Further Pure 1

Lesson 5 – Complex Numbers

WiltshireNumbers

What types of numbers do we already know? Real numbers – All numbers ( 2, 3.15, π ,√2) Rational – Any number that can be expressed as a

fraction

– (4, 2.5, 1/3) Irrational – Any number that can`t be expressed as a

fraction. ( π ,√2, √3 + 1) Natural numbers – The counting numbers (1, 2, 3, ….) Integers – All whole numbers ( -5, -1, 6) Complex Numbers (Imaginary numbers)

WiltshireComplex Numbers

What is the √(-1)? We define the √(-1) to be the imaginary

number j. (Hence j2 = -1) Note that lots of other courses use the letter

i, but we are going to use j. We can now use this to calculate a whole

new range of square roots. What is √(-144)? Answer is +12j and -12j, or ±12j.

WiltshireComplex Numbers

Now we can define a complex number (z) to be a number that is made up of real and imaginary parts.

z = x + y j Here x and y are real numbers. x is said to be the real part of z, or Re(z). y is said to be the imaginary part of z, or

Im(z).

WiltshireSolving Quadratics

Use the knowledge you have gained in the last few slides to solve the quadratic equation

z2 + 6z + 25 = 0 Remember

Solutiona

acbbz

2

42

12

251466 2

z

WiltshireSolving Quadratics

2

100366 z

2

646 z

2

86 jz

jz 83

WiltshireAddition and Subtraction

To Add and subtract complex numbers all you have to do is add/subtract the real and imaginary parts of the number.

(x1+y1j) + (x2+y2j) = x1 + x2 + y1j + y2j

= (x1 + x2)+ (y1 + y2)j

(x1+y1j) – (x2+y2j) = x1 - x2 + y1j - y2j

= (x1 - x2)+ (y1 - y2)j

WiltshireMultiplying

Multiplying two complex numbers is just like multiplying out two brackets.

You can use the FOIL method. First Outside Inside Last. Remember j2 = -1 (x1+y1j)(x2+y2j) = x1x2 + x1y2j + x2y1j + y1y2j2

= x1x2 + x1y2j + x2y1j - y1y2

= x1x2 - y1y2 + x1y2j + x2y1j

= (x1x2 - y1y2) + (x1y2 + x2y1)j What is j3, j4, j5?

WiltshireMultiplying

Alternatively you could use the box method.

(x1+y1j)(x2+y2j) = (x1x2 - y1y2) + (x1y2 + x2y1)j

x1 y1j

x2 x1x2 x2y1j

y2j x1y2j - y1y2

WiltshireQuestions

If z1 = 5 + 4j z2 = 3 + j z3 = 7 – 2j

Find

a) z1 + z3 = 12 + 2j

b) z1 - z2 = 3 + 3j

c) z1 – z3 = -2 + 6j

d) z1 × z2 = 11 + 17j

e) z1 × z3 = 43 + 18j

WiltshireComplex Conjugates

The complex conjugate of

z = (x + yj) is z* = (x – yj) If you remember the two solutions to the quadratic

from a few slides back then they where complex conjugates.

z = -3 + 8j & z = -3 – 8j In fact all complex solutions to quadratics will be

complex conjugates. If z = 5 + 4j What is z + z* What is z × z*

WiltshireActivity

Prove that for any complex number z = x + yj, that z + z* and z × z* are real numbers.

First z + z* = (x + yj) + (x – yj)

= x + x + yj – yj

= 2x = Real Now z × z* = (x + yj)(x – yj)

= x2 – xyj + xyj – y2j2

= x2 – y2(-1)

= x2 + y2 = Real Now complete Ex 2A pg 50

WiltshireDivision

There are two ways two solve problems involving division with complex numbers.

First you need to know that if two complex numbers are equal then the real parts are identical and so are there imaginary parts.

If we want to solve a question like 1 ÷ (2 + 4j) we first write it equal to a complex number p + qj.

Now we re-arrange the equation to find p and q.(p + qj)(2 + 4j) = 1

jqjp

42

1

WiltshireDivision

Expanding the equation gives2p – 4q + 2qj + 4pj = 1

The number 1 can be written as 1 + 0j So

(2p – 4q) + (2q + 4p)j = 1 + 0j Now we can equate real and imaginary parts.

2p – 4q = 14p + 2q = 0

Solve these equationsp = 1/10 & q = -1/5

Therefore 1 ÷ (2 + 4j) = 0.1 – 0.2j

WiltshireDivision

The second method is similar to rationalising the denominator in C1.

The 20 on the bottom comes from the algebra we proved a few slides back. (x + yj)(x – yj) = x2 + y2

Now see if you can find (3 - 5j) ÷ (2+9j)

Now complete Ex 2B pg 53

jj

j

j

jj 5

1

10

1

20

42

42

42

42

1

42

1

85

3739

814

4527106

92

92

92

53

92

53 jjj

j

j

j

j

j

j

WiltshireArgand Diagrams

Complex numbers can be shown Geometrically on an Argand diagram

The real part of the number is represented on the x-axis and the imaginary part on the y.

-3 -4j 3 + 2j 2 – 2j

Re

Im

WiltshireModulus of a complex number

A complex number can be represented by the position vector.

The Modulus of a complex number is the distance from the origin to the point.

|z| = √(x2+y2) Note |x| = x

Re

Im

y

x

x

y

WiltshireModulus of a complex number

Find

a) |3 + 4j| = 5

b) |5 - 12j| = 13

c) |6 - 8j| = 10

d) |-24 - 10j| = 26

WiltshireSum of complex numbers

z1 + z 2 =

Re

Im

6

7

15

52

1

5

5

2

21

21

2

2

1

1

yy

xx

y

x

y

x

z1 + z2

z1

z2

WiltshireDifference of complex numbers

z2 - z1 =

Now complete Ex 2C pg 57 Re

Im

z2

z1

z2 – z1

1

5

56

27

5

2

6

7

12

12

1

1

2

2

yy

xx

y

x

y

x

WiltshireSets of points in Argand diagram

Re

Im What does |z2 – z1| represent?

If z1 = x1 +y1j

& z2 = x2 +y2j

Then z2 – z1

= (x2 – x1) + (y2 – y1)j

So |z2 – z1|

= √((x2 – x1)2 + (y2 – y1)2)

This represents the distance between to complex numbers

z1 & z2.

(x1,y1)

(x2,y2)

y2- y1

x2- x1

WiltshireExamples

Draw an argand diagram showing the set of points for which |z – 3 – 4j| = 5

Solution First re-arrange the

question |z – (3 + 4j)| = 5

From the previous slide this represents a constant distance of 5 between the point (3,4) and z.

This will give a circle centre (3,4)

Now do Ex 2D pg 60

Re

Im

WiltshireExamples

How would you show the sets of points for which:

i) |z – 3 – 4j)| ≤ 5

ii) |z – 3 – 4j)| < 5

iii) |z – 3 – 4j)| ≥ 5

Im

Re

Re

Im

Re

Im

Re