chapter 1 — complex numbers -...

C o m p l e x n u m b e r s M C 1 2 Q l d - 1 1

Exercise 1A — Operations on and representations of complex numbers 1 a 4u

= 4(3 − 2i) = 12 − 8i b u + v = (3 − 2i) + (2 + 4i) = 5 + 2i c 2u + 3v = 2(3 − 2i) + 3(2 + 4i) = 6 − 4i + 6 + 12i = 12 + 8i d u − v = (3 − 2i) − (2 + 4i) = (1 − 6i) e u × v = (3 − 2i)(2 + 4i) = 6 + 12i − 4i − 8i2 = 14 + 8i f u ÷ v

= 3 2 2 42 4 2 4

i ii i

− −×+ −

= 2

26 12 4 8

4 16i i i

i− − +

−

= 2 1620

i− −

= 2(1 8 )20

i− +

= 110− (1 + 8i)

= −0.1 − 0.8i g u2 = (3 − 2i) (3 − 2i) = 9 − 12i + 4i2 = 5 − 12i

2 a v = − 2 + i

= 2 2( 2) (1)− +

= 5 = ≈ 2.24 b v + 2w = −2 + i + 2(1 + 4i) = −2 + i + 2 + 8i = 9i = 9 c v = −2 − i d 2w v+ = (1 4 2( 2 )i i+ + − +

= (1 4 4 2 )i i+ − +

= ( 3 6 )i− + = −3 − 6i e w

= 1 − 4i

= 2 2(1) ( 4)+ −

= 17 ≈ 4.12

f v v = (−2 + i)(−2 − i) = 4 + 2i − 2i − i2 = 5 g 2 2w w× = 2(1 + 4i) × 2(1 4 )i+ = (2 + 8i)(2 − 8i) = 4 − 16i + 16i − 64i2 = 68 h v−1

= 1v

= 1 22 2

ii i

− −×− + − −

= 22

4i

i− −

−

= 15− (2 + i)

= −0.4 − 0.2i i w−1

= 1 1 41 4 1 4

ii i

−×+ −

= 21 4

1 16ii

−−

= 1 417

i−

= 1 (1 4 )17

i−

≈ 0.06 − 0.24i 3 a z = a + bi

Show that

z−1 = 2zz

LHS z−1 = 1 a bia bi a bi

−×+ −

= 2 2a bi

a b−+

RHS 2zz

= 2 2 2( )

a bi

a b

−

+ = 2 2

a bia b

−+

= LHS

∴ z−1 = 2zz

b zz = 2z

LHS zz = (a + bi) (a − bi) = a2 − b2i2 = a2 + b2

RHS 2z = 2 2 2( )a b+

= a2 + b2 LHS = RHS So zz = 2z

4 a Let z = 3 + 3i

r = 2 2(3) (3)+

r = 18 r = 3 2

Chapter 1 — Complex numbers

M C 1 2 Q l d - 1 2 C o m p l e x n u m b e r s

θ = tan−1 33

(1st quadrant)

θ = 4π

z = 3 2 cis4π

z = 4.24 cis4π

b Let z = 3 − 3i

r = 2 2(3) ( 3)+ −

r = 3 2 θ = tan−1(−1) (4th quadrant)

θ = 4π−

r = 3 2 cis4π−

r ≈ 4.24 cis4π−

c z = −3 + 3i

r = 2 2( 3) (3)− +

r = 3 2 θ = tan−1(−1) (2nd quadrant)

θ = 34π

r = 3 2 cis 34π

r ≈ 4.24 cis 34π

d z = 2 3 + 4i

r = 2 2(2 3) (4)+

r = 12 16+ r = 28 r = 2 7 r ≈ 5.29

θ = tan−1 42 3

(1st quadrant)

θ ≈ 0.86 z ≈ 5.29 cis(0.86) e z = 2 3 − 4i

r = 2 2(2 3) ( 4)+ −

r = 12 16+ r = 28 r = 2 7 r ≈ 5.29

θ = tan−1 42 3−

(4th quadrant)

θ = −0.86 z ≈ 5.29 cis(−0.86) f z = 4 + 3i

r = 2 2(4) (3)+

r = 16 9+ r = 25 r = 5

θ = tan−1 34

(1st quadrant)

θ ≈ 0.64 z = 5 cis(0.64) g z = −4 − 3i

r = 2 2( 4) ( 3)− + −

r = 25 r = 5

θ = tan−1 34

(3rd quadrant)

θ = (−π + 0.64) θ = −2.50 z = 5 cis(−2.50) h z = −4 + 3i

r = 2 2( 4) (3)− +

r = 25 r = 5

θ = tan−1 34

−

(2nd quadrant)

θ = (π − 0.64) θ = 2.50 z = 5 cis(2.50) i z = 3 + 4i

r = 2 2(3) (4)+

r = 25 r = 5

θ = tan−1 43

(1st quadrant)

θ = 0.93 z = 5 cis(0.93)

5 a z = 4 cis3π

z = 4 cos sin3 3

iπ π +

z = 4 × 0.5 + i 4 × 0.866 z = 2 + 3.46i

b z = 3 cis6π

z = 3 cos sin6 6

iπ π +

z = 3 × 0.866 + i × 3 × 0.5 z = 2.598 + 1.5i z = 2.6 + 1.5i

c z = 20 cis 23π

z = 2 220 cos sin3 3

iπ π +

z = 20 × −0.5 + i × 20 × 0.866 z = −10 + 17.32i

d z = 8 cis4π−

z = 8 cos sin4 4

iπ π − − +

z = 8 × 0.707 + i × 8 × −0.707 z = 5.66 − 5.66i e z = i cis(π) z = i (cos(π) + i sin(π)) = i (−1 + 0) = −i


f z = 4 cis(1) z = 4(cos(1) + i sin(1)) = 4 × 0.54 + i ×4 × 0.84 = 2.16 + 3.36i g z = 3 cis(2) z = 3(cos(2) + i sin(2)) z = 3 × −0.42 + i × 3 × 0.91 z = −1.26 + 2.73i h z = 2 cis(−1) z = 2(cos(−1) + i sin(−1)) z = 2 × 0.54 + i × 2 × −0.84 z = 1.08 − 1.68i

