1

Expanded Course Notes on Complex Numbers from the Cambridge Text by Arnold and Arnold

By Mr Scovell

2

Complex Numbers

Explanation & Arithmetic of complex numbers.

Why do we need complex numbers?

The need for complex numbers arises when we solve equations that previously have had no solution.

Consider the solution to x2 – 4x + 13 = 0

x = 2

364

To solve this equation we would need to extend the number system to include numbers with negative

squares. To do this we let

i = 1 i2 = -1

Then every real number would have two square roots.

For example: – 4 could be written as 4 i2 then – 4 has two square roots, 2 i and - 2 i

We can now solve more complicated equations

Example: Solve x2 – 4x + 13 = 0

Solution

x =2

364

= 2

1364

=2

164

=2

64 i

= 2 + 3i or 2 – 3i

3

The notation of complex numbers.

Complex numbers are symbolized by the pronumeral z. They comprise of two parts:

z = x + iy where x,y are elements of ℝ

The Real part of z, written Re (z) = x

The Imaginary part of z, written Im (z) = y

Note the description „complex‟ does not imply non-real.

For example: the number 3 is a complex number and can be written as z = 3 + 0i

The number 3i is also a complex number z = 0 + 3i but is considered non real.

The following Venn diagram shows the relationship between:

ℂ ~ The set of complex numbers

ℝ ~ The set of real numbers

ℚ ~ The set of rational numbers

Z ~ The set of integers

M ~ The set of imaginary numbers

4

Arithmetic of complex numbers.

Equality

a + bi = c + di a = c b = d

Conjugate

If z = x + iy then the complex conjugate

z = x – iy

Examples: Find the complex conjugate of

I) 2 – 3i II) i III) 2

Solution

z = 2 + 3i z = 0 - i z = 2 – 0i

Addition z1 + z2

(a + bi) + (c + di) = (a + c) + i(b + d)

Example: If z1 = 3 + 2i and z2 = 4 – 3i evaluate z1 + z2

Solution

z1 + z2 = 3 + 2i + 4 – 3i

= 7 – i

Multiplication z1.z2

(a + bi) (c + di) = (ac - bd) + i(ad + bc)

Example: If z1 = 3 + 2i and z2 = 4 – 3i evaluate z1 z2

Solution

z1 z2 = (3 + 2i)(4 – 3i)

= 18 – i

5

Reciprocals & realising the denominator

The reciprocal of z is z

1 , usually the reciprocal is written in the form x + iy.

This involves a procedure similar to rationalising the denominator.

Example: If z = 4 + 3i find z

1 in the form a + ib

Solution

z

1 =

i34

1

= i34

1

i

i

34

34

= 2916

34

i

i

i

2 = -1 so

= 916

34

i

= 25

34 i

= 25

4 -

25

3i

Square roots of complex numbers

Example: If z2 = 3 + 4i find z

Solution

Let z = x + iy where x,y are ℝ

Then z2 = (x + iy)

2

(x + iy)2

= 3 + 4i

x2 + 2xyi + y

2i2 = 3 + 4i

x2 + 2xyi + y

2(-1) = 3 + 4i

6

x2 – y

2 + 2xyi = 3 +4i

Equating real and imaginary parts

x2 – y

2 = 3 -----1 and

2xy = 4 -----2

Solving simultaneously

in order to get a y2 to substitute into

4x2y

2 = 16

x2y

2 = 4

y2 =

2

4

x

Substitute y2 into

x2 -

2

4

x = 3

x4 – 4 = 3x

2

x4 – 3x

2 − 4 = 0

Let m = x2 then

m2 – 3m – 4 = 0

(m – 4)(m + 1) = 0

m = 4 or m = −1

x2 = 4 x

2 = −1

x = 2 x = i (but x is an ℝ so x i)

If x = 2 y = 1

x = -2 y = -1

z = 2 + i or z = -2 – i

7

Examples:

I If z1 = 2 + 3i and z2 = 1 + i find

a) Re (z1 +z2) b) Im (z1 2z )

Solution

z1 + z2 = (2 +3i) + (1+ i) z1 2z = (2 + 3i)(1 – i)

= 3 + 4i = 2 – 2i + 3i – 3i2

= 2 + i + 3

= 5 + i

Re (z1 +z2) = 3 Im (z1 2z ) = 1

II Show that zz = 2 2x y

Solution

Let z = x + iy

z = x – iy

z z = (x + iy)(x – iy)

= x2 – y

2i2

= x2 + y

2

III [HSC 2000 3 mks]

Find all pairs of integers x and y that satisfy (x + iy)2 = 24 + 10i

Solution

Let z = x + iy

z2 = 24 + 10i You can see that we are just finding the square root of 24 + 10i

Solving

(x + iy)2 = 24 + 10i

8

x2 + 2xyi + y

2i2 = 24 + 10i

x2 – y

2 + 2xyi = 24 + 10i

Equating real and imaginary parts

x2 – y

2 = 24 -------① and

2xy = 10 ------ ②

Rearranging ② to make y2 the subject

xy = 5

x2y

2 = 25

y2 =

2

25

x

Substituting y2 into ①

x2 -

2

25

x = 24

x4 – 25 = 24x

2

x4 – 24x

2 – 25 = 0

Let m = x2

m2 – 24m – 25 = 0

(m – 25)(m + 1) = 0

m = 25 or m = -1

x2 = 25 x

2 = -1

x = 5 x I since x ℝ

When x = 5 y = 1

x = -5 y = -1

i.e z = 5 + i or z = -5 – i

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IV [HSC 1990 2 marks]

Express i

z

53 in the from x + iy , where x and y are real.

