A complex number in the form is said to be in Cartesian form

Complex numbers

What is ?1There is no number which squares to make -1, so there is no ‘real’ answer!

Mathematicians have realised that by defining the imaginary number ,many previously unsolvable problems could be understood and explored.

1i

If , what is:1i

2i 3i 4i

4 100 3

1 i 1

i2 i10 i3

A number with both a real part and an imaginary one is called a complex number

Eg iz 32

Complex numbers are often referred to as z, whereas real

numbers are often referred to as x

The real part of z, called Re z is 2

The imaginary part of z, called Im z is 3

iyxz

Manipulation with complex numbers

WB1 z = 5 – 3i, w = 2 + 2i

Express in the form a + bi, where a and b are real constants,

(a) z2 (b) wz

22 35 iz 93025 i i3016 a)

ii

wz

2235

ii

2222

44

661010

iib) 8

164 i i22

1

Techniques used with real numbers can still be applied with complex numbers:

Expand & simplify as usual, remembering that i2 = 1

An equivalent complex number with a real denominator can be found by multiplying by the complex conjugate of the denominator

If then its complex conjugate is

iyxz

iyxz *

Modulus and argumentiyxz

Re

Im

yx,The complex number can be represented on an Argand diagram by the coordinates

Eg iz 311

311 ,z

Eg iz 22

122 ,z

22 yxz

zargis the angle from the positive real axis to in the range yxz ,

The principal argument

The modulus of z,

Eg 1z 23122

Eg 2z 512 22

1z

2z

12

3

31 tan

31

zarg

....tan 4630211

(2dp) 4602 .arg z

Eg iz 223

223 ,z

3z

3

1

2

Eg 3z 2222 22

4221

tan

43

3

zarg3

3zarg

Remember the definition of arg z

2

1

zz

2

1

zz

WB2 The complex numbers z1 and z2 are given by

Find, showing your working,

(b) the value of

(c) the value of , giving your answer in radians to 2 decimal places.

(a)

iz 821 iz 12

in the form a + bi, where a and b are real,

2

1

zzarg

ii

zz

1

82

2

1

ii

11

118822

ii

2106 i

i53

i53

22 yxz The modulus of is iyxz

22 53 34

Re

Im zargis the angle from the positive real axis to in the range yxz ,

The principal argument

53,

031351 .tan

zarg 112.

WB3 z = 2 – 3i

Find, showing your working,(b) the value of z2,

(c) the value of arg (z2), giving your answer in radians to 2 decimal places.

(d) Show z and z2 on a single Argand diagram.

(a) Show that z2 = −5 −12i. 22 32 iz 9664 ii i125

iz 1252 22 125 13

Re

Im

125 ,

....tan 17615

121

zarg 971.

Re

Im

2z

z

02 cbxaxa

acbbx2

42 If then

We get no real answers because the discriminant is less than zero

12

131444 2

x2

364 i32

This tells us the curve

will have no intersections with

the x-axis

1342 xxy

264 i

Complex rootsIn C1, you saw quadratic equations that had no roots.

Eg 01342 xx Quadratic formula

We can obtain complex roots

though

We could also obtain these roots by completing the square:

01342 xx

01342 2 x

92 2 x

ix 32 ix 32

WB4 z1 = − 2 + i(a) Find the modulus of z1

(b) Find, in radians, the argument of z1 , giving your answer to 2 decimal places.

The solutions to the quadratic equation z2 − 10z + 28 = 0 are z2 and z3 (c) Find z2 and z3 , giving your answers in the form p iq, where p and q are integers.

(d) Show, on an Argand diagram, the points representing your complex numbers

i 2 22 12 5

Re

Im 12,

....tan 4630211

zarg 682.

028102 zz

028255 2 z

35 2 z

35 iz

35 iz

Re

Im

WB6

Given that , where a and b are real constants,

(c) Find the sum of the three roots of f (x) = 0.

(b) Find the three roots of f(x) = 0.

1504423 xxxxf

baxxxxf 23(a) find the value of a and the value of b.

baxxx 23 baxxbxaxx 333 223 bxabxax 333 23

Comparing coefficients of x2 2 a

Comparing coefficients of x0 50 b

05022 xx 05011 2 x 491 2 x

ix 71

ix 71

5023 2 xxxxf 0

03 xeither 3 xor

37171 ii

So sum of the three roots is -1

Problem solving with roots

In C2 you met the Factor Theorem: If a is a root of f(x) then is a factor)( ax

Eg Given that x = 3 is a root of the equation ,(a) write down a factor of the equation,

(b) Given that x = -2 is the other root, find the values of a and b

3x

062 xx

6 1 ba ,

02 baxx

2 x is the other factor

23 xx is the equation factorised

expanding

In FP1 you apply this method to complex roots…

Problem solving with complex roots

ix 32We have seen that complex roots come in pairs:

Eg 01342 xx

This leads to the logical conclusion that if a complex number is a root of an equation, then so is its conjugateiyxz iyxz *

We can use this fact to find real quadratic factors of equations:

WB5 Given that 2 – 4i is a root of the equation z2 + pz + q = 0,where p and q are real constants,(a) write down the other root of the equation,

(b) find the value of p and the value of q.

i42

04242 iziz

0424242422 iiziziz

0168844242 22 iiiizzizzz

02042 zz 20 4 qp ,

Factor theorem: If a is a root of f(x) then is a factor)( ax

WB7 Given that 2 and 5 + 2i are roots of the equation

012 23 dcxxx Rdc ,

(c) Show the three roots of this equation on a single Argand diagram.

