MAT01A1: Complex Numbers (Appendix H)

Dr Craig

14 February 2018

Announcements:

I e-Quiz 1 is live. Deadline is Wed 21 Feb

at 23h59.

I e-Quiz 2 (App. A, D, E, H) opens tonight

at 19h00. Deadline is Thu 22 Feb at23h59.

I Saturday class this week: 09h00 to 12h00

in D1 LAB 110. Please come along if you

would like additional assistance with any

of the topics that we have covered so far.

Lecturers’ Consultation Hours

Monday:

12h00 – 13h30 Ms Richardson (C-503)

Wednesday:

15h00 – 16h00 Ms Richardson (C-503)

Thursday:

11h20 – 12h55 Dr Craig (C-508)

Friday:

11h20 – 12h55 Dr Craig (C-508)

Today’s lecture

I Complex numbers: introduction and basic

operations

I Polar form (including multiplication and

division)

I Powers and roots of complex numbers

Why do we cover complex numbers?

Applications in Physics and Electronics.

Also, studied in mathematics in Complex

Analysis (3rd year module).

Complex numbersA complex number is made up of a realpart and an imaginary part. The imaginary

part involves the square root of −1. We

define i =√−1.

All complex numbers will be of the form:

z = a + bi where a, b ∈ R and i2 = −1

The set of complex numbers is denoted by

C. (Remember, N is the natural numbers, Zis the integers, R is the real numbers.)

Complex numbers in the plane

The complex number z = a + bi is

represented on the complex plane by the

point (a, b). For z = 2 + i we have a = 2

and b = 1.

(a, b) = (2, 1)Im

R

More numbers in the complex plane

2 + 3i

3− 2i

−4 + 2i

−2− 2i

Im

R

Addition and subtraction in CTo add/subtract complex numbers, we

simply add/subtract the real and imaginary

parts separately and then combine them.

Let z = a + bi and w = c + di. Then

z + w = (a + bi) + (c + di)

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

Subtraction:

z − w = (a + bi)− (c + di)

= (a− c) + (b− d)i

Multiplication

Let z = a + bi and w = c + di. Then

z × w = (a + bi)(c + di)

= ac + adi + bci + (bi)(di)

= ac + (ad + bc)i + bd(i2)

= ac + (ad + bc)i + (−1)(bd)

= (ac− bd) + (ad + bc)i

The real part of z × w is ac− bd and the

imaginary part is ad + bc.

Examples

Add: z = 2− 7i and w = −4 + 2i.

Calculate: (1 + i)− (3− 4i).

Multiply: z = 2 + 3i and w = 4− 2i.

The conjugate of a complex number

Consider the complex number z = a + bi.

The complex conjugate of z is the

complex number

z̄ = a− bi

z

z̄

(a, b)

(a,−b)

Im

R

Division of complex numbersTo divide complex numbers we make use of

the complex conjugate of the denominator.

Let z = a + bi and w = c + di. Then

z

w=a + bi

c + di=a + bi

c + di× c− dic− di

=(a + bi)(c− di)

c2 + d2=ac + bd

c2 + d2+bc− adc2 + d2

i

Now that we know how to divide, we can

consider reciprocals of complex numbers:

1

z=

1.z̄

z.z̄=a− bia2 + b2

=

(a

a2 + b2

)+

(−b

a2 + b2

)i

Conjugates and absolute valueProperties of conjugates:

z + w = z̄ + w̄ zw = z̄w̄ zn = z̄n

The absolute value, or modulus, of a

complex number is the distance from the

origin in the complex plane. If z = a + bi

then

|z| =√a2 + b2

We see that

z.z̄ = |z|2

Roots of quadratic equations in C

ax2 + bx + c = 0 x =−b±

√b2 − 4ac

2a

When we allow complex roots as solutions,

we can apply the above formula to cases

when b2 − 4ac < 0. When y > 0, we let√−y = (

√y)i. Thus every quadratic has

complex roots.

Example: solve x2 + x + 1 = 0 for x ∈ C

Complex roots of polynomials

Consider a polynomial of degree n with

coefficients from R. Such a polynomial has

the general form

anxn + an−1x

n−1 + . . . + a1x + a0

A root of a polynomial is a value of x that

makes the polynomial equal to zero.

Every polynomial of degree n has n

complex roots.

Complex roots of polynomials

Every polynomial of degree n has n

complex roots.

(Roots might be repeated, e.g. x2 = 0.)

Terminology:

A complex number z is said to be in

rectangular form when it is written as

z = a + bi. This terminology distinguishes

rectangular form from the one we are about

to introduce: polar form.

