation mapping (0, 1 , oo) to (0, 1 , oo) and so according to the above argument it is equal to the identity transformation. It follows that S = r- 1 o T, that is, S is also the identity D transformation, S z z for all z . =
Mobius Transformations
217
If S and T a re Mobius transformations obeying Szj Tzj for any three distinc t points z1 , z2 and z3 in C00 , then S = T. Corollary 1 4 . 1
=
1 , 2 , 3 , is fixed under the map Proof. We simply note that each Zj , j S o r- 1 so that S o T- 1 z = z for all z, by the theorem. It follows that Sz = Tz for all z . D =
Chapter 1 5
Harmonic Functions
15.1
Harmonic Functions
We recall that if f u + iv is analytic in a domain D then it obeys the Cauchy-Riemann equations there; Ux = Vy and uy = -Vx · Indeed, we have seen that =
f' = Ux + ivx
=
Vy - iu y .
Moreover, we know that f has derivatives of any order and so u and v also possess partial derivatives of any order. It follows that Vx y = Vy x and so we find that Uxx = (ux ) x = (vy )x = (vx) y = -Uyy , that is, u satisfies Laplace's equation
Uxx + Uyy = 0 in the domain D. Definition 15.1 We say that the function cp(x, y) is harmonic in the domain D in IR2 if all partial derivatives of cp up to second order exist and are continuous in D and if cp satisfies Laplace's equation 'Pxx + 'P yy = 0 in D.
We can summarize the remarks above by saying that the real part of a function analytic in a domain D is harmonic there. Note also that if cp is harmonic in D , then the mixed partial derivatives 'Px y and 'Py x are equal in D. It is natural to ask whether every harmonic function is the real part of some analytic function. Let cp(x, y) = log y'x2 + y 2 for (x, y) =f. (0, 0) . Then one is harmonic in this region. However, cp(x, y) = Re Log z
Example 1 5 . 1
checks that
cp
2 19
220
Lecture Notes on Complex Analysis
where z = x + iy and it follows that cp is not the real part of any function analytic in the punctured plane D = C \ {0} . In fact, if f E H(D) and if cp = Re f in D, then f and Log z have the same real part in the cut-plane C \ { z : z + l z l 0 } and so differ by a constant there; f(z) = Log z + a for some constant a E C for all z E C \ { z : z + l z l 0 } . But this is not possible because the left hand side has a limit on the negative real axis (with 0 excluded) whereas the right hand side does not. =
=
15.2
Local Existence of a Harmonic Conjugate
Notice that the domain in the example above is not star-like, and it is the presence of holes which allows for such a counterexample. For star-like domains, every harmonic function is indeed the real part of some suitable analytic function, as we now show. Definition 1 5 . 2 Let cp be harmonic in a domain D. A function '1/J (x, y) is said to be a harmonic conj ugate for cp in D if the function f(x + iy) = cp(x, y) + i'ljJ (x, y) is analytic in D.
The analyticity of f implies that '1/J is also harmonic in D. Furthermore, if w(x, y) is also a harmonic conj ugate for cp in D, then the functions cp + i'ljJ and cp + iw are analytic in D and have the same real part and so they differ by a constant. In other words, if cp possesses a harmonic conj ugate in D, then it is unique to within a constant. We can now prove the main result here.
Suppose that u is harmonic in a star-domain D . Then u possesses a harmonic conjugate there, that is, there is some f analytic in D such that u (x, y) = Re f(z) for all z = x + iy E D .
Theorem 1 5 . 1
Proof. For any z = x + iy E D, set g (z) since u is harmonic in D , it follows that
=
u x (x, y) - iu y (x, y). Now,
so the real and imaginary parts of g satisfy the Cauchy-Riemann equations in D. Furthermore, all partial derivatives are continuous in D (because u is harmonic) and so we deduce that g is analytic in D . By hypothesis, D i s a star-domain and s o g possesses a primitive i n D , that is, there i s G E H(D) such that G' g i n D. =
221
Harmonic FUnctions
Let G = U + iV and set f = u + iV. We shall show that f E H(D) . Indeed, general theory tells us that
G'
=
Ux + iVx
=
Vy - iUy
but we know that
G'
=
g
=
U x - iu y
so that U x = Vy and u y = -Vx . In other words, the real and imaginary parts of f obey the Cauchy-Riemann equations in D. Moreover, the partial derivatives of u and V are continuous in D (because u is harmonic and G D is analytic) and so f E H(D) and the proof is complete. Corollary 1 5 . 1 Let u be harmonic in a domain D . For any z0 E D, there is some r > 0 such that u has a harmonic conjugate in the disc D (z0 , r) . Proof. For given z0 E D, there is some r > 0 such that D (z0 , r) Since D (z0 , r) is star-like, u has a harmonic conj ugate there.