Solution

Let z = a + ib

Then i

z

53 =

i

iba

53

=i

iba

53

i

i

53

53

= 2

2

259

5353

i

bibiaia

= 259

)53(53

abiba

= 34

)53(53 abiba

= 34

53 ba +

34

)53( ab i

V [HSC 2003 2 marks]

Let z = 2+ i and w = 1 – i. Find in the form x + iy,

Solution

(i) z w ii) z

4

z w = (2 + i)(1 + i) z

4 =

i2

4

= 2 + 2i + i + i2 =

i2

4

i

i

2

2

= 1 + 3i = 24

48

i

i

= 5

48 i

= 5

8-

5

4i

Students attempt CAMBRIDGE EXERCISE 21 on p. 31

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Geometrical representation of complex numbers.

The Argand diagram

Complex numbers can be plotted on an Argand diagram.

z = x + iy (x, y)

Example: If z = 3 + 2i , plot z and z on an Argand diagram

Solution

z = 3 + 2i A (3, 2)

z = 3 - 2i B (3, -2)

Radians (background knowledge).

Geometry uses degrees, minutes & seconds to measure an angle.

Another way to measure angles is in radians. A radian is the angle

an arc of 1 unit subtends at the centre of the circle of radius 1 unit.

π radians = 180

1 = radians180

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To convert degrees to radians:

Divide the angle (in degrees) by 180 (usually resulting in a fraction) and multiply by π.

Example: Convert to radians a) 30 b) -120

Solution

= 180

30 =

180

120

= 6

=

3

2

Polar co-ordinates.

The point A with the Cartesian co-ordinates (a, b) on an Argand diagram can also be specified by polar

co-ordinates (r, θ).

(r, θ) : r is the length the point is from the origin, r > 0

θ is the angle from the positive direction of the x-axis to the ray r.

For the point A to have unique polar co-ordinates (r, θ) it is necessary to restrict θ to 2π.

Otherwise points will be indistinguishable. For example (2, 6

) ≡ (2,

6

7). We will make the

restriction -π < θ ≤ π , as illustrated below .

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Examples: Plot on an Argand diagram the following polar co-ordinates:

I) (1, 4

)

II) (-2, 3

2)

III) (1, 3

)

IV) (2, 2

)

V) (1, 6

5)

Modulus – Argument form.

With regard to the polar co-ordinate (r, θ):

r is called the modulus of a complex number and is written │z│

│z│ = │x + iy│ = 22 yx = r

θ is called the argument (or principal argument) and is written as arg z

arg z = tan-1

x

y = θ

A

B

C

D

E

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Examples: Find │z│ and arg z of

Solution

I) z = 2 + 2i │z│ = 22 22

= 22

arg z = tan-1

2

2

= 45

= 4

II) z = √3 + i │z│ = 22

13

= 2

arg z = tan-1

3

1

= 30

= 6

III) z = -√3 + i │z│ = 22

13

= 2

arg z = π - tan-1

3

1

= 180 - 30

= π - 6

= 6

5

(2, 2)

2

2

1

√3

(√3, 1)

1

√3

(-√3, 1)

θ = arg z

14

IV) z = 2 + 4i │z│ = 22 42

= 52

arg z = tan-1

2

4

= 6326‟ (in radians ??)

= tan-1

2

V) z = 1 - i │z│ = 22 )1(1

= 1

arg z = 0 - tan-1

1

1

= 0 - 45

= 4

VI) z = √3 i │z│ = 22

)1(3

= 2

arg z = -π + tan-1

3

1

= -180 + 30

= -π + 6

= 6

5

(4, 2)

4

2

(1, -1)

1

1

1

√3

(-√3, 1)

θ = arg z

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Modulus-Argument form.

From trigonometry

x = r cosθ

y = r sinθ

z = x + iy

= rcosθ + i.rsinθ

= r(cosθ + isinθ) [this is called modulus-argument form]

[can be abbreviated to rcisθ ]

where r = │z│ θ = arg z in radians

Example: [HSC 1997 2 marks] Express √3 – i in modulus-argument form

Solution

r = 22

)1(3

= 2

tanθ = 3

1

θ = 6

So, √3 – i = 2(cos

6

+ isin

6

)

arg z = 6

= 2(cos

6

- isin

6

) *

r

θ

y

x

(r, θ)

x

y

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Vectors.

A vector is a graphical representation of a magnitude and a direction.

Thus z = x + iy

= OP

Magnitude of OP = length OP

= │z│

Direction of OP = θ

= arg z

z as a vector can be written in Cartesian form (x + iy)

Modulus-argument form (rcisθ)

Examples: Sketch the following vectors on an Argand diagram

Solution

I) z = 3 – i II) z = 2 cis3

2

r

θ

y

x

(x, y)

x

y

0

P

y

P (3, -1)

x 0

x

0

y

Q

2

3

2

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Since a vector is just a magnitude and a direction these vectors can be reproduced by translation.

