MEI FP1 Study Resources Complex Numbers 1-4

MEI Further Pure Mathematics 1 Complex Numbers Section 1: Introduction to complex numbers Study Plan Background You have already encountered several number systems in your study of mathematics. Starting with learning to count with the natural numbers, you have progressed through fractions and decimals, negative numbers, and real numbers. Each extension to the number system allows us to solve more problems. For example, the equation x+4=0 has no solution in the natural numbers, but it does in the integers. The equation x² - 2 = 0 has no solution in the rational numbers, but it does have two real solutions, x = 2 and x = - 2 . However, within the real numbers there is no solution to an equation such as z² + 4 = 0. The complex numbers are an extension to the real numbers in which there is a solution to the equation above. It turns out that the complex numbers are the final extension we need for the number system, as all possible polynomial equations have solutions in the complex numbers. In this section you will learn to manipulate complex numbers, and some of the basic terminology associated with complex numbers.

Detailed work plan 1. Read pages 46 – 49 carefully. You will find some help on some of the Activities in the Notes and Examples, as well as an example showing multiplication of complex numbers. 2. The Flash resource Complex numbers and the completed square form shows how the complex roots of a quadratic equation can be related to the graph of the quadratic function. You do not need to know this, but it is interesting extension work. 3. You can practice these techniques using the interactive questions Addition and subtraction of complex numbers, Multiplying complex numbers and Complex conjugates. 4. Exercise 2A Attempt questions 1, 2, 3, 4 and 5. In Question 2 you need only check the roots for the first one or two equations, as this is rather time consuming.

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MEI FP1 Complex numbers Section 1 Study plan 5. Read pages 50 – 53. This covers two important techniques: equating real and imaginary parts, and dividing complex numbers. There are two additional examples in the Notes and Examples, showing how these techniques can be used. 6. You can practice division using the interactive questions Dividing complex numbers. You can also test yourself on some of the techniques in this section using the Flash resource Working with complex numbers. 7. Exercise 2B Attempt at least half of the parts of questions 1 and 2, and also questions 3-9. If you have time try Questions 10 and 11 (you need to have covered the binomial expansion in C1 to do Question 11). You might also like to try the enrichment questions 12 and 13 – these are beyond the syllabus but you have all the tools you need to do them.

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MEI Further Pure Mathematics 1 Complex Numbers Section 1: Introduction to complex numbers Notes and Examples These notes contain subsections on  The growth of the number system  Quadratic equations with complex roots  Working with complex numbers  Equating real and imaginary parts  Dividing complex numbers

The growth of the number system Look at Figure 2.1. This type of diagram is called a Venn diagram (you may have met these before if you have studied any Statistics) and it shows the relationships between sets, in this case sets of numbers. This diagram deals with the real numbers, which include all numbers which you have come across until now. Notice that the positive and negative integers (whole numbers) are subsets of the rational numbers. This means that all integers are also rational numbers, but there are other rational numbers which are not integers, such as 32 or  117 . Similarly, all rational numbers are real numbers, but there are other real numbers which are not rational, such as

3 and π.

In this chapter you will see that the real numbers are also a subset of a larger set called the complex numbers. You will be looking at numbers which lie outside the set of real numbers. Look at Activity 2.2. You may find it difficult to think of a problem which would lead to equation (v). One possible answer is the following: You have two thermometers, A and B, one of which has been calibrated incorrectly, so that the reading on thermometer A is always 7 degrees below the reading on thermometer B. For which readings on thermometer A is the product of the two readings zero? Don‟t spend too long thinking about (vi). Read on!

Quadratic equations with complex roots When you first learned to solve quadratic equations using the quadratic formula, you found that some quadratic equation had no real solutions. However, using complex numbers you can find solve all quadratic equations.

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MEI FP1 Complex nos Section 1 Notes & Examples Example 1 Solve the quadratic equation x² + 6x + 13 = 0 Solution Using the quadratic formula with a = 1, b = 6, c = 13

b  b 2  4ac 2a 6  36  4 113  2 1 6  16  2 6  4i  2  3  2i The solutions of the equation are x = -3 + 2i and x = -3 – 2i x

Notice that the quadratic equation in Example 1 has two complex solutions which are a pair of complex conjugates. All quadratic equations with real coefficients have two solutions: either two real solutions (which could be a repeated solution) or two complex solutions which are a pair of complex conjugates.

The Flash resource Complex numbers and the completed square form shows how the complex roots of a quadratic equation relate to the graph of the quadratic function. You do not need to know this work, but it is interesting extension work.

The next example shows how you can find a quadratic equation with roots at particular complex values. A quadratic equation with roots at x = a and x = b can be written as (x – a)(x – b) = 0, and this also applies to situations where the roots are complex numbers.

Example 2 Find the quadratic equation which has roots at x = 4 + 2i and x = 4 – 2i. Solution (x – (4 + 2i))(x – (4 – 2i)) = 0 (x – 4 – 2i)(x – 4 + 2i) = 0 (x – 4)² – (2i)² = 0 x² – 8x + 16 + 4 = 0 x² – 8x + 20 = 0

The two middle terms cancel out since this expression is of the form

(x – a)(x + a)

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MEI FP1 Complex nos Section 1 Notes & Examples Working with complex numbers In all calculations with complex numbers, remember to use the fact that j² = -1. When you multiply two complex numbers: (x + yj)(u + vj) you get one term (xu) involving just real numbers, two terms in j (xvj and yuj) and one term in j² (yvj²) which is equal to yv × -1 = -yv. So you end up with two real terms and two terms in j.

Example 3 Find (i) (2 + 3j)(4 + j) (ii) (3 – 2j)(2 + 5j) Solution (i) (2 + 3j)(4 + j) = 8 + 2j + 12j + 3j² = 8 + 14j – 3 = 5 + 14j (ii) (3 – 2j)(2 + 5j) = 6 + 15j – 4j – 10j² = 6 + 11j + 10 = 16 + 11j

using j² = -1

Make sure that you have grasped the terminology in this section, in particular that you know what is meant by the real part and the imaginary part of a complex number, and what is meant by a complex conjugate. Complex conjugates are very useful and you should take particular note of Activity 2.4.

You can practise calculations with complex numbers using the interactive questions Addition and subtraction of complex numbers, Multiplying complex numbers and Complex conjugates.

Equating real and imaginary parts Examples 2.2, 2.3 and 2.4 demonstrate two very important techniques which are essential tools for working with complex numbers. The first, shown in Example 2.2, is known as “equating real and imaginary parts”. This is based on the fact that for two complex numbers to be equal, then the real parts must be equal and the imaginary parts must be equal. So one equation involving complex numbers can be written as two equations, one for the real parts, one for the imaginary parts. Example 4 below shows one application of equating real and imaginary parts. There are some problems of this type in Exercise 2B Question 2.

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MEI FP1 Complex nos Section 1 Notes & Examples

Example 4 Find real numbers a and b with a > 0 such that (a + bj)² = 16 – 30j. Solution (a + bj)² = 16 – 30j a² + 2abj + b²j² = 16 – 30j a² + 2abj – b² = 16 – 30j Equating imaginary parts:

2ab = –30  b  

Equating real parts:

a² – b² = 16 225 a 2  2  16 a 4 a  225  16a 2

Substituting: Multiplying through by a²:

15 a

a 4  16a 2  225  0 This is a quadratic in a² and can be factorised: (a² – 25)(a² + 9) = 0 Since a is real, a² + 9 cannot be equal to zero. Since a > 0, the only possible solution is a = 5. 15 15 b    b    3 a 5 a = 5, b = –3

Notice that in this example you are finding the square root of a complex number. The square root of 16 – 30j is 5 – 3j. You can find the other square root of this complex number by allowing a to be negative, so a = -5 and then b = 3. So the other square root is -5 + 3j. As with real numbers, one square root is the negative of the other.

Dividing complex numbers The second important technique shown in this section is demonstrated in Examples 2.3 and 2.4. It is difficult to work with complex numbers in the denominator of an expression, so it can be very useful to be able to make the denominator real. This is sometimes known as “realising the denominator” of a fraction. The technique hinges on the fact that the product of a complex number and its conjugate is real (see Activity 2.4). So by multiplying both numerator and denominator of an expression by the conjugate of the denominator, you make the denominator real, without changing the value of the expression (multiplying the top and bottom of a fraction by the same thing is equivalent to multiplying by 1, which leaves the value unchanged).

You can practice this technique using the interactive questions Dividing complex numbers.

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MEI FP1 Complex nos Section 1 Notes & Examples

Example 5 below shows how an equation involving complex numbers can be solved by using this technique of dividing complex numbers. There are some similar problems in Exercise 2B Question 4.

Example 5 Solve the equation (3 – 2j)(z – 1 + 4j) = 7 + 4j Solution (3 – 2j)(z – 1 + 4j) = 7 + 4j 7  4j z 1 4 j  3 2j (7  4 j)(3  2 j)  (3  2 j)(3  2 j) 21  14 j  12 j  8  94 13  26 j  13  1 2 j z = 1 + 2j – (-1 + 4j) = 2 – 2j

Divide both sides by 3 – 2j

Multiply top and bottom by 3 + 2j [this makes the denominator („bottom‟) real]

Subtract -1 + 4j from each side

Alternatively, you could solve this equation by equating real and imaginary parts. Put z = x + yj, multiply out and equate real and imaginary parts to get two equations in x and y. You can then solve these as simultaneous equations.

You can test yourself on the techniques in this chapter using the Flash program Working with complex numbers.

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Further Pure Mathematics 1 Complex Numbers Section 1: Introduction to complex numbers Crucial points 1. Always simplify j² Remember that when you are working with complex numbers, you should always simplify j² to -1. 2. Remember that zz* is always real In particular, remember that you can use this in dividing complex numbers. 3. Make sure that you know the condition for equality For two complex numbers to be equal, the real parts must be equal and the imaginary parts must be equal.

