P3 Chapter P3 Chapter 1616

CIE Centre A-level CIE Centre A-level Pure MathsPure Maths

© Adam Gibson

COMPLEX NUMBERSCOMPLEX NUMBERS

Girolamo Cardano: 1501-1576

A colourful life!

3 0x px q

The “depressed cubic”;other cubics can be expressedin this form.

3 2ax bx cx d

COMPLEX NUMBERSCOMPLEX NUMBERS

Cardano “stole” the method from Tartaglia. Finding x you mustwrite:

2 3

3 2 , 2 4 27 3 3

q q p p au x u

u

can be less than zerobut the solutionis a real number!

Square roots of negative numbers can be useful,just as negative numbers themselves.

COMPLEX NUMBERSCOMPLEX NUMBERS

By introducing one number 2 1i

we can solve lots of new problems, and makeother problems easier.

It can be multiplied, divided, added etc. just as anyother number; but the equation above is the onlyextra rule that allows you to convert between i andreal numbers.

Start by noticing that

4 4 1 2i So the square root ofany negative numbercan be expressed in termsof i.

COMPLEX NUMBERSCOMPLEX NUMBERS

Therefore we can solve equations like:

2 8x The answer is 2 2x i

But what is 1 3i 1 3i

The two types of number cannot be “mixed”.

Numbers of the form , k i k are called imaginary numbers (or “pure imaginary”)

Numbers like 1, 2, -3.8 that we used before are calledreal numbers.

When we combine them together in a sum we havecomplex numbers.

COMPLEX NUMBERSCOMPLEX NUMBERS

COMPLEX NUMBERSCOMPLEX NUMBERS

To summarize,

z a bi

•a and b are real numbers•a is the “real part” of z; Re(z) •b is the “imaginary part” of z; Im(z)•The sum of the two parts is called a “complex number”

COMPLEX NUMBERSCOMPLEX NUMBERS

Adding and subtracting complex numbers:

1

2

(2 3 )

(4 9 )

z i

z i

1 2z z 6 6i

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

For addition and subtraction the real and imaginaryparts are kept separate.

COMPLEX NUMBERSCOMPLEX NUMBERS

Multiplying and dividing complex numbers:

1

2

(2 3 )

(4 9 )

z i

z i

1 2z z

2

(2 3 ) (4 9 )

2 4 (2 9 ) (3 4) (3 9 )

8 18 12 ( 27 )

35 6

i i

i i i i

i i i

i

( ) ( ) ( ) ( )a bi c di ac bd bc ad i

Notice how, for multiplication, the real and imaginaryparts “mix” through the formula i2 = -1.

COMPLEX NUMBERSCOMPLEX NUMBERS

Multiplying and dividing complex numbers:

1

2

(2 3 )

(4 9 )

z i

z i

1

2

z

z (2 3 )

(4 9 )

i

i

(2 3 ) (4 9 )

(4 9 ) (4 9 )

i i

i i

2

2

8 18 12 (27 )

4 4 36 36 ( 9 9 )

i i i

i i i

19 30 19 30

97 97 97

ii

Read through Sections

16.1 and 16.2 to makesure you understand the basics.

Rememberthis trick!!

COMPLEX CONJUGATESCOMPLEX CONJUGATES

Now that we have introduced complex numbers, we can view the quadratic solution differently.

2 4

2

b b acx

a

Now there are always two solutions, albeit they can berepeated real solutions.

If the equation has no real roots, it must have two complex roots.If one complex root is 1 8i what is the other?

1 8i These two numbers are called“complex conjugates”.

COMPLEX CONJUGATESCOMPLEX CONJUGATES

What are the solutions to 2 6 21 0x x ?

3 2 3i

If we write 3 2 3z i

Then the complex conjugate is written as * 3 2 3z i

* means conjugate

Calculate the following:

*

*

*

z z

z z

zz

22

6 2Re( )

4 3 2 Im( )

3 2 3 21

z

i z

2z

This will be discussedlater.

COMPLEX POWERSCOMPLEX POWERS

What happens if we square a complex number z?

2

2 2

2 2

( )( )

( ) (2 )

z x iy

z x iy x iy

x xiy xiy y

x y i xy

*

2*

2 2

( )( )

(2 )

z x iy

z x iy x iy

x y i xy

Compare the two results;they are complex conjugates!

And then squareits conjugate, z*:

* 22 *z z Later we will understandthis result geometrically

COMPLEX POWERSCOMPLEX POWERS

Continuing these investigations further (you may studyin your own time if you wish):

* *

* * *1 2 1 2

nnz z

z z z z

This will be easy to justifylater.