Consider z = 1 + i marked by OP below.

the vector OP ≡ CD ≡ AB

√2

1

1

x

y

0

P

√2

1

1

x

y

0

P

D (-1, -1)

C (-2, -2)

√2

√2

A (2, 1)

B (3, 2)

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Operations with vectors. “tip to tail”

Addition p + q = z

Subtraction p - q = z

Examples: If z1 = 3 – 2i and z2 = -1 + 4i sketch z1, z2, z1 + z2

Solution

z

Z

p

q

q p

p

q

- q

p

x

y

z1

z2

z1 + z2

z2 moved “tip to tail”

19

In the above example, could you easily sketch a vector equivalent to z1 – z2 ?

Example: In the following diagram find vectors if OP = z

a) BP b) CP c) PA d) DP

Solution

Some other vectors we can easily find are

BO = 4 OC = 3i OA = 1

a) BP = BO + OP

= 4 + z

b) CP = OP – OC

= z – 3i

c) PA = OA – OP

= 1 – z

d) DP = DO + OP

= 3i + z

x

y

z1

z2

z1 - z2

x

y

A (1,0)

P

B (-4,0) O

C (0,3)

D (0,-3)

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Multiplication & division

Consider the following vectors & z1.z2

∆OAB Ⅲ ∆OCD arg z3 = arg z1 + arg z2

OA

OD=

OB

OC = arg (z1.z2)

1

3

z

z=

1

2z

│z3│ = │z2││z1│

= │z1.z2│

z3 = z1.z2

Rule: When multiplying ~ multiply moduli and add arguments

When dividing ~ divide moduli and subtract arguments

x

z3

1 B (1, 0)

z2

z1

A

C

D

y

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Example: If z1 = 1 + i and z2 = √3 – i

a) find 2

1

z

z

b) find

2

1argz

z

c) hence find the smallest positive integer n such that if z = i

i

3

1, zn is real and evaluate zn

Solution

a) │z1│ = 22 11 │z2│ = 2

2

)1(3

= √2 = 2

2

1

z

z =

2

1

z

z

= 2

2

b) arg z1 = tan-1

1

1 arg z2 = - tan

-1

3

1

= 4

= -

6

arg z =

2

1argz

z(from pt b) =

4

-

6

= 12

5

x

y

z1

z2

1

1

√3

1 Note the – sign as the arg is

below the x – axis.

Subtracting the arg‟s as two

vectors are being divided.

22

c) If zn is real then arg z

n = 0, π, 2π, 3π, … i.e graphically z

n is

= k π , where k = 0, ± 1, ± 2 …

Now arg zn = n arg z (prove later with de Moivre)

So k π = n 12

5 (from above)

n = k π 12

5

= 5

12k , where k = 0, ± 1, ± 2 …

Since n is an integer > 0 the first value of k that makes n > 0 is k = 5

n = 12

Because zn is real it lies on the x axis.

│z12

│ = │z│12

arg z12

= 5π

=

12

2

2

= 64

1

z12

= 64

1cis5π

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Triangular inequality

Draw any triangle ABC

What can you say about the relationship

between a,b & c?

a + b ≥ c a – b ≤ c [triangular inequality]

Example: If z1 = 3 + 4i and │z2│ = 13

a) find the greatest value of │z1 + z2│.

b) If │z1 + z2│ takes its greatest value, express z2 in the form a + i b

Solution

a) │z1│ + │z2│ ≥ │z1 + z2│ (triangle inequality)

│z1│ = 22 43

= 5

So, 5 + 13 ≥ │z1 + z2│

18 ≥ │z1 + z2│

thus a maximum value of 18 is obtained.

b) │z2│= k│z1│, where k ∈ ℝ z2 = k.z1

13 = k.5 = 5

13(3 + 4i)

k = 5

13 =

5

39 +

5

42i

(when │z1 + z2│ is a max)

A

B C

c

a

b

Students attempt CAMBRIDGE EXERCISE 22 & 23 on p. 43 & 54

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Powers & roots of complex numbers.

Example: Prove De Moivre’s theorem by mathematical induction

Solution

S(n): (cosθ + i sinθ)n = cosnθ + i sinnθ), n = 1, 2, 3, …

S(1): (cosθ + i sinθ)1 = cosθ + i sinθ TRUE

If S(k) is true, (cosθ + i sinθ)k = coskθ + i sinkθ

S(k + 1):

(cosθ + i sinθ)k + 1

= (cosθ + i sinθ)k (cosθ + i sinθ)

= (coskθ + isinkθ) (cosθ + i sinθ)