Polar form

Any complex number z = a + bi can be

considered as a point (a, b). Thus it can also

be represented by polar coordinates as (r, θ).

(a, b)

r

θa

b

Im

R

Now a = r cos θ and b = r sin θ.

Since a = r cos θ and b = r sin θ, any

complex number z = a+ bi can be written as

z = r(cos θ + i sin θ)

where r = |z| =√a2 + b2 and tan θ =

b

a.

The angle θ is called the argument of the

complex number z. We write θ = arg(z).

Note: arg(z) is not unique. If θ = arg(z)

then we also have n.2π.θ = arg(z) where n

is any integer (n ∈ Z).

Finding the argument of z ∈ C

When converting a complex number into

polar form the best method for finding the

argument of z = a + bi is to plot z.

If you use the fact that tan θ =b

athen there

are two possible solutions for θ ∈ [0, 2π].

Examples: z = −1 +√3i and w = 1− i.

Find |z|, arg(z), |w|, arg(w).

(−1,√3)

r

(1,−1)

Im

R

Multiplication and division in polar form

Let z1 = r1(cos θ1 + i sin θ1) and

z2 = r2(cos θ2 + i sin θ2).

We use the addition and subtraction

formulas for sin θ and cos θ. Multiplication in

polar form gives us:

z1z2 = r1r2[

cos(θ1 + θ2) + i sin(θ1 + θ2)]

(Demonstrated in class. Also explained in

the textbook.)

Multiplication in polar form gives us:

z1z2 = r1r2[

cos(θ1 + θ2) + i sin(θ1 + θ2)]

Example: z =√

3 + i and w = −√

3− i.

Calculate z · w using both rectangular form

and polar form.

What about division?

z1z2

=r1r2

[cos(θ1 − θ2) + i sin(θ1 − θ2)

]How can we show that this is true? As an

exercise, calculate

r1(cos θ1 + i sin θ1)

r2(cos θ2 + i sin θ2)× cos θ2 − i sin θ2

cos θ2 − i sin θ2

Powers of complex numbers

We can generalise the multiplication of

complex numbers in polar form to obtain a

formula for taking powers of complex

numbers.

For z = r(cos θ + i sin θ) and n a positive

integer, we have De Moivre’s Theorem:

zn = [r(cos θ + i sin θ)]n = rn(cosnθ + i sinnθ)

Example: find(1 +√

3i)4

.

From powers to roots in CSuppose we want to find the n-th root of

z = r(cos θ + i sin θ). That is, we want the

complex number w = s(cosϕ + i sinϕ) such

that wn = z. From De Moivre’s Theorem we

want

sn(cosnϕ + i sinnϕ) = r(cos θ + i sin θ)

To get this, we need

sn = r and cosnϕ = cos θ and sinnϕ = sin θ

Thus nϕ = θ + 2kπ.

Roots of a complex number: n-th roots

Let z = r(cos θ + i sin θ) and let n be any

positive integer.

Then z has n distinct n-th roots. That is,

for k = 0, 1, 2, . . . , n− 1 the roots are

wk = r1/n

[cos

(θ + 2kπ

n

)+ i sin

(θ + 2kπ

n

)]All of the roots of z lie on the circle of radius

r1/n in the complex plane.

It often helps to think of the argument of a

complex root in the following way:

θ + 2kπ

n=θ

n+ k

(2π

n

)The argument of the first root will simply beθn (because k = 0). Each root after that has

the same modulus (r1/n) but is rotated

anti-clockwise by 2πn . You only need to

calculate solutions up to k = n− 1. If you let

k = n then you will have the same complex

number as w0 but with a different argument.

Example of roots of a complex number

Find the cube roots of z = i.

Note that a = 0 and b = 1. Thus we get

r = 1 and arg(z) = θ =π

2

Solutions to 3√i

w0 =3√

1(

cos(π

6

)+ i sin

(π6

))w1 =

3√

1

(cos

(π

6+

2π

3

)+ i sin

(π

6+

2π

3

))w2 =

3√

1

(cos

(π

6+

4π

3

)+ i sin

(π

6+

4π

3

))Exercise: simplify the angles in w1 and w2

and convert each solution to the form a+ bi.

You will see that w2 is a solution that you

might have found by inspection.

Another exercise:

Take w0, w1 and w2 and cube each of them

using the rectangular form (a + bi). Check

that in each case you get i as the solution.

Also, look at Example 7 on page A62 of the

textbook.