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MEI Further Pure Mathematics 1 Complex Numbers Section 1: Introduction to complex numbers Exercise 1. Find the roots of the following equations: (i) z2 + 25 = 0 (ii) 2 (iii) z – 2z +2 = 0 (iv) 2. In each of the following cases find a) z1 + z2 b) z1 – z2 e) z2* f) z1* + z2*

4z2 + 9 = 0 4z2 + 4z + 5 = 0

c) z1z2 g) z1* - z2*

d) z1* h) z1*z2*

(i) z1 = 2 + 3j; z2 = 1 – 2j (ii) z1 = -2j; z2 = 3 + j What do you notice about the results? 3. Given that z = (a + j)4 where a is real, find values for a such that (i) z is real, (ii) z is wholly imaginary. 4. Given that a + bj is the conjugate of (a + bj)2 find all possible pairs of values for a and b. 5. Simplify and write in the form a +bj: 1 1  (i) 3 2j 3 2j 4 (ii) 3  j  3 j 3 2j  (iii) 1 j 2  j 6. Find values for a and b given that: (i) (a + bj)(2 + j) = a – 3j (ii) (a + j)(4 – bj ) = 3b + 2aj 7. By writing (a + bj)2 = 3 – 4j, find values for a and b and hence find the square roots of 3 – 4j. 8. Find the values of p and q given that one root of the equation z2 + pz + q = 0 is: (i) 2 – j (ii) 1 – 3j (iii) 2j (iv) 5 – 3j 9. Given that

5 2   1 , where a and b are real, find the values of a and b. a  bj 1  3j

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Further Pure Mathematics 1 Complex Numbers Section 1: Introduction to complex numbers Solutions to Exercise 1. (i) z 2 + 25 = 0

z 2 = −25 z = ±5 j (ii) 4z 2 + 9 = 0 9 z2 = − 4 3 z =± j 2

(iii) z 2 − 2 z + 2 = 0 2 ± 4 − 4× 1 × 2 z= 2 2 ± −4 = 2 2±2 j = 2 =1± j (iv) 4z 2 + 4z + 5 = 0 −4 ± 16 − 4 × 4 × 5 z= 8 −4 ± −64 = 8 −4 ± 8 j = 8 1 =−2 ± j

2. (i) (a) (b) (c) (d) (e) (f)

z1 + z2 = 2 + 3 j + 1 − 2 j = 3 + j z1 − z2 = 2 + 3 j − 1 + 2 j = 1 + 5 j z 1 z 2 = (2 + 3 j)(1 − 2 j) = 2 − 4 j + 3 j + 6 = 8 − j z1 * = 2 − 3 j z2 * = 1 + 2 j z 1 * +z 2 * = 2 − 3 j + 1 + 2 j = 3 − j

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Further Pure Mathematics 1 (g) (h) (ii)(a) (b) (c) (d) (e) (f) (g) (h)

z 1 * −z 2 * = 2 − 3 j − 1 − 2 j = 1 − 5 j z 1 * z 2 * = (2 − 3 j)(1 + 2 j) = 2 + 4 j − 3 j + 6 = 8 + j z 1 + z 2 = −2 j + 3 + j = 3 − j z 1 − z 2 = −2 j − 3 − j = −3 − 3 j z 1 z 2 = −2 j(3 + j) = −6 j + 2 = 2 − 6 j z1* = 2 j z2 * = 3 − j z 1 * +z 2 * = 2 j + 3 − j = 3 + j z 1 * −z 2 * = 2 j − (3 − j) = −3 + 3 j z 1 * z 2 * = 2 j(3 − j) = 6 j + 2 = 2 + 6 j

z 1 * +z 2 * = ( z 1 + z 2 ) * z 1 * −z 2 * = ( z 1 − z 2 ) * z 1 * z 2 * = ( z 1 z 2 )* 3. z = (a + j)4

= a 4 + 4a 3 j + 6a 2 j2 + 4a j3 + j4 = a 4 + 4a 3 j − 6a 2 − 4a j + 1 (i) If z is real, 4a 3 − 4a = 0

4a(a 2 − 1) = 0 4a(a + 1)(a − 1) = 0 so a = 0, -1 or 1. (ii) If z is wholly imaginary, a 4 − 6a 2 + 1 = 0 6 ± 36 − 4 = 3±2 2 a2 = 2 a = ± (3 ± 2 2 )

4. (a + b j)* = (a + b j)2

a − b j = a 2 + 2ab j − b 2 Equating imaginary parts:

Equating real parts:

−b = 2ab

b + 2ab = 0 b(1 + 2a ) = 0 b = 0 or a = − 21 a = a2 −b2

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Further Pure Mathematics 1 If b = 0: a = a 2

If a = − 21 :

a(1 − a ) = 0 a = 0 or 1

− 21 =

b2 =

1 4

−b2

3 4

b = ± 21 3 a=b=0 The possible values of a and b are: a = 1, b = 0 a = − 21 , b = ± 21 3

5. (i)

3−2 j+3+2 j 1 1 + = 3 + 2 j 3 − 2 j (3 + 2 j)(3 − 2 j) 6 = 9+ 4 6 = 13

(ii) 3 + j +

4(3 + j) 4 = 3+ j+ 3− j (3 − j)(3 + j) = 3+ j+

4(3 + j) 9+ 1

= 3 + j + 52 (3 + j) = 57 (3 + j) (iii)

2j 3(1 + j) 2 j(2 − j) 3 − = − 1 − j 2 + j (1 − j)(1 + j) (2 + j)(2 − j) = =

3+3 j 4j+2 − 2 5 15 + 15 j − 8 j − 4

10 11 + 7 j = 10

6. (i) (a + b j)(2 + j) = a − 3 j

2a + a j + 2b j − b = a − 3 j Equating real parts:

2a − b = a

a =b

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Further Pure Mathematics 1 a + 2b = −3 3a = −3 a = −1

Equating imaginary parts:

a = -1, b = -1 (ii) (a + j)(4 − b j) = 3b + 2a j

4a − ab j + 4 j + b = 3b + 2a j 4a + b = 3b

Equating real parts:

2a = b −ab + 4 = 2a −2a 2 + 4 = 2a

a2 +a − 2 = 0 (a + 2)(a − 1) = 0 a = −2 or a = 1

Equating imaginary parts:

a = -2, b = -4 or a = 1, b = 2. 7. (a + b j)2 = 3 − 4 j

a 2 + 2ab j − b 2 = 3 − 4 j Equating imaginary parts:

Equating real parts:

2ab = −4 2 b =−

a 2 −b 2 = 3 a2 −

4

a2

a

=3

a 4 − 4 = 3a 2 a 4 − 3a 2 − 4 = 0 (a 2 − 4)(a 2 + 1) = 0 Since a is real, a = ±2 so b = ∓1 The square roots of 3 – 4j are 2 − j and −2 + j .

8. (i) One root is 2 – j so the other root is 2 + j Equation is ( z − 2 + j)( z − 2 − j) = 0

( z − 2)2 + 1 = 0

z 2 − 4z + 4 + 1 = 0 z 2 − 4z + 5 = 0

p = -4, q = 5

(ii) One root is 1 – 3j so the other root is 1 + 3j

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Further Pure Mathematics 1 Equation is ( z − 1 + 3 j)( z − 1 − 3 j) = 0

( z − 1)2 + 9 = 0

z 2 − 2z + 1 + 9 = 0 z 2 − 2 z + 10 = 0

p = -2, q = 10

(iii) One root is 2j so the other root is -2j Equation is ( z − 2 j)( z + 2 j) = 0

z2 + 4 = 0

p = 0, q = 4

(iv) One root is 5 – 3j so the other root is 5 + 3j Equation is ( z − 5 + 3 j)( z − 5 − 3 j) = 0 ( z − 5 )2 + 9 = 0

z 2 − 10 z + 25 + 9 = 0 z 2 − 10 z + 34 = 0

p = -10, q = 34

9.

5 2 + =1 a +b j 1 + 3 j 1 + 3 j − 2 −1 + 3 j 5 2 =1− = = 1+3j 1+3j 1+3j a +b j

a +b j 5

= = = = =

1+3j −1 + 3 j (1 + 3 j)( −1 − 3 j) ( −1 + 3 j)( −1 − 3 j) −1 − 6 j + 9 1+ 9 8 −6 j 10 4−3 j 5

a +b j = 4 − 3 j

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Further Pure Mathematics 1 Complex Numbers Section 1: Introduction to complex numbers Multiple Choice Test 1) (3 + 4j) – (2 – j) = (a) 5 + 5j (c) 5 + 3j (e) I don’t know

(b) 1 + 5j (d) 1 + 3j

2) [(1 – 2j) + (1 + j)](-3 + j) = (a) -5 + 5j (c) -7 + 5j (e) I don’t know

(b) -7 – j (d) 5 – 5j

3) The roots of the equation z² + 6z + 10 = 0 are (a) 3 + j, 3 – j (c) -3 + 2j, -3 – 2j (e) I don’t know

4) Given that p + qj = 12 5 ,q = 169 169 12 5 (c) p = ,q = 119 119 (e) I don’t know

(a) p =

(b) 3 + 2j, 3 – 2j (d) -3 + j, -3 – j

1 , the values of p and q are given by 12 − 5j

12 5 ,q = − 169 169 12 5 (d) p = ,q = − 119 119

(b) p =

5) Which of the following complex numbers is not equal to the others? (a) 2 – 3j 13 2 − 3j (e) I don’t know

(c)

13 2 + 3j 3+ 2j (d) j

(b)

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Further Pure Mathematics 1 6) z =

3+ 4j . The complex number w which satisfies the equation zw = 1 is 2 − 3j

6 + 17 j 25 −6 − 17 j (c) w = 5 (e) I don’t know

−6 − 17 j 25 6 + 17 j (d) w = 5

(a) w =

(b) w =

7) The solution of the equation (3 – j)(z + 4 – 2j) = 10 + 20j is (a) z = 1 – 3j (c) z = 46 – 52j (e) I don’t know

(b) z = -3 + 9j (d) z = 5 + 5j

8) Which of the following is NOT true? (a) j4 = 1 1 3 +j =0 j (e) I don’t know

(c)

1 − j=0 j3 1 (d) 2 = j2 j

(b)

9) The values of a and b (with a > 0) which satisfy (a + bj)² = 5 + 12j are (a) a = 3, b = -2 (c) a = 2, b = 3 (e) I don’t know

(b) a = 3, b = 2 (d) a = 2, b = -3

10) Which of the following statements is NOT true? z is real z* (c) zz* is real (e) I don’t know

(a)

(b) z + z* is real (d) z – z* is pure imaginary

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Further Pure Mathematics 1 Complex Numbers Section 2: The Argand diagram Study Plan Background If you want to place a complex number on a number line, you have a problem. Is 1 + j larger or smaller than 1? Clearly this kind of question just does not make sense. In this section you will learn about the Argand diagram, which provides a way of representing complex numbers geometrically, in two dimensions, in the same way that a number line can represent the real numbers in one dimension. You will also learn how you can use an Argand diagram to represent a set of complex numbers described by an inequality.