= 2cos cos cos sin cos sin sin sink i k i k i k

= cos cos sin sin cos sin cos sink k i k k

= cos sink i k

= cos(k + 1)θ + i sin(k + 1)θ TRUE

If S(1) is True, S(k) is ture & S(k + 1) is true then

(cosθ + i sinθ)n = cosnθ + i sinnθ) by induction

De Moivre’s theorem (cosθ + i sinθ)n = cosnθ + i sinnθ

i.e. show (cosθ + i sinθ)k + 1

= cos(k + 1)θ + i sin(k + 1)θ

25

Example: Show that nn zz , where n is an integral

Solution

Let z = r(cosθ + i sinθ)

then z = r(cosθ – i sinθ)

zn = nr (cosθ + i sinθ)

n

nz = nr (cosθ − i sinθ)n

LHS nz = nr (cosθ − i sinθ)

n

= nr (cosnθ − i sinnθ) [De Moivre‟s theorem]

RHS nz = r

n(cosθ - i sinθ)

n

= rn

(cosnθ - i sinnθ) [De Moivre‟s theorem]

LHS = RHS

26

Example: Express (√3 + i)8 + (√3 - i)8 in the form a + i b

Solution

Let z = √3 + i z = √3 – i

Now │z│ = 22

13 arg z = tan-1

3

1

= 2 = 30

= 6

So z = r (cosθ + isinθ) z = r (cosθ – i sinθ)

= 2 (cos6

+ isin

6

) = 2 (cos

6

– i sin

6

)

Now z8 = 2

8 (cos

6

+ i sin

6

)8

= 2

8 (cos

3

4 + i sin

3

4) [De Moivre‟s theorem]

And from previous example nn zz

So z8 = 8z = 2

8(cos

6

- i sin

6

)8

= 28 (cos

3

4 - i sin

3

4) [De Moivre‟s theorem]

Now (√3 + i)8 + (√3 - i)

8 = z

8 + z

8 = 2

8 (cos

3

4 + i sin

3

4) + 2

8 (cos

3

4 - i sin

3

4)

= 28 cos

3

4 + 2

8 cos

3

4

= 256 - 05 + 256 - 05

= - 256 + 0i

1

√3

(√3, 1)

27

Example: a) By expressing cos4θ, sin4θ in terms and powers of cosθ and sinθ show that

tan4θ =

42

3

tantan61

tan4tan4

Solution

Let z = cosθ + isinθ

z4 = (cosθ + isinθ)

4

= cos4θ + isin4θ [De moivre‟s theorem]

and z4 = (cosθ + isinθ)

4 =

n

k 0

nCk an-k bk

=

4

0k

nCk (cosθ)

n-k (i sinθ)

k [polynomial expansion]

= 4C0(cosθ)

4 (i sinθ)

0 +

4C1(cosθ)

3 (i sinθ)

1 +

4C2(cosθ)

2 (i sinθ)

2 +

4C1(cosθ)

1 (i sinθ)

3 +

4C4(cosθ)

0 (i sinθ)

4

= cos4θ + 4cos

3θ i sinθ + 6cos

2θ i

2sin

2θ + 4cosθ i

3 sin

3θ + i

4 sin

4θ

= cos4θ + 4cos

3θ i sinθ + 6cos

2θ (-1)sin

2θ + 4cosθ (-1i) sin

3θ + (+1)sin

4θ

= cos4θ + 4icos

3θsinθ - 6cos

2θ sin

2θ - 4icosθsin

3θ + sin

4θ

= cos4θ - 6cos

2θ sin

2θ + sin

4θ + i(4cos

3θsinθ - 4cosθsin

3θ) [in the form x + i y ]

Equating real and imaginary parts

cos4θ = cos4θ - 6cos

2θ sin

2θ + sin

4θ and

sin4θ = 4cos3θsinθ - 4cosθsin

3θ

tan4θ =

4224

33

sinsincos6cos

sincos4sin4cos

=

42

3

tantan61

tan4tan4 when divided thru by cos

4θ

28

b) Hence solve the equation t4 + 4t3 – 6t2 – 4t + 1 = 0

Solution

Rearranging the equation

t4 + 4t

3 – 6t

2 – 4t + 1 = 0

t4 – 6t

2 + 1 = 4t − 4t

3 dividing thru by LHS

1 = 16

4424

3

tt

tt

Let t = tanθ then

16

4424

3

tt

tt =

42

3

tantan61

tan4tan4 = tan4θ = 1

solving tan4θ = 1 let v = 4θ

tan v = 1

= 4

,

4

5,

4

9,

4

13, …

But v = 4θ = 4

,

4

5,

4

9,

4

13, …

θ = 16

,

16

5,

16

9,

16

13, … finding a general rule

θ = 16

)14( k where k = 0, 1, 2, 3 …

29

Examining solutions remembering that the solution t = tanθ

k 0 1 2 3 4 5 6

θ

16

16

5

16

9

16

13

16

17

16

21

16

25

t 019… 149… -502… -066… 019… 149… -502…

We can see that for values above k = 3 the solution repeats.

