Complex Numbers

If we wish to work with , we need to extend the set of real numbers

1

Definitions

i is a number such that i2 = -1 i

C is the set of numbers Z, of the form where a and b are real numbers.

ibaz

a is called the real part of Z and we write a = R(z) of a = Re(z)

b is called the imaginary part of Z and we write b = i(z) or b = Im(z)

Given biaz 1 and dicz 2

Addition is defined by: )()(21 dicbiazz

Multiplication is defined by: ))((21 dicbiazz

idbca )()(

2bdibciadiac ibcadbdac )()(

We may write a + bi or a + ib, whichever we find more convenient

a) Given findizandiz 3423 21

(i) 21 zz (ii) 21zz

(i) )34()23(21 iizz

i57

21 zz (iii)

(ii) )34()23(21 iizz

i 1

(iii) )34)(23(21 iizz 268912 iii

i176

b) Solve the equation 0522 zz

Using the quadratic formulaa

acbbz

242

22042 z

216

22

16 11

2

i21

Page 90 Exercise 1Questions 1, 2, 3, 6, 7, 8

Complex Conjugate

When , then its complex conjugate is denoted bybiaz

biaz

Note: 22 bazz

This is useful when we wish to carry out a division.

a) Calculate)32()24(

ii

)32()32(

)32()24(

)32()24(

ii

ii

ii

9464128 2

iii

13814 i

14 8

13 13i

b) Calculate 5 12i

Let 5 12a bi i where ba,

Then ibia 125)( 2 iabiba 125222

Equating parts we get: 122 ab6 ab

ba

6

also 522 ba

56 2

2

b

b

536 2

2 b

b24 536 bb

0365 24 bb

0365 24 bb

0)4)(9( 22 bb

942 orb

Since 2, bb

33 ora

iorii 2323125

Page 91 Ex 2

Questions 1(a), (b), (c), 2(c), (e)3(a), (b), (f), 5(a), (b)

TJ Exercise 2. TJ Exercise 1 - if needed.

Argand DiagramsThe complex number is represented on the plane by the point P(x,y). The plane is referred to as “The Complex Plane”, and diagrams of this sort are called Argand Diagrams.

yixz

Any point on the x-axis represents a purely Real Number

Any point on the y-axis represents a purely imaginary number

is represented by z x iy OP ��������������

is considered a vector which has been rotated of the x-axis.OP��������������

The length of , r, is called the modulus of z denoted OP z��������������

y

x

r y

x

p

The size of the rotation is called the amplitude or argument of z. It is often denoted Arg z. This angle could be 2 where is any integer.n n

We refer to the value of Arg z which lies in the range -< asthe principal argument. It is denoted arg z, lower case ‘a’.

2 2r x y

1tan ( )y x

By simple trigonometry: cosx r siny r

Thus can be re written as

(cos sin )

x iy z

z r i

This is referred to as the Polar form of z.

y

x

r y

x

p

a) Find the modulus and argument of the complex number 3 4z i

2 23 4 5z 1tan (4 3)Arg z n radians

0.927 n radians

Since (3,4) lies in the first quadrant, n = 0

arg 0.927 3z radians to significant figures

b) Find the modulus and argument of the complex number 3 4z i

2 23 4 5z 1tan ( 4 3)Arg z n radians

0.927 n radians

Since (-3,-4) lies in the third quadrant, n = -1arg 0.927z

2.21 3radians to significant figures

0 0c) Express 2 2 in the form (cos sin )z i r i

2 22 2r z

2 2

1 02arg tan ( ) 452z (2,2) is in Q1

0 02 2(cos 45 sin 45 )z i

Page 94 Exercise 3Questions 3a, b, d, e, i

6a, b, f7a, b, c

Loci-Set of points on the complex planea) Given , draw the locus of the point

which moves on the complex plane so that

(i) 4 (ii) 4

z x iy

z z

2 24 4z x y 2 2 16x y

This is a circle, centre the origin radius 4

-4 4

4

-4

y

x-4 4

4

-4

y

x

(i) (ii)

b) If

( ) Find the equation of the locus 2 3

( ) Draw the locus on an argand diagram.

z x iy

i z

ii

( ) 2 3 2 3a z x iy

2 2( 2) 3x y

2 2( 2) 9x y

This is a circle centre (2, 0) radius 3 units.

-1 5

y

x

c) if , find the equation of the locus

arg z3

z x iy

1arg tan ( )3 3

yz x

tan 3y

x

3yx

3y x

This is a straight line through the origin gradient 3

Page 96 Exercise 4 Questions 1a, b, d, f, j 3a, b, 4a, c

TJ Exercise 7

Polar Form and Multiplication

The polar form of z is (cos sin )z r i

1 2 1Consider z where (cos sin ) andz z a A i A

2 (cos sin )z b B i B

1 2 (cos sin )(cos sin )z z ab A i A B i B 2(cos cos sin sin cos sin cos )ab A B i A B i A B iSinA B

(cos cos sin sin (sin cos cos sin ))ab A B A B i A B A B

(cos( ) sin( ))ab A B i A B

1 2 1 2Hence, z z z z 1 2 1 2( )Arg z z Arg z Arg z and

Note arg(z1z2) lies in the range (-, ) and adjustments have to be made by adding or subtracting 2 as appropriate if Arg(z1z2) goes outside that range during the calculation.

Note: cos sin cos sini i

Also, if (cos sin ) thenz r i

1 1(cos sin )i

z r

1 1 1So, and arg arg

zz

z z

11

2 2

1zz

z z

11 2

2

Hence z

z zz

11 2

2

zArg Arg z Arg z

z

) 3(cos sin ) 4(cos sin )3 3 2 2

a Simplify i i

12(cos sin )3 2 3 2

i

5 512(cos sin )

6 6i

Now turn to page 96 Exercise 5 Questions 1 and 2.