6 See answers above 7 See answers above 8 a u × v

= 4 cis4π

× 8 cis 23π

= (4 × 8) cis 24 3π π +

= 32 cis 3 812 12π π +

= 32 cis 1112

π

b u ÷ v

= 4 cis

428 cis3

π

π

= 48

cis 24 3π π −

= 12

cis 3 812 12π π −

= 12

cis 512

π−

c v ÷ u

=

28 cis3

4 cis4

π

π

= 84

cis 23 4π π −

= 2 cis 8 312 12π π −

= 2 cis 512π

d u2

= 4 cis4π

× 4 cis4π

= 16 cis4 4π π +

= 16 cis2π

e v3

= 83 cis 233π ×

= 512 cis(2π) = 512 (cos(2π) + i sin(2π)) = 512

9 a Let z = 2 + i

r = 2 22 1+ r = 5 ≈ 2.24

tan(θ) = 12

θ = tan−1 12

θ = 0.46 In polar form z = 2.24 cis(0.46)

b 2 i+ = 12z

z = 2.24 cis(0.46)

12z =

12 12.24 cis 0.46

2 ×

12z = 1.5 cis(0.23)

c 2 i+ = 1.5 cis(0.23) = 1.5 (cos(0.23) + i sin(0.23)) = 1.46 + 0.34i

10 a Let z = 1 + 2i

r = 2 21 2+ r = 5 ≈ 2.24 tan(θ) = 2 θ = tan−1(2) θ = 1.12 z = 2.24 cis(1.12)

z = 12z =

12 12.24 cis 1.12

2 ×

12z = 1.5 cis(0.56)

12z = 1.5 cos(0.56) + 1.5 sin(0.56)i

12z = 1.27 + 0.80i

12

1

z = 1 1.27 0.80

1.27 0.80 1.27 0.80i

i i−×

+ −

= 2 21.27 0.80

(1.27 0.80 )i−

+

= 1.27 0.802.2529

i−

= 0.56 − 0.35i

b i = 12i =

12(0 1 )i+

=

12

cis 2π

= cis4π

= cos sin4 4

iπ π +

= 0.71 + 0.71i

c Let z = 3 12 2

i−

r = 2 23 1

2 2 − +

= 3 14 4

+

= 1


tan(θ) =

123

2

= 13

(4th quadrant)

θ = 6π−

z = cis6π−

z32 = cis 326π− ×

z32 = 16cis3

π−

z32 = 12 4cis3 3

π π− −

= 4cis3π−

= 2cis3π π π − < θ ≤

= 2 2cos sin3 3

iπ π +

= −0.5 + 0.866i 11

12

13

14

Operation on z Geometric relationship to z a z Reflection in x-axis b iz Anticlockwise rotation of 90° c −z Rotation of 180° d −iz Clockwise rotation of 90°

Investigation — Complex numbers and matrices 1 a z = 3 + 4i A = 3I + 4H

= 1 0 0 1

3 40 1 1 0

− +

= 3 0 0 40 3 4 0

− +

= 3 44 3

−

b A2 = 3 4 3 44 3 4 3

− − ×

= 3 4 ( 4) 4 3 ( 4) ( 4) 34 3 3 4 4 ( 4) 3 3

× + − × × − + − × × + × × − + ×

= 9 16 12 12

12 12 16 9− − −

+ − +

= 7 24

24 7− − −

This corresponds to the complex number a + bi where

aI + BH = 7 24

24 7− − −

aI + BH =1 0 0 10 1 1 0

a b−

+

= a bb a

−

So a bb a

−

= 7 24

24 7− − −

Therefore a = −7 and b = 24, giving the complex number −7 + 24i

z2 = (3 + 4i)(3 + 4i) = 9 + 12i + 12i + 16i2 = −7 + 24i So A2 corresponds to the complex number z2

c A−1 = 3 414 33 3 ( 4) 4

−× − − ×

= 3 414 39 16

−+

= 3 414 325

−

A−1 corresponds to the complex number c + di where

cI + dH = 3 414 325

−

cI + dH = 1 0 0 10 1 1 0

c d−

+

= c dd c

−

So c dd c

−

= 3 414 325

−

c dd c

−

=

3 425 25

4 325 25

−

Therefore c = 325

and d = 425−

So A−1 corresponds to the complex number 3 425 25

i−

z−1 = (3 + 4i)−1

= 1 3 43 4 3 4

ii i

−×+ −


= 23 4

9 16ii

−−

= 3 425

i−

= 3 425 25

i−

So the complex number corresponding to A−1 is the inverse of z, z−1.