Detailed work plan 1. Read the section “Representing complex numbers geometrically”. There is an example of finding the modulus of a complex number in the Notes and Examples, and some further notes on representing addition and subtraction of complex numbers using vectors. 2. There is also a Flash resource on The Argand diagram to look at, and a Geogebra resource Addition in the Argand diagram. 3. Exercise 2C Attempt all the questions. 4. Read pages 58 – 60, on sets of points in the Argand diagram. This is fairly straightforward and is summed up in the Notes and Examples. Pay careful attention to the Discussion point which follows Example 2.5. If you have time look at the enrichment example (2.6) as well. 5. You can also look at the Flash resource Investigation of loci and the interactive spreadsheet Investigating loci, both of which look at loci in the Argand diagram. In both of these you can look at sets of points of the form |z – a| = r, and also at sets of points of the form |z – a| = |z – b|, which is enrichment work. 6. Exercise 2D Attempt Questions 1 and 2. If you have time try the starred questions 3 and 4 and the enrichment question 5.

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MEI Further Pure Mathematics 1 Complex Numbers Section 2: The Argand diagram Notes and Examples These notes contain subsections on  Representing complex numbers geometrically  Sets of points in an Argand diagram

Representing complex numbers geometrically In Activity 2.1 at the start of this chapter, you looked at the relationships between numbers in two different ways. You saw that sets of numbers can be represented in a Venn diagram, with some sets of numbers being a subset of another set: e.g. the integers are a subset of the rational numbers. You were also asked to represent numbers on a number line. You have probably used number lines from a very early stage in your mathematical development. Even irrational numbers can be placed on a number line: for example, 2 can be expressed to as many decimal places as you like. You then found that all the types of number that you have met so far can be considered to be a subset of a larger set of numbers: the complex numbers. This can be represented on the Venn diagram by a larger set encircling the set representing the real numbers. However, if you want to place a complex number on the number line, you have a problem. Is 1 + j larger or smaller than 1? Clearly this kind of question just does not make sense. The Argand diagram provides a way of representing complex numbers geometrically, in the same way that a number line can represent the real numbers. The modulus of a complex number z is the distance of the point representing z from the origin on the Argand diagram. Notice that this definition also holds for real numbers on the number line: the modulus (or absolute value) of a real number is its distance on the number line from zero. In the same way, |z – w| (or |w – z|) is the distance of the point representing z from the point representing w. This also holds for real numbers on the number line: the distance of a real number x from a real number y on the number line is |x – y| (or |y – x|). For example, the distance between 2 and -3 on the number line is |2 – (-3)| = 5. Look at the Flash resource The Argand diagram. As you move the point around, the complex number represented by the point is given, and you can also see the modulus. Ignore the part about the argument – this will be introduced in the next section.

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MEI FP1 Complex nos Section 2 Notes and Examples Example 1 Given that z = 2 + 5j and w = 3 – j, find (i) |z| (ii) |w| (iii) |z – w| Solution (i)

z  22  52  29

(ii)

w  32  (1)2  10

(iii)

z – w = 2 + 5j – (3 – j) = -1 + 6j

z  w  (1)2  62  37

As well as thinking of a complex number z = x + yj as a point with coordinates  x (x, y), you can also think of it as a vector   . This could be a position vector  y (a vector from the origin to the point (x, y)) but it can be any vector (sometimes called a directed line segment) parallel to this. Im

z z z

Re

All the vectors on this diagram represent the complex number z = 2 + 3j. Notice that it is the vector itself that is labelled z, not the point at the end of it.

z

Don’t worry if you don’t know very much about vectors (they are covered in some depth in C4) – all you need to know here is how to represent the addition and subtraction of vectors on a diagram. Im Addition of two complex numbers Here z1 = 3 + j and z2 = 1 + 2j. You can see from the diagram z 1 +z 2 that z1 + z2 = 4 + 3j, as you z2 would expect from adding z1 z1 and z2 together. Re

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MEI FP1 Complex nos Section 2 Notes and Examples Subtraction of two complex numbers You can think of subtraction in two different ways: either by thinking of z1 – z2 as adding together the vectors z1 and –z2 (shown in the diagram on the left) or by going from the point z2 to the point z1 (shown in the diagram on the right). In either case, with z1 = 3 + j and z2 = 1 + 2j, you can see that the vector z1 – z2 is given by 2 – j, the result you would expect from subtracting the complex number z2 from z1. Im Im

z2

z1 z1-z2

-z2

z1-z2 z1

Re

Re

You can see how complex addition works geometrically using the Geogebra resource Addition in the Argand diagram.

Sets of points in an Argand diagram This section is fairly straightforward. It can be summed up as follows:  Read |z – (a + bj)| as ‘the distance from z to the point a + bj [note |z – a + bj| can be written |z – (a – bj)|, which is the distance from z to the point (a – bj)].  All sets of points given by |z – (a + bj)| = r can be represented by a circle, centre a + bj, radius r.  |z – (a + bj)| ≤ r represents the circle and its interior.  |z – (a + bj)| < r represents the interior of the circle (but not the circle itself). In this case you should draw the circle as a dotted line, to show that it is not included in the set of points.  |z – (a + bj)| ≥ r represents the circle and its exterior.  |z – (a + bj)| > r represents the exterior of the circle (but not the circle itself). Again, you should draw the circle as a dotted line, to show that it is not included in the set of points. You may like to read through the enrichment example as well. This looks at sets of points given by |z – (a + bj)| = |z – (c + dj)|. You can look at the Flash resource Investigation of loci, which shows both circles and the type of locus shown in the enrichment example. There is also a loci spreadsheet which works in a similar way – use the slide bars to change the variables and see the effect on the Argand diagram.

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Further Pure Mathematics 1 Complex Numbers Section 2: The Argand diagram Crucial points 1. Make sure that you can plot complex numbers correctly on the Argand diagram Remember in particular that the points z and z* are reflections of each other in the x axis, and that the points z and –z are rotations of each other through 180° about the origin. 2. Make sure that you know how to find the modulus of a complex number In particular, don’t forget to take the square root! 3. Remember the significance of the modulus in the Argand diagram The modulus of a complex number z, |z|, is the distance of the point representing z from the origin, and |z – w| is the distance between the points representing z and w. 4. Make sure that you know how to show addition and subtraction in the Argand diagram You need to understand that a complex number can be represented not only by a point in the Argand diagram, but alternatively by a vector. 5. Know the form of a circle in the Argand diagram You need to recognise that any set of points of the form |z – (a + bj)| = r represents a circle, centre a + bj, radius r.

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Further Pure Mathematics 1 Complex Numbers Section 2: The Argand diagram Exercise 1. Given that z1 = 3 + 2j and z2 = 4 – j, represent z1, z2, z1 + z2 and z1 – z2 on an Argand diagram. 2. Given that z1 = 12 + 5j and z2 = -3 + 4j, verify that z1 + z2 ≤ z1 + z2 . Explain geometrically using an Argand diagram why z1 + z2 ≤ z1 + z2 is always true. 3. Given that z1 = 12 + 5j and z2= 3 – 4j verify that z1 − z2 ≥ z1 − z2 . With reference to an Argand diagram give a geometric explanation of this result. 4. Given that z = 2 + j show on an Argand diagram z, z*, jz, and jz*. What transformation describes the relationship between (i) z and z* (ii) z and jz. 5. Draw an Argand diagram showing the set of points z for which the given condition is true. (i) z − 1 + j = 1 (ii)

z − 2 − 3j < 4

(iii) 1 < z < 2 6. Given that z is a complex number such that ⏐z – j⏐ = 1, find the greatest and least values of ⏐z + 1⏐.

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Further Pure Mathematics 1 Complex Numbers Section 2: Loci in the Argand diagram Solutions to Exercise Im

1.

z1

z1 – z2 z1 + z2 Re

z2

2. z 1 = 12 + 5 j ⇒ z 1 = 12 2 + 5 2 = 13

z 2 = −3 + 4 j ⇒ z 2 = 3 2 + 4 2 = 5 so z 1 + z 2 = 18

z1 + z2 = 9 + 9 j ⇒ z1 + z2 =

92 + 92 = 9 2 = 12.7 (1 d.p.)

so z 1 + z 2 ≤ z 1 + z 1 .

Im P1

z1 O

z1 + z2

z2

P2

P

Re

The vector representing z 1 + z 2 is the diagonal of a parallelogram whose sides are formed by the vectors representing z 1 and z 2 . So OP = z 1 + z 2

OP1 = P2 P = z 1 OP2 = P1 P = z 2 For any triangle, the length of one side is always shorter than the sum of the lengths of the other two sides. For triangle OPP1, OP ≤ OP1 + P1 P

z1 + z2 ≤ z1 + z2

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Further Pure Mathematics 1 3. z 1 = 12 + 5 j ⇒ z 1 = 12 2 + 5 2 = 13

z 2 = 3 − 4 j ⇒ z 2 = 3 2 + 42 = 5 so z 1 − z 2 = 8

z1 − z2 = 9 + 9 j ⇒ z1 − z2 =

92 + 92 = 9 2 = 12.7 (1 d.p.)

so z 1 − z 2 ≥ z 1 − z 1 . Im P1

z1 O

z1 – z2 z2

P2

Re

The vector representing z 1 − z 2 is the third sides of a triangle whose other two sides are formed by the vectors representing z 1 and z 2 . So P1 P2 = z 1 − z 2 OP1 = z 1 OP2 = z 2 For any triangle, the length of one side is always shorter than the sum of the lengths of the other two sides. For triangle OPP1, OP1 ≤ OP2 + P1 P2

z1 ≤ z2 + z1 − z2 z1 − z2 ≥ z1 − z2 Im

4. z = 2 + j

z* = 2 − j jz = j(2 + j) = 2 j − 1 = −1 + 2 j jz * = j(2 − j) = 2 j + 1 = 1 + 2 j

(i) z* is obtained by reflecting z in the real axis. (ii) jz is obtained by rotating z through 90° anticlockwise about the origin.