Hence the solutions are

t = tan16

, tan

16

5, tan

16

9, tan

16

13

30

Example: Show that zn + z -n = 2cosnθ

Let z = cosθ + i sinθ

zn = (cosθ + i sinθ)

n

= cosnθ + i sinnθ [De Moivre‟s theorem]

z -n

= (cosθ + i sinθ)-n

= ni )sin (cos

1

= nin sincos

1

= nin sincos

1

nin

nin

sincos

sincos realizing the denominator

=

nin

nin22 sincos

sincos

= cosnθ - i sinnθ

So zn + z

–n = cosnθ + i sinnθ + cosnθ - i sinnθ

= 2cosnθ

{students can prove z n

– z –n

= 2i sinnθ }

31

Example: Show that sin3θ

= ¼ (3sinθ – sin3θ) *reasonably hard question

Method 1. Using the above results

Let z = cosθ + i sinθ

z1 – z

–1 = 2i sinθ [from identity z

n – z

–n = 2i sinnθ ]

(z1 – z

–1)3 = (2i sinθ)

3

= 8i3 sin

3θ

= - 8i sin3θ

z3 – z

–3 = 2i sin3θ [from identity z

n – z

–n = 2i sinnθ ]

Now (z1 – z

–1)3 =

n

k 0

nCk an-k bk

=

3

0k

nCk (z)

n-k (-z

-1)k [polynomial expansion]

= 3C0 (z)

3 (-z

-1)0 +

3C1 (z)

2 (-z

-1)1 +

3C2 (z)

1 (-z

-1)2 +

3C3 (z)

0 (-z

-1)3

= z3 - 3 z

2 z

-1 + 3 z

1 z

-2 - z

–3

= z3 - 3z + 3 z

-1 - z

–3

= z3 - z

–3 - 3z + 3 z

-1

= z3 - z

–3 - 3(z - z

–1)

Substituting the identities above

(z1 – z

–1)3 = z

3 - z

–3 - 3(z - z

–1)

- 8i sin3θ = 2i sin3θ - 3(2i sinθ)

- 8i sin3θ = 2i sin3θ - 6i sinθ dividing through by –8i

sin3θ =

i

i

i

i

8

sin6

8

3sin2

32

sin3θ =

4

sin3

4

3sin

sin3θ =

4

1(-sin3θ + 3sinθ)

= 4

1(3sinθ – sin3θ) = RHS

Method 2. Mr. Scovell’s way

Note: Use this (as it is the most consistent method in attacking problems) way unless

Asked to prove z n

– z –n

= 2i sinnθ or zn + z

-n = 2cosnθ in a previous question

You encounter a problem (see question re: 16cos4θ = 2cos4θ + 8cos2θ + 6)

The method illustrated below is consistent with HSC questions. See Q.2 part d) 2003 HSC.

Let z = cosθ + i sinθ

z3 = (cosθ + i sinθ)

3

= cos3θ + i sin3θ [De Moivre‟s theorem]

also

(cosθ + i sinθ)3 =

n

k 0

nCk an-k bk

=

3

0k

nCk (cosθ)

n-k (i sinθ)

k [polynomial expansion]

= 3C0(cosθ)

3 (i sinθ)

0 +

3C1(cosθ)

2 (i sinθ)

1 +

3C2(cosθ)

1 (i sinθ)

2 +

3C1(cosθ)

0 (i sinθ)

3

= cos3θ + 3cos

2θ i sinθ + 3cosθ i

2sin

2θ + i

3 sin

3θ

= cos3θ + 3cos

2θ i sinθ - 3cosθ sin

2θ - i sin

3θ

= cos3θ - 3cosθ sin

2θ + i 3cos

2θsinθ - i sin

3θ

= cos3θ - 3cosθ sin

2θ + i (3cos

2θsinθ - sin

3θ)

33

So cos3θ + i sin3θ = cos3θ - 3cosθ sin

2θ + i (3cos

2θsinθ - sin

3θ)

Equating imaginary parts

sin3θ = 3cos2θsinθ - sin

3θ

sin3θ = 3cos

2θsinθ – sin3θ

= 3(1 – sin2θ)sinθ – sin3θ

= 3sinθ – 3sin3θ - sin3θ

4sin3θ = 3sinθ – sin3θ

sin3θ =

4

1(3sinθ – sin3θ) = RHS

Students can attempt showing 16cos4θ = 2cos4θ + 8cos2θ + 6

Note: Mr. Scovell‟s method will fail here due to 8cos2θ which won‟t occur in the expansion.

34

Roots of unity & complex roots

Discussion

If z = rcisθ

zn = r

ncisnθ

What about the reverse situation; given zn find z ?

Example: Solve z3 = 1

Let z = r(cosθ + i sinθ)

z3 = r

3(cosθ + i sinθ)

3

= r3 cos3θ + i sin3θ

r3 cos3θ + i sin3θ = 1

r3 cos3θ + i sin3θ = 1 + 0 i

Now │z3│ = 22 01

= 1

│z3│ = 1 then

│z│ = 1

Arg z3 = 0

This means z3 = cos3θ + i sin3θ = cos0 + i sin0

Equating real & imaginary parts

cos3θ = 1 and sin3θ = 0

Solving for θ

3θ = 0, 2π, 4π, 6π, 8π, …

35

θ = 0, 2

3

,

4

3

, …

Finding a general rule

θ = 3

2 k, where k = 0, 1, 2, 3, …

[Note that there a 3 distinct k values to correspond with 3 distinct solutions. We know there are 3

solutions because we are solving z3]