2 a a bb a

−

= 1 0 0 10 1 1 0

a b−

+

= aI + bH

which corresponds to the complex number a + bi

Ta b

b a−

= a bb a

−

= a1 0 0 10 1 1 0

b−

−

= aI − bH which corresponds to the complex number a − bi, the

complex conjugate of a + bi. So the transpose corresponds to the conjugate.

b 1a b

b a

−−

= 2 21 a b

b ca b −+

= 2 2 2 2

2 2 2 2

a ba b a b

b aa b a b

+ +

− + +

= 2 2 2 2

1 0 0 10 1 1 0

a ba b a b

− − + +

= 2 2 2 2 1a bIa b a b

− ++ +

which corresponds to the complex number

2 2 2 2a b i

a b a b−

+ +

(a + bi)−1 = 1 a bia bi a bi

−×+ −

= 2 2 2a bi

a b i−−

= 2 2a bi

a b−+

= 2 2 2 2a b i

a b a b−

+ +

which is the inverse of a + bi So the inverse of the matrix corresponds to the inverse of

its corresponding complex number.

c deta bb a

−

= a × a − b × (−b) = a2 + b2

2a bi+ = 2 2 2( )a b+ = a2 + b2 So the determinant corresponds to the modulus

squared.

Investigation — eiθ 1 Show that eiθ = cos(θ)+ i sin(θ) LΗS = eiθ

= 2 4 2

1 ... ( 1) ...2! 4! (2 )!

nn

nθ θ θ

− + − + − +

+

3 5 2 1

... ( 1) ...3! 5! (2 1)!

nni

nθ θ θθ

+ − + + + − + +

But

cos(θ) = 2 4 2

1 ... ( 1) ...2! 4! (2 )!

nn

nθ θ θ− + − + − +

sin(θ) = 3 5 2 1

... ( 1) ...3! 5! (2 1)!

nn

nθ θ θθ

+− + − + − +

+

So LHS = (cos(θ)) + i(sin(θ)) = cos(θ) + i sin(θ) = RHS, as required

2 Show that cos(θ) = 2

i ie eθ θ−+

RHS = 2

i ie eθ θ−+

= ( )1 ( )2

i ie eθ θ−+

= 12

[(cos(θ) + i sin(θ)) + (cos(−θ)

+ i sin(−θ))]

= 12

(cos(θ) + i sin(θ) + cos(θ) − i sin(θ))

= 12

(2 cos(θ))

= cos(θ) = LHS, as required 3 To express sin(θ) in terms of eiθ and e−iθ we first need to

have an i sin(θ) term, so we multiply and divide by i. We also multiply and divide by 2, because we expect to have had two i sin(θ) terms added together, similarly to question 2.

sin(θ) = 12i

(2i sin(θ))

To have only a sin(θ) term remaining, the cos(θ) terms must cancel out so

sin(θ) = 12i

(cos(θ) + i sin(θ) − cos(θ) + i sin(θ))

= 12i

[(cos(θ) + i sin(θ)) − (cos(θ) − i sin(θ))]

= 12i

[(cos(θ) + i sin(θ)) − (cos(−θ) + i sin(−θ))]

= 12i

(eiθ − e−iθ)

= 2

i ie ei

θ θ−−

4 Show that cos(iθ) = 2

e eθ θ−+

LHS = cos(iθ)

= ( ) ( )

2

i i i ie eθ θ−+

= 2

e eθ θ− +

= 2

e eθ θ−+

= RHS, as required

5 2x = log (2 )ex

e = log (2)exe using the expansion for ex

2x = log (2)exe = 2 3( log (2)) ( log (2))1 log (2) ...

2! 3!e e

ex xx+ + + +

2x = In 2x

e = In 2xe

In 2xe = 2 3( In 2) ( In 2)1 In2

2! 3!x xx+ + + + …


By using the value of loge(2) from a table and by taking many terms from the expansion, an accurate value for 2x can be computed to a certain number of decimal places, depending on the amount of terms used.

Investigation — eiθ and de Moivre’s theorem 1 a Prove that v × w = 1 2 1 2cis( ),r r θ θ+ where v = r1 cis(θ1) and w = r2 cis(θ2) LHS = v × w = [r1(cos(θ1) + i sin(θ1))] × [r2(cos(θ2) + i sin(θ2))]

= 1 21 2( )( )i ir e r eθ θ

= 1 21 2 i ir r e θ θ+

= ( )1 21 2 ir r e θ θ+

= 1 2 1 2cis( )r r θ θ+ = RHS, as required b Prove that v

w = 1 2 1 2cis( ),r r θ θ−

where v = r1 cis(θ1) and w = r2 cis(θ2) LHS = v

w

= 1 1

2 2

cis( )cis( )

rr

θθ

= 1

1

22

i

ir e

r e

θ

θ

= 1 1 2

2

i ir er

θ θ−

= ( )1 1 2

2

ir er

θ θ−

= 11 2

2cis( )r

rθ θ−

= RHS, as required c Prove that vn = 1 1cis ,nr nθ where v = r1 cis(θ1) LHS = vn = (r1 cis(θ1))n

= 11( )i nr e θ

= ( )11

n inr e θ

= ( )11

i nnr e θ

= 1 1cis( )nr nθ = RHS, as required 2 a Show that cos(2θ) = cos2(θ) − sin2(θ) LHS = cos(2θ)

= (2 ) (2 )1 ( )2

i ie eθ θ−+ , since cos(x) = 2

ix ixe e−+

= 2 21 ( )2

i ie eθ θ−+

= 2 21 ( ) ( )2

i ie eθ θ− +

= 2 21 (cos( ) sin( )) (cos( ) sin( ))2

i iθ θ θ θ + + − + −

= 2 21 (cos( ) sin( )) (cos( ) sin( ))2

i iθ θ θ θ + + −

= 2 2 21 (cos ( ) 2 sin( )cos( ) sin ( )2

i iθ θ θ θ+ +

2 2cos ( ) 2 sin( )cos( ) sin( ))i iθ θ θ θ+ − +

= 2 21 (2cos ( ) 2sin ( ))2

θ θ−

= cos2(θ) − sin2(θ) = RHS, as required

b Show sin(2θ) = 2 sin(θ) cos(θ) LHS = sin(2θ)