(a) write down the other complex root of the equation.

(b) Find the value of c and the value of d.

i25

ixix 2525

iixixix 252525252 22 101010252525 iiiixxixxx

35102 xx

235102 xxx

703520102 223 xxxxx

705512 23 xxx

70 5 dc ,Re

Im

Problem solving by equating real & imaginary parts

Eg Given that iibai 153where a and b are real, find their values

iiba 1 bbiaia ibaba

Equating real parts: 3 ba

Equating imaginary parts: 5 ba

)(1

)(2

)()( 21 82 a 4 a

)(2 in Sub 54 b 1 b

WB8 Given that z = x + iy, find the value of x and the value of y such that

where z* is the complex conjugate of z.

iyxz

z + 3iz* = −1 + 13i

iyxz *

iyxiiyxizz 33 *

yixiyx 33

ixyyx 33

Equating real parts: 13 yx

Equating imaginary parts: 133 xy

)(1

)(2

31 )( 393 yx )(3

)()( 23 168 y 2 y

)(1 in Sub 16 x 5 x

then

Eg Find the square roots of 3 – 4i in the form a + ib, where a and b are real

2iba 22 2 babia iabba 222

Equating real parts: 322 ba

Equating imaginary parts: 42 ab

)(1

)(2

ba 2)(2324

2 bb

)(1 in sub

24 34 bb

043 24 bb

041 22 bb

1 b as b real2 a)(2 in sub

Square roots are -2 + i and 2 - i

Eg Find the roots of x4 + 9 = 0 ix 32

2iba 22 2 babia iabba 222

Equating real parts: 022 ba

Equating imaginary parts: 32 ab

)(1

)(2

ba 23)(2

0249

2 bb

)(1 in sub

049 4 b

494 b

23 b

232

3 a)(2 in sub

Roots are 23

23

23i 2

32

3

23

i,23

23

23

i, 23

23

23

i,

Re

Im yxz ,

22 yxzr

r

zarg

cosrx

sinry sincos irrz

sincos irz

Modulus-argument form of a complex number

iyxz If and

then

known as the modulus-argument form of a complex number

Eg express in the form iz 2 sincos irz

From previously,

5r6

665 sincos izso

Eg express 43

432 sincos iz

iyxz in the form

and

cosrx 4

32 cossinry

1 4

32 sin1

iz 1

The modulus & argument of a product

2121 zzzz

Eg if andiz 21 iz 312

2121 zzzz 2222 3112

105 25

It can be shown that:

This is easier than evaluating and then finding the modulus…

21zz

iizz 31221 362 ii i55

izz 5521 22 55 50 25

2121 zzzz argargarg

It can also be shown that:

Eg if andiz 21 iz 312

Re

Im

1z

....tan 460211

1zarg ....460

2z ....tan 3203

11

22zarg

....241

........arg 24146021 zz

4

2

1

2

1

zz

zz

It can be shown that:

Eg if andiz 21 iz 312

2

1

2

1

zz

zz

105

21

The modulus & argument of a quotient

212

1 zzzz

argargarg

This is much easier than evaluating and then finding the modulus…

2

1zz

ii

zz

312

2

1

ii

3131

91

362

iii10

7101

izz

107

101

2

1 21072

101 100

502

1

It can also be shown that:

Eg if andiz 21 iz 312

....arg 4601 z....arg 2412 z

........arg 2414602

1 zz

....711

From previously,

WB9 z = – 24 – 7i(a) Show z on an Argand diagram.

(c) find the values of a and b

(d) find the value of

It is given that w = a + bi, a , ℝ b .ℝ

(b) Calculate arg z, giving your answer in radians to 2 decimal places.

Given also that and4w 65warg

zw

Re

Im

724 ,

....tan 28302471

zarg862.

zr zarg

sincos irz where

and

Modulus-argument form

65

654 sincos iw i232

2121 zzzz wzzw

25724 22 z

4w given

100425

w is a root of . Find the values of a and b

Manipulation with complex numbers i

i2235

ii

2222

44

661010

ii8164 i

i221

Re

Im

z

531tan

zarg

Modulus and argument

3435 22

Complex roots bazz 2

iziz 2222 iiziizz 222222222 444422222 iiizzizzz

842 zz

Equating real & imaginary parts qipwz * Find the values of p and q

iiwz 3522 * i164

Equating real parts: 4 pEquating imaginary parts: 16 q

Complex numbers Using: iz 35 iw 22

wz

z

2121 zzzz

2121 zzzz argargarg

Also2

1

2

1

zz

zz

212

1 zzzz argargarg