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jz

jz*

z Re

z*

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Further Pure Mathematics 1 5. (i) z − 1 + j = 1

z − (1 − j) = 1 This is a circle, centre 1 – j, radius 1.

(ii) z − 2 − 3 j < 4

z − (2 + 3 j) < 4 This is the interior of a circle, centre 2 + 3j, radius 4.

(iii) 1 < z < 2 This is the region in between two circles, centres the origin, radii 1 and 2 respectively. 6. The point z lies on a circle, centre (0, 1) and radius 1. The value of |z + 1| is the distance of the point z from the point (-1, 0). The diagram shows the points P and Q which give the least and greatest values of |z + 1| respectively. 3

y

2

1

Q B

P –2

x

A –1

1

2

–1

The distance AB = 1 2 + 1 2 = 2 PB = 1 so AP = 2 − 1 . AQ = 1 + AB = 2 + 1 . The greatest and least values of |z + 1| are

2 + 1 and

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Further Pure Mathematics 1 Complex Numbers Section 2: The Argand diagram Multiple Choice Test 1) The modulus of the complex number z = 2 – 5j is (b) 29

(a) 29 (c) 7 (e) I don’t know

(d)

7

Questions 2 – 4 refer to the Argand diagram below. Im

A Re C

B

2) In the Argand diagram, the point A represents the complex number (a) 3 – 2j (c) 2 – 3j (e) I don’t know

(b) -3 + 2j (d) -2 + 3j

3) In the Argand diagram, the point B represents the complex number (a) 4 + j (c) 1 + 4j (e) I don’t know

(b) -4 – j (d) -1 – 4j

4) In the Argand diagram, the point C represents the complex number (a) 3 – j (c) -3 + j (e) I don’t know

(b) 1 – 3j (d) -1 + 3j

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Further Pure Mathematics 1 Questions 5-7 refer to the Argand diagram below. The point representing the complex number z is shown on the diagram. Im P

Q

z

R Re

V

S U

T

5) The point which represents z* is (a) V (c) Q (e) I don’t know

(b) R (d) T

6) The point which represents jz is (a) U (c) P (e) I don’t know

(b) S (d) Q

7) The point which represents –z is (a) V (c) S (e) I don’t know

(b) R (d) T

8) The set of points for which |z – 2 + 3j| = 4 is (a) a circle, centre -2 + 3j, radius 4 (c) a circle, centre 2 – 3j, radius 4 (e) I don’t know

(b) a circle, centre 2 – 3j, radius 2 (d) a circle, centre -2 + 3j, radius 2

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Further Pure Mathematics 1 Im

9)

Re

The shaded area in the Argand diagram represents the points z for which (a) |z + 2 – j| ≤ 2 (c) |z + 2 – j| < 2 (e) I don’t know 10)

(b) |z – 2 + j| ≤ 2 (d) |z – 2 + j| < 2

Im

Re

The shaded area in the Argand diagram above represents the points z for which (a) 2 ≤ |z + 1 – j| ≤ 6 (c) 2 ≤ |z – 1 + j| ≤ 6 (e) I don’t know

(b) 1 ≤ |z + 1 – j| ≤ 3 (d) 1 ≤ |z – 1 + j| ≤ 3

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MEI Further Pure Mathematics 1 Complex Numbers Section 3: Modulus and argument Study Plan Background In this section you will look at writing a complex number in a different form. Until now you have dealt with complex numbers in the form z = x + yj, so the number is written in terms of the real and imaginary parts. The alternative form is in terms of the modulus (the distance of the point z from the origin in the Argand diagram, which you met in the last section) and the argument (the angle between a line from the origin to the point z, and the real axis). You will also see how sets of points in the Argand diagram can be described in terms of the argument of a complex number.

Detailed work plan 1. Read pages 61 – 66. If you have not yet done the trigonometry work in C2, you can find some extra help HERE on angles measured in radians, and angles greater than 90°. Look in particular at Example 2.7 and make sure that you understand when you need to add or subtract π to the argument. 2. You can use the Flash resource The Argand diagram to investigate the modulus and argument of complex numbers. 3. The enrichment Activity 2.12 is quite time-consuming but gives an introduction to work you will come across if you do FP2. A full solution is given in the Notes and Examples. The Flash resource Multiplication and division in the Argand diagram illustrates this work. 4. Exercise 2E Attempt all the questions. 5. For practice in finding the modulus and argument of a complex number, try the interactive questions Complex numbers: polar form. 6. For some extra practice in manipulating complex numbers in different forms, try the Complex numbers puzzle. 7. Read pages 66 – 68, paying particular attention to Example 2.8. There is an extra example in the Notes and Examples. 8. Exercise 2F Attempt Questions 1 to 3, and if you have time try Question 4.

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MEI Further Pure Mathematics 1 Complex Numbers Section 3: Modulus and argument Notes and Examples These notes contain subsections on  The modulus-argument form of complex numbers  Sets of points using the modulus-argument form  Extension work: multiplication and division

The modulus-argument form of complex numbers You are familiar with describing a point in the plane using Cartesian coordinates. However, this is not the only way of describing the location of a point. One alternative is to give its distance from a fixed point (usually the origin) and a direction (in this case the angle between the line connecting the point to the origin, and the positive real axis). In the context of complex numbers this is usually called the modulus-argument form; in other contexts it is known as using polar coordinates. This is a common method of describing locations in real life: you might say that a town is “50 miles north-west of London”, or when walking in open countryside your map might show you that you need to walk 2 miles on a bearing of 124°. In mathematics there are some situations in which this method of describing points is more convenient than Cartesian coordinates. In this section you will look at complex numbers in the modulus-argument form. There is more about polar coordinates and about the modulus-argument form of complex numbers in FP2. You can explore the modulus and argument of a complex number using the Flash resource The Argand diagram.

To write a complex number z in modulus-argument form, all you need to do is to find the modulus, r, and the argument, θ. Then write the complex number as z = r(cos θ + j sin θ). Finding the modulus is straightforward enough, but finding the argument involves a little more work. It involves using some knowledge of Trigonometry from C2, including radians, and angles greater than 90°. However, if you haven’t covered this work yet, don’t worry. Look at the Trigonometry notes which give some help on these topics. (If you have already done the chapter on Matrices in FP1, the work you did on rotation matrices will have given you some confidence in dealing with angles greater than 90°, though you still may need some help with radians.)

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MEI FP1 Complex nos section 3 Notes and Examples To find the argument of the complex number z = x + yj, you need to find the y y value of arctan . Activity 2.8 shows that using a calculator to find arctan x x gives angles, in radians, between 

  and . This gives the correct angle for 2 2

complex numbers in the first and fourth quadrants (i.e. those with positive real parts), but not for complex numbers in the second and third quadrants (i.e. those with negative real parts). To find an argument in the second quadrant, you need to add π to the answer the calculator gives, and to find an argument in the third quadrant, you need to subtract π from the answer your calculator gives. The Trigonometry Notes give a fuller explanation of this. Activity 2.9 will help you to see whether you have understood this correctly! Activity 2.11 reminds you of the values of sin, cos and tan for the common angles

  

, , (i.e. 30°, 45°, 60°). When you come across these angles, you 6 4 3 are expected to use the exact values.

Example 1 Find the modulus and argument of each of the following complex numbers. (i) z = 4 + 3j (ii) z = –1 + j (iii) z = 1  3j Solution (i) z = 4 + 3j |z| =

(32  42 )  5

Since z lies in the first quadrant, arg z = arctan (ii)

3  0.644 4

z = –1 + j |z| =

(12  12 )  2

Since z lies in the second quadrant, arg z = arctan(-1) +  =  (iii)

 4

 

3 4

z = 1  3j Modulus = 1  3  2 Since z lies in the third quadrant, arg z = arctan 3   

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  

2 3

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MEI FP1 Complex nos section 3 Notes and Examples Sometimes you may want to find a complex number if you are given its modulus and argument. Im You can see from the diagram that z = x + yj x = r cos  y = r sin  r y

 O

x

Re

These relationships allow you to find the real and imaginary parts of a complex number with a given modulus and argument.

Example 2 Find the complex numbers with the given modulus and argument, in the form x + jy. 3 (i) z  3, arg z  4 (ii) z  2, arg z  



6

Solution

3 4 3 1 3 x  3cos  3   4 2 2 3 1 3 y  3sin  3  4 2 2 3 3 z  j 2 2

(i) r = 3,  =

 6  

(ii) r = 2,  = 

3 x  2 cos     2   3 2  6 1   y  2sin     2    1 2  6 z  3j

For practice in finding the modulus and argument of complex numbers, try the interactive questions Complex numbers: polar form.

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MEI FP1 Complex nos section 3 Notes and Examples For practice in all the work on manipulating complex numbers in different forms, try the Complex numbers puzzle.

Sets of points using the modulus-argument form In the previous section, you looked at sets of points defined using the modulus of a complex number. You will now look at sets of points defined using the argument of a complex number. All sets of points of the form arg( z  (a  bj))   consist of a half-line from the point a + bj in the direction θ (see Example 2.8 (i) and (ii)). Sets of points of the form 1  arg( z  (a  bj))  2 consist of the two half-lines arg( z  (a  bj))  1 and arg( z  ( a  bj))  2 and the region between them (see Example 2.8 (iii)). Be careful with sets of points of the form arg( z  (a  bj))   or arg( z  (a  bj))   . Make sure that you know where the region starts and ends. Remember that the possible values of arg z are given by   arg z   , and that any line which is not included in the set of points should be shown as dotted (see Example 1 below).