= (2 ) (2 )1 ( ),2

i ie ei

θ θ−−

Since sin(x) = 1 ( )2

ix ixe ei

−−

= 2 21 ( )2

i ie ei

θ θ−−

= 2 21 ( ) ( )2

i ie ei

θ θ− −

= 2 21 (cos( ) sin( )) (cos( ) sin( ))2

i ii

θ θ θ θ + − − + −

= 2 21 (cos( ) sin( )) (cos( ) sin( ))2

i ii

θ θ θ θ + − −

= 2 2 21 (cos ( ) 2 sin( )cos( ) sin ( )2

i ii

θ θ θ θ+ +

2 2 2cos ( ) 2 sin( )cos( ) sin ( ))i iθ θ θ θ− + −

= 1 (4 sin( )cos( ))2

ii

θ θ

= 2sin(θ) cos(θ) = RHS, as required 3 a ii

i = 0 + i = cos sin2 2

iπ π +

= 2i

eπ

So ii = 2( )i

ieπ

= 2

2i

eπ

= 2eπ−

b e1 − π i = e1e−π i = [ ]cos( ) sin( )e iπ π− + −

= e(−1 + 0i) = −e c log (2 3 2 )e i−

First express 2 3 2i− in polar form Let z = 2 3 2i− = r cis(θ)

r = 2 2(2 3) ( 2)+ −

= 12 4+ = 16 = 4

tan(θ) = 22 3−

tan(θ) = 13

−

θ = 1 1tan3

− −

, (4th quadrant)

= 6π−

So z = 4cis6π−

So log (2 3 2 )e i− = log 4cis6eπ −

= 6log (4 )i

e eπ−

= 6log (4) log ( )i

e e eπ−

+


= log (4) log ( )6e ei eπ−

= log 46eiπ −

Exercise 1B — Factorisation of polynomials in C 1 a z2 + 4

= z2 − 4i2 = z2 − (2i)2 = (z + 2i)(z − 2i) b z2 + 7 = z2 − 7i2 = 2 2( 7 )z i−

= ( 7 )( 7 )z i z i+ − c z2 + 8z + 25 = (z2 + 8z + 16) − 16 + 25 = (z + 4)2 + 9 = (z + 4)2 − 9i2 = (z + 4)2 − (3i)2 = (z + 4 + 3i)(z + 4 − 3i) d z2 − 3z + 4

= 2 9 93 44 4

z z − + − +

= 23 9 16

2 4z − + − +

= 23 7

2 4z − +

= 2 23 7

2 4iz − −

= 223 7

2 2iz

− −

= 3 7 3 72 2 2 2

z i z i

− + − −

e 4z2 − 4z + 17

= 2 14 1 174

z z − + − +

= 214 16

2z − +

= 2

2 212 162

z i − −

= (2z − 1)2 − (4i)2 = (2z − 1 + 4i)(2z − 1 − 4i) f −9z2 + 24z − 32

= 2 8 169 16 323 9

z z − − + + −

= 249 16

3z − − −

= 2

2 43 163

z − − +

= −[(3z − 4)2 − 16i2] = −[(3z − 4)2 − (4i)2] = −(3z − 4 + 4i)(3z − 4 − 4i)

2 a z4 − 81 Let w = z2 = z4 − 81 = w2 − 81 = (w + 9)(w − 9) = (z2 + 9)(z2 − 9)

= (z2 − 9i2)(z + 3)(z − 3) = (z2 − (3i)2)(z + 3)(z − 3) = (z + 3i)(z − 3i) (z + 3)(z − 3) b z4 − 2z2 − 3 Let w = z2 z4 − 2z2 − 3 = w2 − 2w − 3 = (w − 3)(w + 1) = (z2 − 3)(z2 + 1) = 2 2 2 2( ( 3) )( )z z i− −

= ( 3)( 3)( )( )z z z i z i+ − + − c z4 + 20z2 + 64 Let w = z2 = z4 + 20z2 + 64 = w2 + 20w + 64 = (w + 16)(w + 4) = (z2 + 16)(z2 + 4) = (z2 − 16i2)(z2 − 4i2) = (z2 − (4i)2)(z2 − (2i)2) = (z + 4i)(z − 4i)(z + 2i)(z − 2i) d z4 + 3z2 − 10 Let w = z2 z4 + 3z2 − 10 = w2 + 3w − 10 = (w + 5)(w − 2) = (z2 + 5)(z2 − 2) = 2 2 2 2( 5 )( ( 2) )z i z+ − = 2 2 2 2( ( 5 ) )( ( 2) )z i z+ − = ( 5 )( 5 )( 2)( 2)z i z i z z+ − + −

3 a f(z) = z3 − 4z2 + 2z + 28 f(−2) = (−2)3 − 4(−2)2 + 2(−2)+ 28 = −8 − 16 − 4 + 28 = 0 So (z + 2) is a factor of f(z) Let f(z) = (z + 2)(z2 + pz + q) So z3 − 4z2 + 2z + 28 = (z + 2)(z2 + pz + q) z3 − 4z2 + 2z + 28 = z3 + pz2 + qz + 2z2 + 2pz +2q z3 − 4z2 + 2z + 28 = z3 + (p + 2)z2 + (2p + q)z + 2q Equating coefficients gives p + 2 = −4 2q = 28 p = −6 q = 14 So f(z) = (z + 2)(z2 − 6z + 14) = (z + 2)(z2 − 6z + 9 − 9 + 14) = (z + 2)[(z − 3)2 + 5] = (z + 2)[(z − 3)2 − 5i2] = 2 2( 2)[( 3) ( 5 ) ]z z i+ − −