Now substituting what we know to evaluate the solutions of z

│z│ = r = 1 θ = 3

2 k

z = r(cosθ + i sinθ)

z = cos3

2 k + i sin

3

2 k

When k = 0

z1 = cos 0 + i sin0

= 1

When k = +1

z2 = cos3

2 + i sin

3

2

= 2

1 +

2

3i [In the form x + iy]

z1

1

z2

2

3

2

1

3

2 = 120

36

When k = - 1

z3 = cos3

2 + i sin

3

2

= 2

1 -

2

3i

[Note that if you had let k = 0, 1, 2 then]

For k = 2

z = cos3

4 + i sin

3

4

= 2

1 -

2

3i

= z3

Also note that for z3 = 1 there are the equally spaced solutions around the unit circle, namely

1, cos3

2 + i sin

3

2, cos

3

2 + i sin

3

2

z3

2

3

2

1

3

2 = 120

z3

z2

z1

3

2

37

Example: If w is a non-real cube root of unity show that 1 + w + w2 = 0

w3 = 1

w3 – 1 = 0

(w – 1)(w2 + w + 1) = 0

So w – 1 = 0 or w2 + w + 1 = 0

Since w is a NON-REAL root

w ≠ 1 w2 + w + 1 = 0

Example: a) Find the values for which z6 = -64

Let z = r(cosθ + i sinθ)

z6

= r6(cos6θ + i sin6θ) [De Moivre‟s theorem]

z6

= r6(cos6θ + i sin6θ) = - 64 + 0i

│z6│ = 22 064

= 64 = r6

│z│ = 6 64

= 2 = r

Arg z6 = π

Rearranging - 64 + 0i into mod-arg form

z6

= r6(cos6θ + i sin6θ) = 64 (cosπ + i sinπ)

6θ = π, 3π, 5π, …

θ = 6

,

6

3,

6

5, …

arg z6 = π

z6

38

Finding a general rule

θ = 6

+

6

2 k

= 6

)12( k where k = 0, ±1, ±2, - 3

Substituting into z = r(cosθ + i sinθ)

r = 2 θ = 6

)12( k where k = 0, ±1, ±2, - 3

z = 2(cos6

)12( k + i sin

6

)12( k)

When k = 0,

z1 = 2(cos6

+ i sin

6

)

= 2(2

3 +

2

1 i )

= i3

When k = -1

z2 = 2(cos6

+ i sin

6

)

= i3

When k = 1

z3 = 2(cos6

3 + i sin

6

3)

= 2i

39

When k = -2

z4 = 2(cos6

3 + i sin

6

3)

= - 2i

When k = 2

z5 = 2(cos6

5 + i sin

6

5)

= - i3

When k = -3

z6 = 2(cos6

5 + i sin

6

5)

= - i3

Plotting the points

Note: z2 = 1z z4 = 3z z6 = 5z

z1

z2

z3

z4

z5

z6

40

b) Express z6 + 64 = 0 as a product of quadratic factors

z6 + 64 = 0

(z – z1)(z – z2)(z – z3)(z – z4)(z – z5)(z – z6) = 0

From above: z2 = 1z z4 = 3z z6 = 5z

[(z – z1)(z – 1z )] [(z – z3)(z – 3z )] [(z – z5)(z – 5z )] = 0

[z2 – z.z1 - z 1z + z1 1z ] [z

2 – z.z3 - z 3z + z3 3z ] [z

2 – z.z5 - z 5z + z5 5z ] = 0

[z2 – z(z1 + 1z ) + z1 1z ] [z

2 – z(z3 + 3z ) + z3 3z ] [z

2 – z(z5 - 5z ) + z5 5z ] = 0

From above z1 = i3 z3 = 2i z5 = - i3

1z = i3 3z = -2i 5z = - i3

So

[z2 – z( i3 + i3 ) + ( i3 )( i3 )][z

2 – z(2i + -2i) + (2i.-2i)][z

2 – z(- i3 + - i3 ) +

(- i3 ).( - i3 )] = 0

[z2 – 2 3 z + 3 - i

2] [z

2 - 4i

2] [z

2 +2 3 z + 3 - i

2] = 0

[z2 – 2 3 z + 4] [z

2 + 4] [z

2 +2 3 z + 4] = 0

Note: Some questions are more easily answered or require an answer using modulus-argument form

for z1, z2, z3

41

Example: Factorise z5 – 1 into real linear and quadratic factors.

First step would be to solve z5 – 1 = 0

The working for the solutions is not shown here {students can solve for themselves as practice later ~

also in Ex 24 as a q}.

The 5 solutions would be

z1 = 1 when k = 0

z2 = cos5

2 + i sin

5

2 when k = +1

z3 = cos5

2 + i sin

5

2 when k = -1

z4 = cos5

4 + i sin

5

4 when k = 2

z5 = cos5

4 + i sin

5

4 when k = -2

Note: z3 = 2z z5 = 4z

Now z5 – 1 = (z – z1)(z – z2)(z – z3)(z – z4)(z – z5)

= (z – z1)(z – z2)(z – 2z )(z – z4)(z – 4z )

= (z – z1)[z2 – z(z2 + 2z ) + z2 2z ] [z

2 – z(z4 + 4z ) + z4 4z ]

Note the identities: In x + i y [we could have used these identities in the previous Q]

z + z = 2x z. z = x2 + y

2 = 1 in the case where r = 1

r2 otherwise

= (z – 1)[z2 – z(2 cos

5

2 ) +1] [z

2 – z(2 cos

5

4) + 1]

Students attempt CAMBRIDGE EXERCISE 24 on p. 60

42

Curves & regions in the Argand diagram.