Example 3 Draw Argand diagrams to show each of the following sets of points.  3 arg( z  1  2j)  arg( z  2  j)  (i) (ii) 3 4 Solution (i)

arg( z  1  2j) 

 3

means that arg( z  1  2j) can take any value between –π

  , including but not –π. The half-line from –1 – 2j in the direction –π 3 3  is therefore shown dotted, and the half-line from –1 – 2j in the direction is 3 shown as solid. The point –1 – 2j is not included in the region, since arg( z  1  2j) is not defined where z = –1 – 2j, so this point is shown by an open circle. Im and

 3

-1-2j

Re

arg( z  1  2j) 

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MEI FP1 Complex nos section 3 Notes and Examples (ii)

3 3 means that arg( z  2  j) can take any value between 4 4 3 and π, including π but not since the inequality involves > rather than ≥. 4 3 The half-line from 2 + j in the direction is therefore shown dotted, and the 4 half-line from 2 + j in the direction π is shown solid. The point 2 + j is not included in the region, since arg( z  2  j) is not defined where z = 2 + j, so this point is shown by an open circle. arg( z  2  j) 

arg( z  2  j) 

3 4

Im

3 4

2+j Re

Extension work: Multiplication and division The enrichment Activity 2.12 shows one use of the modulus-argument form. The full solution to this Activity is given below. Solution (i) (a) wz  (1  j)(1  3j)

 1  3j  j  3  1  3  (1  3) j w 1 j  (b) z 1  3j 

(1  j)(1  3j) (1  3j)(1  3j)



1  3j  j  3 1 3



1 4

1 

3  (1  3) j



(ii) (a) w  12  12  2

arg w  arctan1 



w is in the first quadrant

4

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MEI FP1 Complex nos section 3 Notes and Examples (b) z  12  3  2





arg z  arctan  3  

 3

z is in the fourth quadrant

(c) wz  (1  3)  (1  3) 2  1  2 3  3  1  2 3  3  8  2 2 2

 1 3   arg( wz )  arctan     12  1 3  w 1 2 (d)  4 (1  3) 2  (1  3) 2  14 8  z 2  1  3  7  w arg    arctan    1  3 z   12 w w (iii) wz  w z ,  z z

wz is in the fourth quadrant

w is in the fourth quadrant z

 w arg( wz )  arg w  arg z, arg    arg w  arg z z (iv) wz  r1r2 (cos 1  jsin 1 )(cos  2  jsin  2 )

 r1r2 (cos 1 cos  2  jcos 1 sin  2  jsin 1 cos  2  sin 1 sin  2 )

 r1r2   cos 1 cos  2  sin 1 sin  2   j  sin 1 cos  2  cos 1 sin  2    r1r2  cos 1   2   jsin 1   2  

This step involves using the compound angle formulae for sine and cosine: sin( A  B)  sin A cos B sin B cos A and

cos( A  B)  cos A cos B sin A sin B . If you have done the matrices chapter, you will have derived these in activity 1.10. You will also meet them in C4.

w r1 (cos 1  jsin 1 )  z r2 (cos  2  jsin  2 ) 

r1 (cos 1  jsin 1 )(cos  2  jsin  2 ) r2 (cos  2  jsin  2 )(cos  2  jsin  2 )



r1 (cos 1 cos  2  jcos 1 sin  2  jsin 1 cos  2  sin 1 sin  2 ) (cos 2  2  sin 2  2 )

cos²θ + sin²θ = 1

r  1   cos 1 cos  2  sin 1 sin  2   j  sin 1 cos  2  cos 1 sin  2   r2 

r1  cos 1  2   jsin 1  2   r2

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MEI FP1 Complex nos section 3 Notes and Examples

wz  r1r2 ,

w r1  w  , arg( wz )  1   2 , arg    1   2 . z r2 z

This result gives a quick and easy way of multiplying and dividing complex numbers in the modulus-argument form. To multiply complex numbers, you multiply the moduli and add the arguments, and to divide the complex numbers you divide the moduli and subtract the arguments.

The geometrical effect of this in an Argand diagram can be seen using the Flash resource Multiplication and division in the Argand diagram. This topic is covered more fully in FP2.

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Further Pure Mathematics 1 Trigonometry needed for Complex numbers In your work on complex numbers, you need to be able to find the argument of a complex number, which may involve dealing with angles greater than 90°. If you haven’t yet done the work on Trigonometry in C2, or the chapter on Matrices in FP1, this may be new to you. However, the notes and practice questions below should give you all the help you need.

Radians It is usual to express the argument of a complex number in radians. Radians are a different way of measuring angles. There are 2π radians in a complete turn, so that 2π radians = 360°. Angles in radians are often, though not always, given in terms of π. To convert degrees to radians, multiply by To convert radians to degrees, multiply by

π 180 180

π

. .

Try these for practice. (Answers at the bottom of the last page). 1. 2.

3.

Convert the following angles to radians, giving your answers in terms of π. (i) 30° (ii) 60° (iii) 45° (iv) 90° (v) 120° (vi) 150° (vii) 75° (viii) 100° Convert the following angles to radians, giving your answers to two decimal places. (i) 64° (ii) 87° (iii) 127° (iv) 59° (v) 216° (vi) 310° (vii) 21° (viii) 178° Convert the following angles, given in radians, to degrees. Give your answers to one decimal place where appropriate. 5π 7π 3π 11π (i) (ii) (iii) (iv) 3 6 2 12 (v) 2 (vi) 3 (vii) 2.4 (viii) 1.75

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Further Pure Mathematics 1 The argument of a complex number

Im

The diagram shows a complex number z = x + yj joined to the origin. The modulus of z is r and the argument is θ. In this case θ is between 0 and

π

2

z = x + yj r

. We say that z is in the first

θ

quadrant.

x

O

y You can see from the diagram that tan θ = x The real and imaginary parts of z are both positive, so tan θ is positive. y To find the value of θ, you just need to find arctan . x

π

2 in the third quadrant.

Re

Im

r

However, there is another possibility for which tan θ is positive. If both the real part and the imaginary part of z are negative, −π < θ ≤ −

y

O

x

. In this case z is

y

y x

θ

Re

r

z = x + yj

y will give x you the corresponding angle in the first quadrant, (see diagram) where x and y are both positive. To find the correct argument, you need to subtract π.

However, as tan θ is positive, using your calculator to find arctan

Next we need to look at the cases where tan θ is negative. One possibility is for the real part of z to be positive and the imaginary part negative, so that

π

≤ θ < 0 . In this case, z is in the fourth 2 quadrant. −

Im

y To find the value of θ, find arctan . Your x calculator should give you a negative angle, in the fourth quadrant as required.

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O

x

θ r

y

Re

z = x + yj

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Further Pure Mathematics 1 The value of tan θ is also negative if the real part of z is negative, but the imaginary part is positive, so that

π

≤ θ ≤ π . In this case z is in the 2 second quadrant.

Im

z = x + yj y

y will give x you the corresponding angle in the fourth quadrant, with x positive and y negative (see diagram). To find the correct angle in the second quadrant, you need to add π.

Using your calculator to find arctan

r θ x

x

O

Re

r

y

Now look at Example 2.7 in the textbook.

Try these for some extra practice. Find the argument of each of the following complex numbers. Remember to give your answers in radians. (i) 1+j (ii) 1–j (iii) -1 + j (iv) -1 – j (v) 1 + 2j (vi) -1 + 2j (vii) -1 – 2j (viii) 1 – 2j (ix) 3 – 4j (x) -2 – 7j (xi) 5 + 3j (xii) -4 + 2j (Answers at the bottom of the next page).

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Further Pure Mathematics 1

Answers to questions Radians 1.

(i) (v)

2. 3.

(i) (v) (i) (v)

π 6 2π 3 1.12 3.77 300° 114.6°

(ii)

π

3 5π (vi) 6 (ii) 1.52 (vi) 5.41 (ii) 210° (vi) 171.9°

(iii) (vii) (iii) (vii) (iii) (vii)

π 4 5π 12 2.22 0.37 270° 137.5°

The argument of a complex number (i) (v) (ix)

π

4 1.11 -0.93

(ii) (vi) (x)

−

π

4 2.03 -1.85

(iii) (vii) (xi)

3π 4 -2.03 0.54

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(iv) (viii) (iv) (viii) (iv) (viii)

π 2 5π 9 1.03 3.11 165° 100.3°

(iv) (viii) (xii)

3π 4 -1.11 2.68

−

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Further Pure Mathematics 1 Complex Numbers Section 3: Modulus and argument Crucial points 1. Be very careful when you find the argument of a complex number Always decide first which quadrant the complex number is in, and y remember that when you have worked out the value of arctan on your x calculator, this will only be correct for complex numbers in the first and fourth quadrant. For the second quadrant, you need to add π, and for the third quadrant you need to subtract π. It’s a good idea to make a rough sketch of the number on an Argand diagram, so you can ‘see’ the argument. E.g.

z = a + bj

arg z

θ

arg z = π − θ ,

⎛b⎞ where θ = arctan ⎜ ⎟ ⎝a⎠

2. Use the modulus-argument form correctly Remember that the modulus-argument form of a complex number must be of the form r (cos θ + jsin θ ) , with r positive. 3. Think about the end point of a set of points When showing sets of points involving the argument on an Argand diagram, remember that for the set of points arg( z − (a + bj)) = θ the point z = a + bj is not included and should be shown by an open circle. 4. Make sure that you know the possible values for the argument Remember that the possible values of arg z are given by −π < arg z ≤ π . Make sure when drawing sets of points of the form arg( z − (a + bj)) ≤ θ or arg( z − (a + bj)) ≥ θ that you use the correct range for the argument.

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Further Pure Mathematics 1 Complex Numbers Section 3: Modulus and argument Exercise 1. Write each of the following in modulus-argument form. 10 (i) −2 3 − 2j (ii) 3−j 2. Write each complex number in the form x + yj.

(i) z = 3, arg z =

π

4

(iii) z = 2, arg z = −

(ii)

z = 6, arg z =

π

2π 3

6

3. Given that z = 1 + 2j, write in modulus–argument form the complex numbers 1 1 (ii) z* (iii) (iv) (i) z z z* What do you notice? 4. Find the moduli and principal arguments of w, z, wz, and

w , given that z

(i) w = 10j, z = 1 + 3j (ii) w = −2 3 + 2j, z = 1 − j What do you notice? 5. Draw an Argand diagram showing the set of points z for which the following conditions are true: 2π (i) arg( z + 2) = − 3 2π (ii) arg( z + 2 + j) = − 3 π (iii) < arg( z − 2) < π 3 6. Find the greatest and least values of arg z if ⏐z + 2j⏐= 1. 7. Given that z = cos θ + j sin θ find arg (z + 1). (Hint: draw an Argand diagram.) 8. Find a complex number z whose argument is

π 4

and which satisfies the equation

z +2+ j = z −4+ j .