= ( 2)( 3 5 )( 3 5 )z z i z i+ − + − −

The three factors of f(z) are z + 2, 3 5z i− + and 3 5z i− −

b f(z) = z3 + z2 + 2z − 4 f(1) = (1)3 + (1)2 + 2(1) − 4 = 1 + 1 +2 − 4 = 0 so (z − 1) is a factor of f(z) Let f(z) = (z − 1)(z2 + pz + q) So z3 + z2 + 2z − 4 = (z − 1)(z2 + pz + q) z3 + z2 + 2z − 4 = z3 + pz2 + qz − z2 − pz − q z3 + z2 + 2z − 4 = z3 + (p − 1)z2 + (q − p)z − q Equating coefficients gives p − 1 = 1 −4 = −q p = 2 q = 4 So f(z) = (z − 1)(z2 + 2z + 4) = (z − 1)(z2 + 2z +1 − 1 + 4) = (z − 1)[(z + 1)2 + 3] = (z − 1)[(z + 1)2 −3i2]


= 2 2( 1)[( 1) ( 3 ) ]z z i− + − = ( 1)( 1 3 )( 1 3 )z z i z i− + + + −

The three factors of f(z) are z − 1, 1 3z i+ + and 1 3z i+ −

c f(z) = z3 − z2 − z + 10 f(−2) = (−2)3 − (−2)2 + 2 + 10 = −8 − 4 + 12 = 0 So (z + 2) is a factor of f(z) Let f(z) = (z + 2)(z2 + pz +q) So z3 − z2 − z + 10 = (z + 2)(z2 + pz + q) z3 − z2 − z + 10 = z3 + pz2 + qz + 2z2 +2pz + 2q z3 − z2 − z + 10 = z3 + (p + 2)z2 + (2p + q)z + 2q Equating coefficients gives p + 2 = −1 2q = 10 p = −3 q = 5 So f(z) = (z + 2)(z2 − 3z + 5)

= 2 9 9( 2) 3 54 4

z z z + − + − +

= 23 9 20( 2)

2 4z z

− + + − +

= 23 11( 2)

2 4z z

+ − +

= 2

23 11( 2)2 4

z z i + − −

= 223 11( 2)

2 2z z i

+ − −

= 3 11 3 11( 2)2 2 2 2

z z i z i

+ − + − −

The three factors of f(z) are z + 2, 3 112 2

z i− + and

3 112 2

z i− −

d f(z) = 2z3 + 3z2 − 14z − 15 f(−1) = 2(−1)3 + 3(−1)2 − 14(−1) − 15 = −2 + 3 + 14 − 15 = 0 So (z + 1) is a factor of f(z) Let f(z) = (z + 1)(2z2 + pz + q) So 2z3 + 3z2 − 14z − 15 = (z + 1)(2z2 + pz + q) 2z3 + 3z2 − 14z − 15 = 2z3 + pz2 + qz + 2z2 + pz +q 2z3 + 3z2 − 14z − 15 = 2z3 + (p + 2)z2 + (p + q)z + q Equating coefficients gives p + 2 = 3 q = −15 p =1 So f(z) = (z + 1)(2z2 + z − 15) = (z + 1)(2z − 5)(z + 3) The three factors of f(z) are z +1, 2z − 5 and z + 3 e f(z) = z4 − 2z2 − 16z − 15 f(−1) = (−1)4 − 2(−1)2 − 16(−1) − 15 = 1 − 2 + 16 − 15 = 0 So (z + 1) is a factor of f(z) f(3) = (3)4 − 2(3)2 − 16(3) − 15 = 81 − 18 − 48 − 15 = 0 So (z − 3) is a factor of f(z) Let f(z) = (z + 1)(z − 3)(z2 + pz + q) So z4 − 2z2 − 16z − 15 = (z + 1)(z − 3)(z2 + pz + q) z4 − 2z2 − 16z − 15 = (z2 − 3z + z − 3)(z2 + pz + q) z4 − 2z2 − 16z − 15 = (z2 − 2z − 3)(z2 + pz + q) z4 − 2z2 − 16z − 15 = z4 + pz3 + qz2 − 2z3 − 2pz2

− 2qz − 3z2 − 3pz − 3 z4 + 0z3 − 2z2 − 16z − 15 = z4 + (p − 2)z3 + (q − 2p − 3)z2

+ (−2q − 3p)z − 3q

Equating coefficients gives p − 2 = 0 −3q = −15 p = 2 q = 5 So f(z) = (z + 1)(z − 3)(z2 + 2z + 5) = (z + 1)(z − 3)(z2 + 2z + 1 − 1 + 5) = (z + 1)(z − 3)[(z + 1)2 + 4] = (z + 1)(z − 3)[(z + 1)2 − 4i2] = (z + 1)(z − 3)[(z + 1)2 − (2i)2] = (z + 1)(z − 3)(z + 1 + 2i)(z + 1 − 2i) The four factors of f(z) are z + 1, z − 3, z + 1 +2i and

z + 1 − 2i. f f(z) = z6 − 1 Let w = z3 Then z6 − 1 = w2 − 1 = (w + 1)(w − 1) So f(z) = (z3 + 1)(z3 − 1) Let f(z) = g(z)h(z) such that g(z) = z3 + 1 and

h(z) = z3 − 1 g(z) = z3 + 1 g(−1) = (−1)3 + 1 = −1 + 1 = 0 So (z + 1) is a factor of g(z) Let g(z) = (z + 1)(z2 +pz + q) So z3 + 1 = (z + 1)(z2 +pz + q) z3 + 1 = z3 + pz2 + qz + z2 + pz + q z3 + 0z2 + 0z + 1 = z3 + (p + 1)z2 + (p + q)z +q Equating coefficients gives p + 1 = 0 q = 1 p = −1 So g(z) = (z + 1)( z2 − z +1)