Let us now look at question 3 on page 99.

(cos sin )3 3

z r i

2 2 2 2( ) (cos sin )

3 3a z zz r i

3 2 3( ) (cos sin )b z z z r i

4 3 4 2 2( ) (cos sin )

3 3c z z z r i

4 2

3 3Because

This leads to the pattern: If (cos sin )z r i

then (cos sin )n nz r n i n

De Moivre’s Theorem

If (cos sin )z r i

then (cos sin )n nz r n i n

a) Given find 1 3z i 2 5 7( ) ( ) ( )i z ii z iii z

1 3z

2

3tan

1

3 2 cos sin

3 3z i

22 2

( ) 4 cos sin3 3

i z i

5 5 5 5( ) 2 cos sin 32 cos sin

3 3 3 3ii z i i

7 7 7 7( ) 2 cos sin 128 cos sin

3 3 3 3iii z i i

1 34 2 2 3

2 2i i

16 16 3i

64 63 3i

b) Given find 2z i 4.z

4 1z

5

1tan

2 0.464 to 3 decimal places

5 cos0.464 sin 0.464z i

4

4 5 cos1.856 sin1.856z i 25 0.281 0.960 7 24i i

Round your answer to the nearest integer

Page 101 Exercise 6 questions 1 to 3, 4g, h, i, j.

Roots of a complex number

1

2

21

5 5Consider cos sin

6 6

and cos sin6 6

10 10cos sin cos sin cos sin

6 6 3 3 3 3

z i

z i

z i i i

22 cos sin

3 3z i

It would appear that if2 cos sin

3 3z i

then 2z z

5 5cos sin and cos sin the solutions being radians apart.

6 6 6 6z i i

2 cos sin3 3

z i 5 5

cos sin and cos sin 6 6 6 6

z i i

y

x

2z

6

56

The solutions are radians apart, 2

or think of it as radians apart.2

1z

2z

By De Moivre’s theorem, when finding the nth root of a complex number we are effectively dividing the argument by n. We should therefore study arguments in the range (-n, n) so that we have all the solutions in the range (-, ) after division.

n1If cos sin then the n solutions of the equation z are given byz r i z

1

1

2 2cos sin where 0,1,2,...., 1n k k

z r i k nn n

The position vectors of the solution will divide the circle of radius r, centre the origin, into n equal sectors.

3a) Solve the equation 4 4 3.z i

3 16 (16 3) 64 8z 3 1 4 3arg( ) tan

4 3z

3 8 cos sin3 3

z i

For k = 013

2 0 2 03 38 cos sin 2 cos sin

3 3 9 9z i i

For k = 113

2 1 2 1 7 73 38 cos sin 2 cos sin3 3 9 9

z i i

3 16 (16 3) 64 8z 3 1 4 3arg( ) tan

4 3z

3 8 cos sin3 3

z i

For k = 213

2 2 2 2 13 133 38 cos sin 2 cos sin3 3 9 9

z i i

5 52 cos sin

9 9i

5 52 cos sin

9 9i

7 7 5 5cos sin , cos sin , cos sin ,

9 9 9 9 9 9z i i i

5b) Solve the equation 1z

5 1z arg 0z

2 21 cos sin , for 0,1,2,3,4.

5 5k k

z i k

0, gives cos0 sin 0 1k i

2 21, gives cos sin

5 5k i

4 42, gives cos sin

5 5k i

4 43, gives cos sin

5 5k i

2 24, gives cos sin

5 5k i

Page 106 Exercise 7: Question 2 plus a selection from 1

Polynomials

If the root of a ploynomial equation is non-real, ie of the form cos sin ,

sin 0, then its conjugate cos sin is also a root.

r i

r r i

A polynomial of degree n will have complex roots. n

In 1799 Gauss proved that every polynomial equation with complex coefficients, f(z) = 0, where z C, has at least one root in the set of complex numbers. He later called this theorem the fundamental theorem of algebra. In this course we restrict ourselves to real coefficients but the fundamental theorem still applies since real numbers are also complex.

A polynomial of degree with real coefficients, can be reduced to a

product of real linear factors and real irreducible quadratic factors which

will have complex solutions.

n

a) Show that 1 2 is a root of the equationz i

4 3 2( ) 6 18 30 25f z z z z z

b) Hence find all the other roots.

1 2z i 1 4 5z arg 1.107z radians

5 cos1.107 sin1.107z i We need to find z2, z3 and z4

And substitute them into the Original equation.

22 5 cos2.214 sin 2.214 3 4z i i

33 5 cos3.321 sin3.321 11 2z i i

44 5 cos 4.428 sin 4.428 7 24z i i

( ) 7 24 6 11 2 18 3 4 30 1 2 25f z i i i i

7 24 66 12 54 72 30 60 25i i i i

0

Thus 1 2 is a root.z i

If 1 2 is a root, then the conjugate 1 2 is also a root. z i i

Thus 1 2 and z- 1-2i are complex factors of .z i z

Multiplying the complex factors to find the real quadratic

1 2 1 2z i z i 2 22 1 2 2 2 4z z zi z i zi i i 2 2 5z z

Using Division2 4 3 22 5 6 18 30 25z z z z z z

2z

4 3 22 5z z z 34z 213z 30z

4z

3 24 8 20z z z 25 10 25z z

5

25 10 25z z

0

Hence the complimentary real factor is 2 4 5z z

4 16-20Using

2

2 i

Hence all four roots are: 1 2 , 1 2 , 2 , 2 .i i i i

Page 108 Exercise 8 Questions 2, 3, 4, 5 and 6

Review on Page 110