Discussion.

Consider z to be a complex number that satisfies Re (z) = 3

On the Argand diagram Re z would be a locus as illustrated below.

All the dotted lines represent the vectors z where Re z = 3. There are infinite amount of vectors

terminating along the bold line.

Sketching a curve

Example: a) Sketch the curve defined by the equation Im (z – 1 + 3i) = 4

Let z = x + iy

Then z – 1 + 3i = x + iy - 1 + 3i

= x - 1 + i( y + 3)

Since Im (z – 1 + 3i) = 4

( y + 3) = 4

y = 1

All the dotted vectors z have Im z = 1 and terminate along the bold line. You don‟t have to draw the

dotted vectors. They are there simply to illustrate the concept.

│z│

1 P

3

43

b) Find the minimum value of │z│ for which Im (z – 1 + 3i) = 4 occurs and state the vector z

for this condition.

z will have a minimum value when z is perpendicular to y = 1

z = 0 + i

= i

Example: Sketch the region in the Argand diagram defined by

6 ≤ Re [(2 – 3i) z ] < 12 and Re z . Im z ≥ 0

Let z = x + iy

Then [(2 – 3i) z ] = (2 – 3i)( x + iy)

= 2x + 2iy – 3ix – 3i2y

= 2x + 3y + i (2y – 3x)

So Re [(2 – 3i) z ] = 2x + 3y

We need to graph 6 ≤ 2x + 3y < 12

And the intersection of

Re z . Im z = x . y

Which is the graph of x . y ≥ 0

[Note xy ≥ 0 are all points in the first quadrant]

│z│

1

2x +3y = 6

2

3

2x +3y = 12

The shaded area represents

6 ≤ 2x + 3y < 12

and the intersection with

x.y ≥ 0

44

Example: z satisfies │z – 1 + 2i│ = │z + 3│

a) Sketch the locus of the point P representing z

ALGEBRAIC SOLUTION

Let z = x + iy

Then │z – 1 + 2i│ = │z + 3│ becomes

│ x + iy – 1 + 2i│ = │ x + iy + 3│

│x – 1 + i(y + 2)│ = │x + 3 + iy│

Finding the modulus of each side

22 )2()1( yx = 22)3( yx

(x – 1)2 + (y + 2)

2 = (x + 3)

2 + y

2

x2 – 2x + 1 + y

2 + 4y + 4 = x

2 + 6x + 9 + y

2

4y = 8x + 4

y = 2x + 1

GEOMETRICAL SOLUTION

Let the point P be the locus of z

Rearranging │z – 1 + 2i│ = │z + 3│

│z – (1 - 2i)│ = │z – (- 3)│

│ z – z1 │ = │ z – z2│

Geometrically this means that the distance of z

from the point (1, -2) is equal the distance from

the point (-3, 0).

Since z is equidistant from the two points the locus of z

must be the perpendicular bisector of the two points.

Note: the vectors z – z1 & z – z2

Start from the respective points and end along the bold line.

If they have trouble with the above geometry- can use below to reinforce the geometrical concepts.

1

x

y

x

y

(1, -2)

-3

P

z – z2

z – z1

45

z starts from the origin and terminates in infinite places along the dotted line

z2 starts from the orgin and ends at the point (-3, 0)

z – z2 is found with vector subtraction

b) Find the minimum value of │z│

The minimum value of │z│ is the perpendicular distance from the origin to the locus.

d = 22

11

ba

cbyax

= 22 )1(2

1)0(1)0(2

= 5

1 units

x

y

-3

P

z – z2

z2

z

x

y

-3

P

z2

z z – z2

x

1

y

46

Discussion

Sketch the region defined by │z│ ≤ 2

The shaded region defines all vectors with a modulus ≤ 2

Example: If z satisfies │z – 2 – 2i│ = 2

a) Sketch the locus of point P representing z on an Argand diagram.

b) Find the maximum value for │z│

c) Find the range of values for arg z

ALGEBRAIC SOLUTION

a) Let z = x + iy

│z – 2 – 2i│ = 2

│x + iy – 2 – 2i│ = 2

│x – 2 + i(y – 2 )│ = 2

z z

z

2 -2

-2

2

z z

47

│ x – 2 + i(y – 2 )│= 22 )2()2( yx = 2

(x – 2)2 + (y – 2)

2 = 2

This is a circle with centre (2, 2) and a radius of 2 units

GEOMETRICAL SOLUTION

Let z = 2 +2i

│z – 2 – 2i│ = 2

│z – (2 + 2i)│ = 2

│z – z1│ = 2

Constructing the answer in parts

z1 is a vector starting at the origin and

terminating at (2, 2)

Since we are dealing with the modulus │z – z1│

(2, 2)

x

y

2 zmax

z

zmin

(2, 2)

z1

48

y

(2, 2)

x

2

zmax

1

1

(3, 3)

The vector z – z1 has a distance of 2 units

z – z1 will be a vector originating at (2, 2) and meeting vector z 2 units away (in all directions).

b) The maximun value for │z│

z has a maximum modulus when z

is furthest from the origin. Remember z

starts at the origin and terminates at any

point on the circle. The maximum

value of │z│ occurs at the point (3, 3).