= 2 1 1( 1) 14 4

z z z + − + − +

= 21 3( 1)

2 4z z

+ − +

= 2

21 3( 1)2 4

z z i + − −

= 221 3( 1)

2 2z z i

+ − −

= 1 3 1 3( 1)2 2 2 2

z z i z i

+ − + − −

h(z) = z3 − 1 h(1) = (1)3 − 1 = 0 So (z − 1) is a factor of h(z) Let h(z) = (z − 1)(z2 + rz + s) So z3 − 1 = (z − 1)(z2 + rz + s) z3 − 1 = z3 + rz2 + sz − z2 − rz − s z3 + 0z2 + 0z − 1 = z3 + (r − 1)z2 + (s − r)z − s Equating coefficients gives r − 1 = 0 −s = −1 r = 1 s = 1 So h(z) = (z − 1)(z2 + z +1)

= 2 1 1( 1) 14 4

z z z − + + − +

= 21 3( 1)

2 4z z

− + +

= 2

21 3( 1)2 4

z z i − + −

= 221 3( 1)

2 2z z i

− + −


= 1 3 1 3( 1)2 2 2 2

z z i z i

− + + + −

Since f(z) = g(z)h(z), we can now write

f(z) = 1 3 1 3( 1)2 2 2 2

z z i z i

+ − + − −

1 3 1 3( 1)2 2 2 2

z z i z i

− + + + −

= 1 3( 1)( 1)2 2

z z z i

+ − − +

1 3 1 3 1 32 2 2 2 2 2

z i z i z i

− − + + − −

The six factors of f(z) are z + 1, z − 1, 1 32 2

z i− + ,

1 32 2

z i− − , 1 32 2

z i+ + and 1 32 2

z i+ −

4 a 1 + i is a zero of P(z) = z3 + 4z2 − 10z + 12 Let z1 = 1 + i, then z2 = 1 − i is another zero Let z3 be the third zero. Then P(z) = (z − z1)(z − z2)(z − z3) = (z − 1 − i)(z − 1 + i)(z − z3) = [(z − 1)2 + 1](z − z3) = (z2 − 2z + 1 + 1)(z − z3) = (z2 − 2z + 2)(z − z3) so z3 + 4z2 − 10z + 12 = (z2 − 2z + 2)(z − z3) Comparing the left- and right-hand sides: 12 = 2 × −z3 z3 = −6 Therefore the two zeros required are 1 − i and −6 b −2 + i is a zero of P(z) = 2z3 + 9z2 +14z + 5 Let z1 = −2 + i, then z2 = −2 − i is another zero Let z3 be the third zero. Then P(z) = 2(z − z1)(z − z2)(z − z3) = 2(z + 2 − i)(z + 2 + i)(z − z3) = 2[(z + 2)2 + 1](z − z3) = 2(z2 + 4z + 4 + 1)(z − z3) = 2(z2 + 4z + 5)(z − z3) So 2z3 +9z2 + 14z + 5 = 2(z2 + 4z + 5)(z − z3) Comparing the left- and right-hand sides: 5 = 2 × 5 × − z3 z3 = 1

2−

Therefore the two zeros required are −2−i and 12

−

c 4 − i is a root of P(z) = z3 − 10z2 + 33z − 34 Let z1 = 4 − i, then z2 = 4 + i is another root let z3 be the third root, then P(z) = (z − z1)(z − z2)(z − z3) = (z − 4 + i) (z − 4 − i) (z − z3) = [(z − 4)2 + 1](z − z3) = (z2 − 8z + 16 + 1)(z − z3) = (z2 − 8z + 17)(z − z3) So z3 − 10z2 + 33z − 34 = (z2 − 8z + 17)(z − z3) Comparing the left- and right-hand sides: −34 = 17 × −z3 z3 = 2 Therefore the two roots required are 4 + i and 2

5 2z4 − 4z3 + 21z2 − 36z + 27 If z − 3i is a factor, then z +3i is a factor Let P(z) = 2z4 − 4z3 + 21z2 − 36z + 27, then P(z) has factors (z − 3i) and (z + 3i) Let P(z) = (z − 3i) (z + 3i)(2z2 + pz + q) = (z2 + 9)(2z2 + pz + q) = 2z4 + pz3 + qz2 + 18z2 + 9pz +9q = 2z4 + pz3 + (q + 18)z2 + 9pz +9q

So 2z4 − 4z3 + 21z2 − 36z + 27 = 2z4 + pz3 + (q + 18)z2 + 9pz + 9q Equating coefficients gives p = −4 9q = 27 q = 3 So P(z) = (z − 3i)(z + 3i)(2z2 − 4z + 3) = 2(z − 3i)(z + 3i)(z2 − 2z + 3

2)

= 2(z − 3i)(z + 3i)((z − 1)2 + 12

)

= 2(z − 3i)(z + 3i) ( )12 22

( 1)z i− −

= 2(z − 3i)(z + 3i) 1 11 12 2

z i z i − − − +

Therefore the remaining factors are z + 3i, 1 11 , 12 2

z i z i − − − +

and 2.