15.3
0 and f E H (D (zo , r) ) such that u = Re f in D (z0 , r) . Let g = e f . Then g E H(D(z0 , r)) . Moreover, for any given z = x + iy E D (z0 , r) ,
l g(z) l
=
eu( x ,y )
�
e u( x o,yo )
=
l g (zo) l
and so g is constant in D (z0 , r) by the Maximum Modulus Principle. In particular, e f ( z) = g(z) = g (zo) = ef( zo ) and so (f(z) - f(z0)) /27ri E Z for all z E D (zo , r) . Since f is continuous, this means that f is also constant in the disc D (z0 , r) and so f is constant on D, by the Identity Theorem. But then u Re f is constant on D , a contradiction. We conclude that u has no local maxima. By replacing u by - u in the above, we see that u also has no local D minima. =
222
Lecture Notes on Complex Analysis
Suppose that u i s harmonic and non-constant i n IR2 • Then u is neither bounded from above nor from below.
Corollary 1 5 . 3
First we note that, by the theorem, there is some function f analytic in 0 be given. Then there is N such that l fn (z) - f(z ) l c: for any z E aT , whenever n > N. Hence
0 is such that D (z0 , 2 R) 0. Then, by Cauchy's Integral Formula
I f ( k l (z) - fn( k ) (z) I =
- fn(w) I � 1 f(w) dw I (w - z) k+l 2m
"/
where we take 1 to be the circle 1(t) = z0 + � Re21rit, 0 � t � 1 . Now, for all sufficiently large n, l f(w) - fn (w) l c: for all w in the compact set tq. Moreover, for all such w, lw - z l 2:: ! R. Hence, by the Basic Estimate, theorem 7.1 , we see that the right hand side above is bounded by
f locally uniformly on the domain D . Suppose that zo E D is a zero of f of order m, then for any R > 0 there is a disc D(z0 , r) , with r < R, and N E N such that for all n > N, fn has exactly m zeros in D(zo, r) .
Theorem 16.4
Let R > 0 b e given. Since fn ---> f locally uniformly, there is some disc D (z0, p) such that fn ---> f uniformly on D(zo , p). Then, arguing as above, we deduce that there is 0 < r < mi n {p, R} and a > 0 such that 1/ (z) l � a for all z E C = { l z - zo l = r} . Moreover , we may assume that zo is the only zero of f inside the circle C (because the zeros of f are isolated) . The uniform convergence on compact sets implies that there is N E N such that if n > N then Proof.
lf(z) - fn ( z) l < a for all z E C. Hence
1/1
�a>
I f - fn l
on C and so, by Rouche's Theorem, f and fn have the same number of 0 zeros inside the circle C, namely m.
Local Properties of Analytic FUnctions
16.3
229
Vitali's Theorem
Definition 1 6 . 2 A sequence Un) of functions is said to be locally uni formly bounded on a domain D if for each z0 E D there is some r > 0 with D(zo , r) 0 such that l fn (z) l < M for all n and all z E D(z0 , r) . Note that both r and M may depend on z0•
The result of interest in this connection is Vitali's Theorem ( which we will not prove here ) , as follows.
Let Un) be a sequence offunctions analytic in a domain D. Suppose that Un) is locally uniformly bounded in D and that there is some set A in D such that A has a limit point in D and such that Un(z)) converges for all z E A. Then there is a function f such that fn ---7 f locally uniformly in D (so that, in particular, f E H (D)).