Examine the argument on page 228. This is a key idea, although you don’t have to understand the proof.

Non-real roots of polynomialsNon-real roots of polynomials with real coefficientswith real coefficients

always occur in conjugate pairs.always occur in conjugate pairs.

COMPLEX POWERSCOMPLEX POWERS

Tasks

Find all the roots of the following two polynomials:

4 2b) 1z z

3 2a) 7 65z z z

The first example can be attacked using the factortheorem.Examining +/-1,+/-5,+/-13 gives one root as -5.Equating coefficients therefore gives:

3 2 27 65 ( 5)( 4 13)z z z z z z

COMPLEX POWERSCOMPLEX POWERS

3 2 7 65 ( 5)( 2 3 )( 2 3 )z z z z z i z i

The roots are therefore -5, 2-3i, 2+3i.

4 2b) 1z z

The second example looks simpler but is, in a way, more difficult.First set w = z2. 2

2 2

1 0

1 3

21 3 1 3

or 2 2 2 2

w w

w

i iz z

It seems we haveto find the squareroot of a complexnumber!

COMPLEX POWERSCOMPLEX POWERS

Algebra is not the best way to do it, but let’s try anyway.

( )( )

z a ib

a ib a ib z

The next step is important tounderstand. It is called“equating real and imaginary parts”.

2 2

Re :

Im :

2

a b x

ab y

22

2

4

ya

by

b xb

Simultaneousequations.Let’s apply it to our problem.

COMPLEX POWERSCOMPLEX POWERS

2

2 2

22

4 2

2

1 3

2 21 3

( )( )2 2

1 3 , 2

2 29 1

16 2

16 8 9 0

8 24 1 3 11 or

32 4 4 2

iz

ix iy x iy

x y xy

xx

x x

x

COMPLEX CONJUGATESCOMPLEX CONJUGATES

Special properties of complex conjugates:

z

z*

* 2Re( )z z z * 2Im( )z z z

What is *zz ?

How do we knowit must be a realnumber?

Im( )z

Re( )z

COMPLEX CONJUGATESCOMPLEX CONJUGATES

*

(cos sin )

(cos sin )

z r i

z r i

*

2* * 2

arg( ) 0

or

zz

zz z z z r

2*zz z This is a very important result.

1 2 1 2arg( ) arg( ) arg( )z z z z

COMPLEX CONJUGATESCOMPLEX CONJUGATES

How many complex roots do the following polynomialshave?

10

2 3

2 5 4

A 3 4

B 65 63

C 3 4 18 13

z z

z z z

z z z

10

3

5

See page 229. We always have n roots for a polynomial of degree n. If the coefficients are real numbers, then we also know that any non-real roots occur in complex conjugate pairs.

If 1-8i is a root of polynomial B, what are the other roots?

POLAR COORDINATE FORMPOLAR COORDINATE FORM

zIm( )z

Re( )z

r

cosr

sinr

The modulus is the length of the line from 0+0i to the number z, i.e. r.The argument is the angle between the positive real axis and that line, by convention we use

POLAR COORDINATE FORMPOLAR COORDINATE FORM

Find, to 3 s.f. the modulus and argument of the followingcomplex numbers:

4

7 6

i

i

arg( )z 2 24 1 17r

(cos sin )z r i

To find θ we have two equations:

4 1cos , sin

17 17

0.245

modulus

POLAR COORDINATE FORMPOLAR COORDINATE FORM

4

7 6

i

i

2 26 7 17 5 9.22 (3 s.f.)z r

7 6cos , sin

17 5 17 5

arg( ) 2.43 (3 s.f.)z

7 6 9.22(cos(2.43) sin(2.43))z i i

EXPONENTIAL FORMEXPONENTIAL FORM

( ) cos( ) sin( )y x x i x

or

( ) cos( ) sin( )z i

dy

dx sin cosx i x ( sin cos )i i x x

iy

Which function does this?dy

kydx

EXPONENTIAL FORMEXPONENTIAL FORM

kxy Ae

So (not proof but good enough!)

(cos sin ) iz r i re

If you find this incredible or bizarre, it means youare paying attention.

Substituting gives “Euler’s jewel”:

1 0ie which connects, simply, the 5 mostimportant numbers in mathematics.

EXPONENTIAL FORMEXPONENTIAL FORM

We can write any complex number in this form ire

As before, r is the modulus and θ is the argument.

Examples:

2

6

1 3

i

i

i

2

2

3

0

2

6

2

i

i

i

i

e

e

e

e

Do you see how easy itis to calculate powers?

Find 10

1 3i

23

1

1024ie