6 P(z) = z2 + (3 + 2i)z + 6i P(−2i) = (−2i)2 + (3 + 2i)(−2i) + 6i = 4i2 − 6i − 4i2 + 6i = 0, as required P(−3) = (−3)2 + (3 + 2i)(−3) + 6i = 9 − 9 − 6i + 6i = 0, as required So −2i and −3 are the zeros pf P(z) E 7 P(z) = z3 − (1 + 2i)z2 + 2(1 + i)z − 2 P(z) = (z − 1) Q(z), where Q(z) is a polynomial Let Q(z) = z2 + pz + q. Then P(z) = (z − 1)(z2 + pz + q) = z3 + pz2 + qz − z2 − pz − q = z3 + (p − 1)z2 + (q − p)z − q So z3 − (1 + 2i)z2 + 2(1 + i)z − 2

= z3 + (p − 1)z2 + (q − p)z − q Equating coefficients gives P − 1 = − (1 + 2i) −q = −2 P = 1 − 1 − 2i q = 2 P = −2i So Q(z) = z2 − 2iz + 2 Q(i) = (i)2 − 2i(i) + 2 = −1 + 2 + 2 = 3 D 8 P(z) is a polynomial of degree 4 so Let P(z) = (z − z1)(z − z2)(z − z3)(z − z4) ai and bi are roots, so let z1 = ai and z2 = bi If ai is a root so is −ai, and if bi is a root so is −bi, so let

z3 = −ai and z4 = −bi The term that does not contain z is given by z1 × z2 × z3 × z4 = ai × bi × −ai × −bi = a2 b2 i4 = a2b2 E 9 P(z) = z3 + 2z2 − 6z + a, P(1 − i) = 0 P(1 − i) = (1 − i)3 + 2(1 − i)2 − 6(1 − i) + a = (1)3 − 3(1)2(i) + 3(1)(i)2 − (i)3 + 2(1 − 2i + i2) − 6

+ 6i + a = 1 − 3i − 3 + i + 2 − 4i − 2 − 6 + 6i + a = −8 − 7i + 7i + a = a − 8 = 0 So a = 8 C 10 a Let P(z) = z3 + 3z2 + az + 8

If (z + 2) is a factor of P(z), P(−2) = 0 So P(−2) = (−2)3 + 3(−2)2 + a(−2) + 8 = 0 −8 + 12 − 2a + 8 = 0 −2a = −12 a = 6


b Let P(z) = z3 + az2 + z − 4 If (z + i) is a factor of P(z), P(−i) = 0 So P(−i) = (−i)3 + a(−i)2 + (−i) − 4 = 0 i − a − i − 4 = 0 −a = 4 a = −4 c Let P(z) = 2z3 + 3z2 +8z + a If (z + 1 − 2i) is a factor of P(z), P(−1 + 2i) = 0 So P(−1 + 2i) = 2(−1 + 2i)3 + 3(−1 + 2i)2 + 8(−1 + 2i) + a = 0 2[(−1)3 + 3(−1)2(2i) + 3(−1)(2i)2 + (2i)3] + 3(1 − 4i + 4i2) − 8 + 16i + a = 0 2(−1 + 6i + 12 − 8i) + 3(1 − 4i − 4) − 8 + 16i + a = 0 2(11 − 2i) + 3(−3 − 4i) − 8 + 16i + a = 0 22 − 4i − 9 − 12i − 8 + 16i + a = 0 5 + a = 0 a = −5 d Let P(z) = z3 − 2z2i + az − 32i If 2i is a root of P(z), P(2i) = 0 P(2i) = (2i)3 − 2(2i)2i + a(2i) − 32i = 0 −8i + 8i + 2ia − 32i = 0 2ia = 32i a = 16

11 a Let P(z) = z3 + az2 + 8z + b If −3 and 2 are roots, P(−3) = 0 and P(2) = 0 P(−3) = (−3)3 + a(−3)2 + 8(−3) + b = −27 + 9a − 24 + b = 9a + b − 51 = 0 So 9a + b = 51 [1] P(2) = (2)3 + a(2)2 + 8(2) + b = 8 + 4a + 16 + b = 4a + b + 24 = 0 So 4a + b = −24 [2] Subtracting [2] from [1] gives 9a − 4a = 51 + 24 5a = 75 a = 15 Substituting back into [1] gives 9(15) + b = 51 135 + b = 51 b = −84 b Let P(z) = z4 + az3 + bz2 − 7z + 12 If 4 and 1 are zeros of P(z), P(4) = 0 and P(1) = 0 P(4) = (4)4 + a(4)3 + b(4)2 − 7(4) + 12 = 256 + 64a + 16b − 28 + 12 = 64a + 16b + 240 = 0 So 64a + 16b = −240 4a + b = −15 [1] P(1) = (1)4 + a(1)3 + b(1)2 − 7(1) + 12 = 1 + a + b − 7 + 12 = a + b + 6 = 0 So a + b = −6 [2] Subtracting [2] from [1] gives 4a − a = −15 + 6 3a = −9 a = −3 Substituting back into [1] gives 4(−3) + b = −15 −12 + b = −15 b = −3 c Let P(z) = z3 + aiz2 + bz − 12i If 2i and 3i are roots of P(z), P(2i) = 0 and P(3i) = 0 − P(2i) = (2i)3 + ai(2i)2 + b(2i) − 12i = −8i − 4ia + 2ib − 12i = −4ia + 2ib − 20i = 0 So −4ia + 2ib = 20i 2a − b = −10 [1] P(3i) = (3i)3 + ai(3i)2 + b(3i) − 12i = −27i − 9ia + 3ib − 12i = −9ia + 3ib − 39i = 0