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Further Pure Mathematics 1

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Further Pure Mathematics 1 Complex Numbers Section 3: Modulus and argument Solutions to Exercise 1. (i)

z = −2 3 − 2 j is in the third quadrant.

(2

z =

3 ) + 2 2 = 12 + 4 = 4 2

π 5π ⎛ 2 ⎞ ⎛ 1 ⎞ arg z = arctan ⎜ ⎟ − π = arctan ⎜ ⎟ −π = −π = − 6 6 ⎝2 3⎠ ⎝ 3⎠ 5π ⎞ ⎛ ⎛ 5π ⎞⎞ z = 4 ⎜ cos ⎛⎜ − ⎟ + j sin ⎜ − ⎟⎟ ⎝ 6 ⎠⎠ ⎝ ⎝ 6 ⎠ 10 ( 3 + j ) 10 ( 3 + j ) 5 10 5 = = = 3+ j 3 − j ( 3 − j )( 3 + j ) 4 2 2 This is in the first quadrant.

(ii) z =

z =

5 2

(

3 ) + 12 = 2

5 2

4=5

⎛ 1 ⎞ π arg z = arctan ⎜ ⎟= ⎝ 3⎠ 6 π π z = 5 ⎛⎜ cos + j sin ⎞⎟ 6 6⎠ ⎝

2. (i)

r = 3, θ =

π 4

π

3 4 2 π 3 y = r sin θ = 3 sin = 4 2 3 3 z= + j 2 2

x = r cos θ = 3 cos

(ii) r = 6, θ =

=

2π 3

2π ⎞ 1 ⎟ = 6 × − = −3 2 ⎝ 3 ⎠ 2π ⎞ 3 y = r sin θ = 6 sin ⎛⎜ =3 3 ⎟ = 6× 2 ⎝ 3 ⎠ z = −3 + 3 3 j

x = r cos θ = 6 cos ⎛⎜

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Further Pure Mathematics 1 (iii) r = 2, θ = −

π 6

π 3 = 3 x = r cos θ = 2 cos ⎛⎜ − ⎞⎟ = 2 ×

2 ⎝ 6⎠ π 1 y = r sin θ = 2 sin ⎛⎜ − ⎞⎟ = 2 × − = −1 2 ⎝ 6⎠ z = 3− j

z =1+2 j

3. (i)

z = 12 + 22 = 5 2 z is in the first quadrant, so arg z = arctan ⎛⎜ ⎞⎟ = 1.11 (3 s.f.) ⎝1⎠ z = 5 ( cos 1.11 + j sin 1.11 ) (ii) z * = 1 − 2 j

z * = 12 + 22 = 5

−2 ⎞ ⎟ = −1.11 (3 s.f.) ⎝ 1 ⎠

z* is in the fourth quadrant, so arg z * = arctan ⎛⎜ z = 5 ( cos( −1.11) + j sin( −1.11)) 1

(iii)

z 1

z 1

=

1 5

12 + 22 =

5 1 = 5 5

is in the fourth quadrant, so arg

z

z=

(iv)

1−2 j 1−2 j 1 = = 1 + 2 j (1 + 2 j)(1 − 2 j) 5

=

1 ( cos( −1.11) + j sin( −1.11)) 5

−2 ⎞ = arctan ⎛⎜ ⎟ = −1.11 (3 s.f.) z ⎝ 1 ⎠ 1

1+2 j 1+2 j 1 1 = = = 5 z * 1 − 2 j (1 − 2 j)(1 + 2 j) 1 1 5 1 = 12 + 22 = = 5 5 z* 5 1 2 1 = arctan ⎛⎜ ⎞⎟ = 1.11 (3 s.f.) is in the first quadrant, so arg z* z* ⎝1⎠ 1 z= ( cos 1.11 + j sin 1.11 ) 5 1

z

=

1 1 1 1 1 = = and arg z = arg = − arg z * = − arg . z* z z* z* z

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Further Pure Mathematics 1 4. w = 10 j, z = 1 + 3 j

w = 10 argw =

π 2

z = 1+3 =2 ⎛ 3⎞ π z is in the first quadrant so arg z = arctan ⎜ ⎟= ⎝ 1 ⎠ 3

wz = 10 j(1 + 3 j) = 10 j − 10 3 = −10 3 + 10 j wz = 10 3 + 1 = 20

π 5π ⎛ 1 ⎞ wz is in the second quadrant so arg(wz ) = arctan ⎜ ⎟+π = − +π = 6 6 ⎝− 3⎠ 10 j 10 j(1 − 3 j) 10 j + 10 3 5 3 + 5 j w = = = = 1+3 2 z 1 + 3 j (1 + 3 j)(1 − 3 j) w 5 = 3+1 = 5 2 z w w ⎛ 1 ⎞ π is in the first quadrant so arg ⎛⎜ ⎞⎟ = arctan ⎜ ⎟= z ⎝z ⎠ ⎝ 3⎠ 6

wz = w z ,

w w = z z

w arg(wz ) = argw + arg z , arg ⎛⎜ ⎝z

5. (i)

arg( z + 2) = −

⎞ = argw − arg z ⎟ ⎠

2π 3

This is a half-line, starting at -2, at an 2π below the positive real axis. angle of 3

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Further Pure Mathematics 1 (ii) arg( z + 2 + j) =

π 2

This is a half-line, starting at -2 – j, at an angle of

(iii)

π 3

π

above the positive real axis.

2

-2 - j

≤ arg( z − 2) ≤ π

The boundaries of this region are two half-lines, both starting at 2, one at an angle to

π

π

3

to the

3 positive real axis, and the other on the negative real axis.

6. z + 2 j = 1 is a circle, centre -2j and radius 1.

1

y x

From diagram, sin θ =

θ = 30°

1 2

–2

–1

θθ

1

2

–1

The greatest value of arg z is − 2π The least value of arg z is − 3

π

–2

3 –3 –4

7.

sin θ cos θ + 1 2 sin 21 θ cos 21 θ = 2 cos 2 21 θ = tan 21 θ ⇒ φ = 21 θ arg( z + 1) = 21 θ tan φ =

sin θ

1

z φ

z+1 cos θ

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Further Pure Mathematics 1 8.

z +2 + j = z −4+ j ( x + 2)2 + ( y + 1)2 = ( x − 4)2 + ( y + 1)2

x 2 + 4 x + 4 = x 2 − 8 x + 16 12 x = 12 x =1 π y = x tan θ = x tan = x 4 When x = 1, y = 1

so the complex number is 1 + j.

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Complex Numbers Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must find two expressions that are equivalent to one another. To build up the puzzle, place the edges on which equivalent expressions are written together, so that the triangles are joined along this edge. Begin by just finding pairs of matching expressions and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

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(4

j)(4 − 3

j) + 3

5

1 + 2

10

2x

4j( 3 −

6j − 12

2

x − 4

+ 5

= 0

(3

(9

j) 3 +

(8 − j) 7 −

3 +

7j) j) 5 +

3 − 8j 5 + 2j

x

189 − 23j

(2 − 3j)(1 + 4j)(5 + 2j)

2− 3 + 3j 2j

(1 ) j + 2

x 4 −

± j

7 − 5j 3 + j

8j − 12 j 2

36j

+ 2

= 0

1

7 −

3j

−7

29j

6

4

3(2j − 4)

3 j)(1 − 3 j)

1 . 6 −

2 . 2 j

)(3 j −

j) 5 +

(3

j)(1 4 +

cos π 4 +j sin π 4

j)(5 7 −

29 +

(5 − 3j)(4 + 3j)

2

(1 +

− 1 − 46j 29

(6

53

+

(2 ) j 3

j) 2 +

3j

(9 − 6j) + (4 − 3j)

2 ± j

13

− 4 − 6j

(2 4 − j 2j + 7

6j −

1

25

j) 3 −

7 2 − j) − 2

π π + j sin 3 3

12j

6j 3 −

(7 j) − + 4

2 cos

−j

5j)

−

(3 − 6j) 2

28 + 12j

−5

8j −

15

(1

+

+ j

3j

2 6 − j) 9 −

6j

1

−

7j 15j − 26 53

2

(2

8

7j

4j) (2 −

6 0 +

4

(3 +

27 j

x 2 − 2x + 2 = 0

2

53j

(1 + 4j) 2 +

13 − 9j

23 +

Further Pure Mathematics 1 Complex Numbers Section 3: Modulus and argument Multiple Choice Test 1) The principal argument of the complex number -1 + (a)

(c)

2π 3

(b) −

π

(d) −

3 (e) I don’t know

3 j is

2π 3

π

3

2) The principal argument of the complex number 2 – 2j is (a) − (c)

π

π

(b) −

4

(d)

4 (e) I don’t know

3π 4

3π 4

3) The complex number -3 – 4j can be written in modulus-argument form as (a) 5(cos (-2.21) + j sin(-2.21)) (c) -5(cos (-2.21) + j sin(-2.21)) (e) I don’t know

(b) 5(cos (0.923) + j sin(0.923)) (d) 5(cos (-0.923) + j sin(-0.923))

4) The complex number -2 + 5j can be written in modulus-argument form as (a)

29(cos( −1.19) + jsin(−1.19))

(c) 29(cos(1.19) + jsin(1.19)) (e) I don’t know

(b)

29(cos(1.95) + jsin(1.95))

(d)

29(cos( −1.95) + jsin( −1.95))

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Further Pure Mathematics 1 5π 5π ⎞ ⎛ 5) The complex number 3 ⎜ cos + jsin ⎟ is equal to 6 6 ⎠ ⎝ 3 3 3 + j 2 2 3 3 3 (c) − + j 2 2 (e) I don’t know

3 3 3 − j 2 2 3 3 3 (d) − j 2 2

(a) −

(b)

6) The complex number with modulus 2 and argument -1.5 is (a) 1.99 – 0.14j (c) –0.14 + 1. 99j (e) I don’t know

(b) 0.14 – 1.99j (d) –1.99 +0.14j

7) The modulus and principal argument of the complex number z = −2 ( cos α − jsin α ) (where 0 < α ≤ π2 ) are, respectively, (a) 2, -α (c) -2, α (e) I don’t know