Theorem 1 6 . 5 (Vitali's Theorem)
App endix A
S ome Result s from Real Analysis
We collect here some of the basic results from real analysis that we have needed. They all depend crucially on the Completeness Property of R We begin with some definitions. A.l
Completeness of IR
Definition A.3 A non-empty subset S of IR is said to be bounded from above if there is some M E IR such that a � M for all a E S. Any such number M is called an upper bound for the set S. The non-empty subset S of IR is said to be bounded from below if there is some m E IR such that m � a for all a E S. Any such number m is called a lower bound for the set S. A subset of IR is said to be bounded if it is bounded both from above and from below. Suppose S is a non-empty subset of IR which is bounded from above. The number M is the least upper bound of S ( lub S ) if
( i ) a � M for all a E S ( i.e . , M is an upper bound for S ) . ( ii ) If M' is any upper bound for S, then M � M ' . I f S is a non-empty subset o f IR which i s bounded from below, then the number m is the greatest lower bound of S ( glb S ) if
( i ) m � a for all a E S ( i .e . , m is a lower bound for S ) . ( ii ) If m' is any lower bound for S, then m' � m. Note that the least upper bound and the greatest lower bound of a set S need not themselves belong to S. They may or they may not. The least 231
232
Lecture Notes on Complex A nalysis
upper bound is also called the supremum (sup) and the greatest lower bound is also called the infimum (inf) . Evidently, if M is an upper bound for S, then so is any number greater than M. It is also clear that M is an upper bound for any non-empty subset of S. In particular, sup S is an upper bound for any such subset of S. Note that if M = lub S, then there is some sequence ( xn ) in S such that Xn ---7 M, as n ---7 oo . (Indeed, for any n E N, the number M - � fails to be an upper bound for S and so there is some Xn E S such that Xn > M - � · Hence Xn obeys M - � < Xn ::; M which demands Xn ---7 M.) Analogous remarks apply t o lower bounds and glb S. sup[O, 1] = 1 = sup(O, 1) and inf[O, 1] = 0 = inf(O, 1 ) . Example A.l Note that (0, 1 ) has neither a maximum element nor a minimum element. The essential property which distinguishes IR from Q is the following. The Completeness Property of IR Any non-empty subset of IR which is bounded from above possesses a least upper bound.
A consequence of this property, for example, is that any positive real num ber possesses a square root. In particular, thanks to this we can be confident that y'2 exists as a real number . (It is given by sup{ x : x2 < 2 } .)
If (an ) is an increasing sequence of real numbers and bounded from above, then it converges.
Proposition A.3 is
By hypothesis, { an : n E N } is bounded from above. Let K = lub { an : n E N } . We claim that an ---7 K as n ---7 oo . Le t c: > 0 b e given. Since K i s an upper bound for { an : n E N } , it follows that a n ::; K for all n. On the other hand, K - c: < K and K is the least upper bound of { an : n E N } and so K - c: is not an upper bound for { a n : n E N } . This means that there is some aj , say, with aj > K - c: . But the sequence (an ) i s increasing and so an � ai for all n > j . Hence an > K - c: for all n > j . We have shown that Proof.
K - c: < an
::;
K < K + c:
for all n > j and so the proof is complete.
D
Any sequence (bn ) in IR which is decreasing and bounded from below must converge. Corollary A . l
Some Results from Real A nalysis
Proof.
Just apply the above result to the sequence an = -bn .
233
D
In fact, bn converges to the greatest lower bound of { bn : n E N } . A.2
Balzano-Weierstrass Theorem
Theorem A.6 (Bolzano-Weierstrass Theorem) Any bounded sequence of real numbers possesses a convergent subsequence. Proof.
Suppose that M and m are upper and lower bounds for ( an ) ,
We construct a certain bounded decreasing sequence and use the fact that this converges to its greatest lower bound and so drags a suitable subse quence of ( an ) along with it. To construct the first element of the auxiliary decreasing sequence, let M1 = lub { an : n E N } . Then M1 - 1 is not an upper bound for { an : n E N } and so there must be some n 1 , say, in N such that
Next, we construct M2 as follows. Let M2 = lub { an : n > n 1 } so that M2 :-::; M1 . Moreover, M2 - � is not an upper bound for { an : n > n1 } and so there is some n 2 > n1 such that
M2 - � < an 2 : N imply that Proof.
In particular , for any j >
N,
l ai l � l aj - a N+l l + l a N+l l < 1 + l a NH I It follows that if M
=
·
1 + max{ l a 1 l , l a 2 l , . . . , l a N+l l } , then
for all k E N, i.e. , (an ) is bounded. To show that (an ) converges , we note that by the Bolzano-Weierstrass Theorem , (an ) has some convergent subsequence, say an k � a, as k � oo . We show that an � a. Let c > 0 be given. Then there is k0 E N such that k > k0 implies that
l an k - a l < � €
Since (an ) is a Cauchy sequence , there is m > No imply that
N0 such that both n > No and
I an - a m I < � €
Let
•
•
N = max { ko , No } . Then
I an - al � I an - an k I + l a n k - a l < � € + � € = €
whenever n >
N. Thus an � a as k �
oo as
required.
D
Remark A. 2 Note that a convergent sequence is necessarily a Cauchy sequence. Indeed , if an � a, then the inequality
shows that (an ) is a Cauchy sequence.