So z = 3 + 3i

(2, 2)

x

y

│z – z1│= 2

z

z1

z – z1

49

c) The range of values for arg z is found by COD ≤ arg z ≤ AOD

Finding COD & AOD

OB = 2 + 2 = 22

BOD = 45

= 4

AOB = BOC = sinα = 22

2

= 2

1

= 6

Now COD = BOD – BOC (α )

= 6

-

4

= 12

And AOD = BOD + AOB

= 6

+

4

= 12

5

the range of values for arg z is 12

≤ arg z ≤

12

5

x

(2, 2)

y

A

z

z1

2

B

C

0

z

2

D

22

α

50

Discussion

Consider the Argand diagram below and the equation of the locus z

arg z = 3

The gradient of OP is = tan3

= - 3

the equation of the locus is y = - 3 x where x >0 [in the form y = mx + b]

Note since the locus has an argument z ≠ 0 as the vector z = 0 has no argument.

Example: Shade in the region in the Argand diagram defined by

4

≤ arg z ≤

2

OR* │z│< 2

* Note OR and not AND. This means you shade in the union of both regions and not the intersection.

z

-3

P

O

4

2

2

-2

-2

51

Example: Sketch the locus

a) arg (z – 1 – 2i) = 3

When dealing with arguments a GEOMETRICAL SOLUTION is sometimes easier.

GEOMETRICAL SOLUTION

arg (z – 1 – 2i) = 3

arg (z – [1 + 2i]) = 3

Let z1 = 1 + 2i [see z1 below]

arg (z – z1) = 3

Note the point from which the locus

originates is (1, 2). The locus then

heads in the direction of 3

.

Note that it does not include the

point A

z1 z - z1

3

z

z

A

52

b) i) sketch arg (z – 1 – 2i) = 6

arg (z – 1 – 2i) = 6

arg (z – [1 + 2i]) = 6

Let z1 = 1 + 2i [see z1 below]

arg (z – z1) = 6

ii) What is the Cartesian equation of the locus?

We know a point (1, 2) and the gradient can be found using

m = tanα y – y1 = m(x – x1)

= tan6

y – 2 =

3

1(x – 1)

= 3

1 y =

3

1x -

3

1+ 2

y = 3

1x +

3

132 where x > 1

z1

z - z1

6

z

A

53

Example: z is a complex number that satisfies

2 ≤ │z + 3│≤ 3 and 0 ≤ arg (z + 3) ≤ 3

Find the area & perimeter of the region so formed by these conditions.

GEOMETRICAL SOLUTION

2 ≤ │z + 3│≤ 3

2 ≤ │z - (- 3)│≤ 3

Let z1 = - 3

2 ≤ │z – z1│≤ 3 is a circle with centre (-3, 0) and radii 2 & 3 respectively

0 ≤ arg (z + 3) ≤ 3

0 ≤ arg (z - z1) ≤ 3

is the direction of the vector z – z1 from the point (-3, 0)

(-3, 0)

3

-1 0

54

Area = 6

1π (R

2 – r

2) Perimeter =

6

12πR +

6

12πr + 1 + 1

= 6

(3

2 – 2

2) =

6

12π(3 + 2) + 2

= 6

5 units

2 = 2 +

3

5 units

Example: If z satisfies arg (z + i) = arg (z – 1), sketch the locus of P representing z in the

Argand diagram. [reasonably hard question]

GEOMETRICAL SOLUTION

arg (z + i) = arg (z – 1)

The vectors (z + i) starts from the point (0, -1)

(z – 1) starts from the point (1, 0)

The vector z could terminate at the points P1, P2, & P3 however…

A (0, -1)

B (1, 0) O

P1

z

P2

P2

-i

+1

55

Case ① P1

If P1 represents z then

AP = z + i

BP = z – 1

And arg AP = arg BP [both vectors head is a positve direction]

[the direction itself does not matter so long as it is the same direction for both vectors]

The interval containing P1 is a possible locus of arg (z + i) = arg (z – 1)

Case ② P2

If P2 represents z then

AP = z + i

BP = z – 1

But arg AP ≠ arg BP [the vectors arguments are opposite]

[AP is in a positive direction while Bpis in a negative]

The interval with P2 can not be a possible locus of arg (z + i) = arg (z – 1)

Case 3 P3

If P3 represents z then

AP = z + i

BP = z – 1

And arg AP = arg BP [both vectors head is a negative direction]

The interval containing P3 is a possible locus of arg (z + i) = arg (z – 1)

Students attempt CAMBRIDGE EXERCISE 25 on p. 71

56

Formula test.

57

58

59

60