(b) 2, π – α (d) 2, α + π

Im

8)

1+2j π 4

Re

The bold half-line in the diagram above shows the set of points z for which (a) arg( z − 1 − 2 j) =

π 4

(c) arg( z + 1 + 2 j) = −

(b) arg( z − 1 − 2 j) = −

π 4

(d) arg( z + 1 + 2 j) =

π

π

4

4

(e) I don’t know

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Further Pure Mathematics 1 Im

9)

π 3

-1+j Re

The shaded area in the diagram above shows the set of points z for which 2π 3 2π (c) 0 ≤ arg( z + 1 − j) ≤ 3 (e) I don’t know

2π 3 2π (d) arg( z + 1 − j) ≤ 3

(a) 0 ≤ arg( z − 1 + j) ≤

(b) arg( z − 1 + j) ≤

Im

10)

-3-2j

Re π 4

The shaded area in the diagram above shows the set of points z for which

(a) arg( z + 3 + 2 j) ≤ (c) arg( z + 3 + 2 j) ≥

π 4

π

4

(b) arg( z − 3 − 2 j) ≤ (d) arg( z − 3 − 2 j) ≥

π 4

π

4

(e) I don’t know

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Further Pure Mathematics 1 Complex Numbers Section 4: Complex numbers and equations Study Plan Background At the start of this chapter, complex numbers were introduced as a means of giving solutions to equations such as x² + 10 = 0, which have no solutions in the real numbers. You can now find the solutions of any quadratic equation, and you have seen that the complex roots of a quadratic equation are complex conjugates. In this section you will look at the solutions of cubic and higher order equations. It turns out that any polynomial equation of order n has exactly n roots, including complex roots and repeated roots.

Detailed work plan 1. Read pages 69 – 71, paying careful attention to Examples 2.9 and 2.10. There is some background information in the Notes and Examples, and some web links if you would like to find out about the formulae for solving cubic and quartic equations. In this section you need to be able to use the factor theorem and to factorise polynomials when you know a linear or quadratic factor. If you have covered the work on the factor theorem in C1, this should be no problem. If you haven’t, read the extra notes on the factor theorem, or look at chapter 3 of the AS Pure Mathematics textbook. 2. The PowerPoint presentations Factorising cubics and Factorising quartics show examples of solving polynomials with complex roots. 3. Exercise 2G Attempt at least three of Questions 1-6. Questions 7-11 are examination style questions. Attempt at least three of these.

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MEI Further Pure Mathematics 1 Complex Numbers Section 4: Complex numbers and equations Notes and Examples These notes contain subsections on  Complex numbers and equations  Solving equations with complex roots

Complex numbers and equations You now know from your work on complex numbers that every quadratic equation has exactly two solutions, if you count repeated roots and complex roots. There are two possibilities:

The graph crosses the x axis twice. There are two real distinct roots. If the graph just touches the axis, then the root is repeated, but this still counts as two roots.

The graph does not cut the x axis. There are two complex roots which are a conjugate pair.

For a cubic equation, there are also two possibilities:

The graph cuts the x axis three times. In the diagram there are three real distinct roots. However, two of the roots could be the same, in which case the graph would touch the axis at one of the turning points, or all three roots could be the same, in which case there would be a point of inflection on the x axis.

Here the graph cuts the x axis only once. There is one real root, and there is also a conjugate pair of complex roots.

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MEI FP1 Complex nos Section 4 Notes and Examples You can see graphically that a cubic equation must have at least one real root. If the term in x³ is positive, then for large positive values of x the value of the function is large and positive, and for large negative values of x the value of the function is large and negative. (If the term in x³ is negative, this is reversed). So all cubic graphs must cut the x axis at least once. Of course, the real root may not be an integer or even a rational number, so you may not be able to find it! However, any cubic equations you meet in this section will have a simple real root, so that you can solve it. A formula does exist for solving all cubic equations, but it is extremely complicated. Find out more here and here. For quartic equations, there are three possibilities:

There may be four real roots, some of which may be repeated.

There may be two real roots, and a conjugate pair of complex roots.

There may be no real roots. In this case, there are two conjugate pairs of complex roots.

There is, again, an extremely complicated formula for the roots of a quartic equation. Find out more here. For higher degree equations, there are no general formulae to find the roots. It is not simply that no-one has managed to find them yet: it was proved by Galois (a very interesting character) that no such formulae exist for any polynomial equations higher than quartic. If there happens to be one or more integer root which you can find by trial and error, it may be possible to solve a higher degree equation by algebraic methods. Otherwise, there are numerical methods which provide approximate solutions. As the textbook states, the main difficulty in proving that any polynomial equation of degree n has exactly n roots is proving that any polynomial has at least one root. If you assume that a polynomial of degree n has at least one root, then you can express the polynomial as a product of a linear factor and a polynomial of degree n-1. Then, since the assumption that any polynomial has at least one root also holds for the new polynomial of degree n-1, then you can express this polynomial as a product of a linear factor and a polynomial of degree n-2. And so on, until the polynomial has been factorised into n linear

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MEI FP1 Complex nos Section 4 Notes and Examples factors, giving n roots. (This applies even if the roots are irrational or complex). For example, if you know one root of a quintic equation, you can express it as the product of a linear factor and a quartic factor. Then since a quartic equation must have at least one root, you can express the quartic factor as the product of a linear factor and a cubic factor. Since a cubic equation has at least one root, you can express the cubic factor as the product of a linear factor and a quadratic factor, which can be factorised using the quadratic formula. So, if we can prove that all polynomial equations have one root, then we can prove that a polynomial equation of degree n has exactly n roots using the method above. (This is an example of proof by induction, in which you show that if a statement is true for n, then it is also true for n+1. The proof has been stated very informally here: you will learn about proof by induction in Chapter 5). Proving that all polynomial equations have at least one root is much more difficult: there are a number of approaches, all well beyond ‘A’ level. You can use a graphical approach to show that all polynomials of odd degree have at least one root, as described above for cubics, however, this is not a rigorous proof!

Solving equations with complex roots In practice, the situations you are likely to encounter include 

cubics where you are given one complex root. In this case you can deduce a second complex root which is the conjugate of the first, and use these two roots to find a quadratic factor of the cubic. You can then factorise the cubic into the quadratic factor and a linear factor (by inspection or polynomial division) and deduce the third (real) root from the linear factor. Example 2.9 deals with a problem of this type.



cubics where you are given the real root (or told that an integer root exists, which you can find by trial and error). In this case you can factorise the cubic into a linear factor and a quadratic factor, by inspection or polynomial division, and then use the quadratic formula to find the other two roots.



quartics where you are given a complex root. In this case you can again deduce a second complex root which is the conjugate of the first, and find a quadratic factor. You can then factorise the quartic into two quadratics, and use the quadratic formula to find the other two roots (which could be real or complex). Example 2.10 deals with a problem of this type.



quartics where you are given one or two real roots, or told that they exists. Find the real roots by trial and error if you need to, then factorise

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MEI FP1 Complex nos Section 4 Notes and Examples the quartic into the two known linear factors and a quadratic factor, which you can then use to find the other two roots.

If you have already covered the work on the Factor Theorem in C1, you will be familiar with the techniques of factorising a polynomial by inspection or by polynomial division. If not:  read the additional notes on the factor theorem  look at the PowerPoint presentations Factorising cubics (which demonstrates the different methods of factorising a cubic when you know a linear factor) and Factorising quartics (which shows you how to factorise the quartic equation in Example 2.10 into two quadratic factors).

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Further Pure Mathematics 1 The factor theorem The factor theorem states that: f(a) = 0 ⇔ (x – a) is a factor of f(x) For example, suppose the cubic polynomial f(x) is defined by f(x) = x³ – 2x² – x + 2 You can work out the value of f(x) for different values of x. f (1) = 13 − 2 ×12 − 1 + 2 = 1 − 2 − 1 + 2 = 0 f (2) = 23 − 2 × 22 − 2 + 2 = 8 − 8 − 2 + 2 = 0 f (3) = 33 − 2 × 32 − 3 + 2 = 27 − 18 − 3 + 2 = 8 f (−1) = (−1)3 − 2(−1) 2 + 1 + 2 = −1 − 2 + 1 + 2 = 0 f (−2) = (−2)3 − 2(−2) 2 + 2 + 2 = −8 − 8 + 2 + 2 = −12 f (−3) = (−3)3 − 2(−3) 2 + 3 + 2 = −27 − 18 + 3 + 2 = −41

You can see that the polynomial is zero when x = 1, 2 or -1. (In fact these are the only values of x for which the polynomial is zero.) The factor theorem tells us that (x – 1), (x – 2) and (x + 1) are factors of f(x). Hence f(x) = (x – 1)(x – 2)(x + 1). Usually, if you need to factorise a cubic equation, you find one factor by trial and error, and then factorise the cubic into a linear factor and a quadratic factor. This is shown below.

Example Solve the equation 2x³ – 5x² – 4x + 3 = 0

Try some different values for x until you find one which makes the polynomial zero

Solution f(x) = 2x³ – 5x² – 4x + 3 f(1) = 2×1³ – 5×1² – 4×1 + 3 = 2 – 5 – 4 + 3 = -4 f(2) = 2×2³ – 5×2² – 4×2 + 3 = 16 – 20 – 8 + 3 = -9 f(3) = 2×3³ – 5×3² – 4×3 + 3 = 54 – 45 – 12 + 3 = 0 so (x – 3) is a factor. The cubic can be factorised into the 2x³ – 5x² – 4x + 3 = 0 (x – 3)(2x² + x – 1) = 0 (x – 3)(2x – 1)(x + 1) = 0 The solution is x = 3, ½, or -1

For more help, look at C1 chapter 3.

linear factor (x – 3), and a quadratic factor. There is a PowerPoint presentation showing how this is done.

Now the quadratic can be factorised.

Further Pure Mathematics 1 Complex Numbers Section 4: Complex numbers and equations Crucial points 1. Remember that complex roots of polynomial equations with real coefficients always occur in conjugate pairs This means that if you know one complex root, then you know another one. Note: this does not apply if the coefficients of the equation are not real! 2. Make sure that you can divide a polynomial by a linear or quadratic factor If you find this difficult, there are two PowerPoint presentations in this section which demonstrate factorisation of polynomials by inspection or algebraic long division. 3. Check your work carefully It is easy to make mistakes in the algebra when solving polynomial equations.