235
Some Results from Real Analysis
A.3
Comparison Test for Convergence of Series
The above result enables us to conclude that various series converge even though we may not know their sum. Theorem A.8 (Comparison Test) Suppose that ao , a1 , . . . and b0 , b1 , . . . are sequences in IR such that 0 ::; an ::; bn for all n = 0 , 1 , 2 , . . . .
If the series L::= o bn converges, then so does the series L::= o an .
Proof. Let Sn = L::�=O a k and Tn = L:: �=O b k be the partial sums. Then we see that for n > m
n
n
k = m+l
k = m+l
If L:: := o bn is convergent , then (Tn ) is a Cauchy sequence and so therefore D is ( Sn ) . Hence L::= o an converges.
A.4
Dirichlet's Test
Theorem A.9 (Dirichlet's Test) Suppose a0 , a 1 , is a sequence in IR such that the partial sums Sn = L:: �=O a k are bounded and suppose � 0 is a decreasing positive sequence in IR such that Yo � Yt � Y 2 � Yn 1 0, as n ---7 oo . Then L::= o an Yn is convergent. •
·
·
•
•
·
The proof uses a rearrangement trick ( called "summation by parts" ) . Let Tn = L:: �= O a kYk · Then, for n > m ( by straightforward verification ) , Proof.
Tn - Tm =
n
n
k = m+l
k = m+ l
L a kYk = Sn Yn + l - SmYm+ l + L Sk ( Yk - Yk+ t ) .
Let c: > 0 be given. By hypothesis , there is M > 0 such that I Sn l ::; M , for all n and since Yn 1 0 , as n ---7 oo , there is N E N such that 0 ::; Yn < t: / 2M,
236
for all
Lecture Notes on Comple,; Analysis
n
> N. Hence, if n , m > N,
ITn - Tm l
:S
I Sn Yn +l l + I SmYm+l l +
n
L I Sk l ( Yk - Yk+ I )
k = m+l n :S M Yn +l + M Ym+l + L M(yk - Yk+ l ) k = m+ l = M Yn + l + M Ym+l + M( Ym+ l - Yn + l ) = 2M Ym+l < e.
It follows that (Tn) is a Cauchy sequence in R and therefore convergent. A.5
E �= O an Yn is D
Alternating Series Test
The Alternating Series Test is an immediate consequence ,
as
follows.
Theorem A . l O (Alternating Series Test) Suppose x0 � x1 � x2 � · · · � 0 is a decreasing positive sequence in R such that Xn 1 0 as n � oo .
Then the alternating series xo -x1 +x2 -x3+ · · = E�=0( - l ) n xn converges. n Proof. Set an = ( - l ) and Yn = Xn · Then the partial sums Sn = n L:: ;=O an = � ( 1 + ( - l) ) are bounded so we can apply Dirichlet 's Test to
obtain convergence , as stated. A.6
D
Continuous Functions on [a, b] Attain their Bounds
Next , we look at properties of continuous functions.
Suppose that the function f : [a, b] � R is continuous. f is bounded on [a, b] and achieves its maximum and minimum on Then [a , b] .
Theorem A. l l
We argue by contradiction. Suppose that f is continuous on [a, b] but is not bounded from above. This means that for any given M whatsoever , there will be some x E [a, b] such that f (x) > M. In particular , for each n E N there is some an , say, with an E [a, b] such that f(a n ) > n. The sequence (an)nEN lies in the bounded interval [a , b] and so , by the Bolzano-Weierstrass Theorem , it has a convergent subsequence (an k )kEN , Proof.
Some Results from Real Analysis
237
say; a nk ---+ a as k ---+ oo. Since a � an k � b for all k, it follows that a � a � b. By hypothesis, f is continuous at a and so ank ---+ a implies that f(a nk ) ---+ f ( a ) . But any convergent sequence is bounded (mimic the argument above for Cauchy sequences) . This is a contradiction, so we conclude that f must be bounded from above. To show that f is also bounded from below , consider g = -f. Then g is continuous because f is. The argument j ust presented, applied to g , shows that g is bounded from above. But this j ust means that f is bounded from below , as required. We must now show that there is a: , f3 E [a, b] such that f ( x ) � f ( a ) and f ( x ) � f(/3) for all x E [a , b] . We know that f is bounded so let M = sup{ f ( x ) : x E [a, b] } . Then f ( x) � M for all x E [a , b] . We show that there is some a E [a , b] such that f ( a ) = M. To see this , suppose there is no such a: . Then f(x) < M and so, in particular , M - f is continuous and strictly positive on [a , b] . It follows that h 1/(M - f) is continuous and positive on [a, b] and so is bounded , by the first part. Therefore there is some constant K such that 0 < h � K on [a , b] , that is, =
0