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Further Pure Mathematics 1 Complex Numbers Section 4: Complex numbers and equations Exercise 1. Write in the form f(z) = 0, where f(z) is a polynomial of degree 4 with real coefficients, the equation having (3 + j) and (1 + 3j) as two of its roots 2. Find the real root of the equation z3 + z + 10 = 0, given that one complex root is 1 – 2j. 3. Given that 1 + j is a root of the equation z3 – 2z + k = 0 find the other roots and a value for k. 4. Show that z = -1 + j is a root of the equation z4 – 2z3 – z2 + 2z + 10 = 0 and find the remaining roots 5. Given that one root of the equation z3 + az + b = 0 is p + qj where a and b are real and b is not zero, prove that i) ii) iii)

2p(p2 + q2) = b 3p2 – q2 = -a p is a root of the equation 8x3 + 2ax – b = 0.

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Further Pure Mathematics 1 Complex Numbers Section 4: Complex numbers and equations Solutions to Exercise 1. 3 + j is a root, so 3 – j is also a root. Therefore a quadratic factor is ( z − 3 − j)( z − 3 + j) = ( z − 3)2 + 1 = z 2 − 6 z + 10

1 + 3j is a root, so 1 – 3j is also a root. Therefore a quadratic factor is ( z − 1 − 3 j)( z − 1 + 3 j) = ( z − 1)2 + 9 So the equation is ( z 2 − 6 z + 10)( z 2 − 2 z + 10) = 0

= z 2 − 2 z + 10

z 4 − 8 z 3 + 32 z 2 − 80 z + 100 = 0 2. 1 – 2j is a root, so 1 + 2j is also a root. So a quadratic factor is ( z − 1 + 2 j)( z − 1 − 2 j) = ( z − 1)2 + 4

z 3 + z + 10 = ( z 2 − 2 z + 5 )( z + 2)

= z 2 − 2z + 5

The real root is z = -2.

3. (1 + j)2 = 1 + 2 j − 1 = 2 j

(1 + j)3 = 2 j(1 + j) = 2 j − 2 = −2 + 2 j Substituting into z 3 − 2 z + k = 0 : −2 + 2 j − 2(1 + j) + k = 0 −2 + 2 j − 2 − 2 j + k = 0

k =4 1 + j is a root, so 1 – j is also a root. So a quadratic factor is ( z − 1 − j)( z − 1 + j) = ( z − 1)2 + 1

z 3 − 2 z + 4 = ( z 2 − 2 z + 2)( z + 2)

= z 2 − 2z + 2

so the other two roots are 1 – j and -2.

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Further Pure Mathematics 1 4. z = −1 + j

z 2 = ( −1 + j)2 = 1 − 2 j − 1 = −2 j z 3 = −2 j( −1 + j) = 2 j + 2 = 2 + 2 j z 4 = (2 + 2 j)( −1 + j) = −2 − 2 = −4 Substituting into z 4 − 2 z 3 − z 2 + 2 z + 10 : −4 − 2(2 + 2 j) − ( −2 j) + 2( −1 + j) + 10

= −4 − 4 − 4 j + 2 j − 2 + 2 j + 10 =0 so -1 + j is a root. Since -1 + j is a root, -1 – j is also a root So a quadratic factor is ( z + 1 − j)( z + 1 + j) = ( z + 1)2 + 1 = z 2 + 2z + 2 z 4 − 2 z 3 − z 2 + 2 z + 10 = ( z 2 + 2 z + 2)( z 2 − 4z + 5 ) The other factors are the roots of the quadratic equation z 2 − 4z + 5 = 0 4 ± 16 − 4 × 1 × 5 z= 2 4 ± −4 = 2 4±2 j = 2 =2± j So the other roots are -1 – j, 2 + j and 2 – j.

5. z = p + qj

z 2 = ( p + qj)2 = p 2 + 2 pqj − q 2 = p 2 − q 2 + 2 pqj z 3 = ( p 2 − q 2 + 2 pqj)( p + qj) = p( p 2 − q 2 ) + (2 p 2q + q( p 2 − q 2 )) j − 2 pq 2 = p 3 − 3 pq 2 + (3 p 2q − q 3 ) j Substituting into z 3 + az + b = 0 : p 3 − 3 pq 2 + (3 p 2q − q 3 ) j + a( p + qj) + b = 0 (i) Equating imaginary parts: 3 p 2q − q 3 + aq = 0 3 p 2 − q 2 = −a (ii) Equating real parts: p 3 − 3 pq 2 + ap + b = 0

p 3 − 3 pq 2 + p( −3 p 2 + q 2 ) + b = 0 −2 p 3 − 2 pq 2 = −b 2 p( p 2 + q 2 ) = b

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Further Pure Mathematics 1 (iii) From (i), q 2 = 3 p 2 + a Substituting into the result from (ii): 2 p( p 2 + 3 p 2 + a ) = b 2 p(4 p 2 + a ) = b 8 p 3 + 2ap − b = 0 so p is a root of the equation 8 x 3 + 2 ax − b = 0

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Further Pure Mathematics 1 Complex Numbers Section 4: Complex numbers and equations Multiple Choice Test 1) How many roots does the equation x4 + 3x² – 4 = 0 have? (a) 0 (c) 1 (e) I don’t know

(b) 2 (d) 4

2) Which of the following groups of numbers could be the roots of a polynomial equation with real coefficients? (i) (iii)

3, 4, 5 1 – j, j, 1

(a) (i), (iii) and (iv) (c) (i) and (iv) (e) I don’t know

(ii) (iv)

1 + j, 2, 4 2, 1 + j, 1 – j

(b) (i) only (d) all of them

3) Which of the following polynomials, when multiplied out, has real coefficients? (a) (z – 2 – j)(z – 4)(z – 2 + j) (c) (z – 3 + 4j)(z + 3 + 4j) (e) I don’t know

(b) (z – 2 – j)(z – 3)(z – 4 + 2j) (d) (z – j)(z + j)(z – 1)(z + 1)(z – 2j)

4) 1 – 2j and 3 + j are two roots of a quartic equation with real coefficients. The other two roots are (a) 1 + 2j and 3 – j (c) -1 – 2j and -3 + j (e) I don’t know

(b) -1 + 2j and -3 – j (d) 1 – 2j and 3 - j

5) The quartic equation referred to in Question 4 is (a) z4 – 8z³ + 27z² – 50z + 50 = 0 (c) z4 – 8z³ + 12z² + 50 = 0 (e) I don’t know

(b) z4 – 8z³ + 17z² + 2z – 24 = 0 (d) z4 – 8z³ + 12z² – 24 = 0

6) 2 + j is a root of z³ – z² – 7z + 15 = 0. The other roots are

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Further Pure Mathematics 1 (a) 2 – j, –3 (c) 2 + j, 3 (e) I don’t know

(b) 2 – j, 3 (d) 2 + j, 2 – j

7) The real root of z³ – 4z² + 14z – 20 = 0 is 2. The other roots are (a) 2 + 3j, 2 – 3j (c) 1 + 3j, 1 – 3j (e) I don’t know

(b) –1 + 3j, –1 – 3j (d) –2 + 3j, –2 – 3j

8) 1 + 2j is a root of the cubic equation z³ + az² + bz + 5. The values of a and b are (a) a = 1, b = 3 (c) a = –1, b = 3 (e) I don’t know

(b) a = 1, b = –1 (d) a = –1, b = –1

9) –2 + j is a root of the equation z4 + 2z³ – z² – 2z + 10 = 0. The other roots are (a) 2 – j, 1 + j, 1 – j (c) –2 – j, 2 – j, 2 + j (e) I don’t know

(b) –2 – j, 1 + j, 1 – j (d) –2 – j, 1 + 2j, 1 – 2j

10) The equation z4 + z³ + 2z² + 4z – 8 = 0 has two real roots. The roots of the equation are (a) –1, 2, 2j, –2j (c) 1, –2, 1 + j, 1 – j (e) I don’t know

(b) –1, 2, 1 + j, 1 – j (d) 1, –2, 2j, –2j

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Further Pure Mathematics 1 Chapter Assessment Complex Numbers

1.

The complex number α is given by α = –2 + 5j. (i) Write down the complex conjugate α*. [1] (ii) Find the modulus and argument of α. (iii) Find

[3]

α +α * in the form a + bj. α

[3]

2.

(i) Given that w = 1 + 2j, express w², w³ and w4 in the form a + bj. [5] (ii) Given that w is a root of the equation z + pz + qz − 6 z + 65 = 0 , find the values of p and q. [5] (iii) Write down a second root of the equation. [1] (iv) Find the other two roots of the equation. [6] 4

3.

3

2

(i) Show that z1 = 2 + j is one of the roots of the equation z² – 4z + 5 = 0. Find the other root, z2. [3] 1 1 4 (ii) Show that + = . z1 z2 5 [3] (iii) Show also that Im (z1² + z2²) = 0 and find Re (z1² – z2²). (iv) Find in the form r (cos θ + jsin θ ) , the complex numbers z1, z1², and z1³.

[4] [7]

(v) Plot the three complex numbers z1, z1², and z1³ on an Argand diagram. [3]

© MEI, September 2001

Further Pure Mathematics 1 4.

(a) Solve the equation z² + 2z + 10 = 0. Find the modulus and argument of each root. [4] (b) Complex numbers α and β are given by π π⎞ 5π 5π ⎞ ⎛ ⎛ α = 2 ⎜ cos + jsin ⎟ , β = 4 2 ⎜ cos + jsin ⎟ 8 8⎠ 8 8 ⎠ ⎝ ⎝ (i) Write down the modulus and argument of each of the complex numbers α and β. Illustrate these two complex numbers on an Argand diagram. (ii)

[3] Indicate a length on your diagram which is equal to β − α , and show that β − α = 6 .

[3] (iii) On your diagram, draw and label (A) the locus L of points representing complex numbers z such that z −α = 6 , (B)

[3] the locus M of points representing complex numbers z such that 5π . arg( z − α ) = 8 [3]

© MEI, September 2001