So −9ia + 3ib = 39i 3a − b = −13 [2] Subtracting [1] from [2] gives 3a − 2a = −13 + 10 a = −3 Substituting into [1] gives 2(−3) − b = −10 −6 − b = −10 b = 4

12 a Let P(z) = z3 − 2iz2 + aiz + b If (z + 1) is a factor of P(z), P(−1) = 0 P(−1) = (−1)3 − 2i(−1)2 + ai(−1) + b = −1 − 2i − ai + b = (b − 1) − (a + 2)i = 0 + 0i Therefore, b − 1 = 0 a + 2 = 0 b = 1 a = −2 b Let P(z) = az3 − 3z2 + biz + 12i If (z − i) is a factor, P(i) = 0 P(i) = a(i)3 − 3(i)2 + bi(i) + 12i = −ai + 3 − b + 12i = (3 − b) − (a − 12)i = 0 + 0i Therefore, 3 − b = 0 a − 12 = 0 b = 3 a = 12 c Let P(z) = z3 + aiz2 + 2iz + (1 + i)b If (z + 2i) is a factor, P(−2i) = 0 P(−2i) = (−2i)3 + ai(–2i)2 + 2i(–2i)+(1+i)b = 8i – 4ai + 4 + b + bi = (4 + b) + (8 + b − 4a) = 0 + 0i Therefore 4 + b = 0 8 + b − 4a = 0 b = −4 8 − 4 − 4a = 0 −4a = −4 a = 1

13 Since complex roots occur in conjugate pairs, an odd number of roots must have at least one real root.

14 Let P(z) = (z − z1)(z − z2)(z − z3) 2 and 4i are zeros, so let z1 = 2 and z2 = 4i. Since 4i is a zero, −4i must also be a zero, so let z3 = −4i So P(z) = (z − 2)(z − 4i)( z + 4i) = (z − 2)(z2 + 16) = z3 + 16z − 2z2 − 32 = z3 − 2z2 + 16z − 32 15 a P(z) = z3 + 3z2 + 36z + 108

ai is a solution to P(z), so P(ai) = 0 P(ai) = (ai)3 + 3(ai)2 + 36(ai) + 108 = −a3i − 3a2 + 36ai + 108 = (108 − 3a2) + (36a − a3)i = 0 + 0i So 108 − 3a2 = 0 36a − a3 = 0 a2 − 36 = 0 a(a2 − 36) = 0 a = ± 6 a = 0 a2 = 36 a2 = ± 36 a must be such that both real and imaginary parts equal

zero, so we discard the a = 0 solution. Therefore a = ± 6. b P(z) = z3 + 6iz2 − 11z − 6i If ai is a solution to P(z), P(ai) = 0 P(ai) = (ai)3 + 6i(ai)2 − 11ai − 6i = −a3i − 6a2i − 11ai − 6i = −(a3 + 6a2 + 11a + 6)i = 0i So a3 + 6a2 + 11a + 6 = 0 Let f(a) = a3 + 6a2 + 11a + 6 f(−1) = (−1)3 + 6(−1)2 + 11(−1) + 6 = −1 + 6 − 11 + 6 = 0 So a + 1 is a factor of f(a) Let f(a) = (a + 1)(a2 + pa + q) = a3 + pa2 + qa + a2 + pa + q = a3 + (p + 1)a2 + (p + q)a + q So a3 + 6a2 + 11a + 6 = a3 + (p + 1)a2 + (p + q)a + q Equating coefficients gives

p + 1 = 6 q = 6 p = 5 So f(a) = (a + 1)(a2 + 5a + 6) = (a + 1)(a + 2)(a + 3) = 0 = a = −1, a = −2, a = −3

16 Let P(z) = z3 + i P(i) = (i)3 + i = −i + i = 0 So (z − i) is a factor of P(z) Let P(z) = (z − i)[z2 + (a + bi)z + (c + di)], where a, b, c, d ∈ R P(z) = (z − i)[z2 + (a + bi)z + (c + di)] = z3 + (a + bi)z2 + (c + di)z − iz2

− i(a + bi)z − i(c + di) = z3 + (a + bi)z2 + (c + di)z − iz2 + (b − ai)z + (d − ci) = z3 + (a + bi − i)z2 + (c + di + b − ai)z + (d − ci) = z3 + [a + (b − 1)i]z2 + [(c + b) + (d − a)i]z + (d − ci) So z3 + 1 = z3 + [a + (b − 1)i]z2 + [(c + b) + (d − a)i]z

+ (d − ci) Equating coefficients gives a + (b − 1) = 0 + 0i d − ci = 0 + i so a = 0 b − 1 = 0 d = 0 −c = 1 b = 1 c = −1 Therefore P(z) = (z − i)[z2 + (0 + i)z + (−1 + ci)] = (z − i)(z2 + iz − 1) 17 a P(z) = z4 − (1 + 3i)z3 + 3(i − 1)z2 + (7 + i)z − 4 − i

Show P(1) = 0 P(1) = (1)4 − (1 + 3i)(1)3 + 3(i − 1)(1)2 + (7 + i)(1)

− 4 − i = 1 − 1 + 3i + 3i − 3 + 7 + i − 4 − i = 0 + 0i = 0, as required b P(z) = (z − 1) Q(z) Let Q(z) = [z3 + (a + bi)z2 + (c + di)z + (e + fi)], where a, b, c, d, e, f ∈ R So P(z) = (z − 1)[z3 + (a + bi)z2 + (c + di)z + (e + fi)] = z4 + (a + bi)z3 + (c + di)z2 + (e + fi)z − z3

− (a + bi)z2 − (c + di)z − (e + fi) = z4 + (a + bi − 1)z3 + [(c + di) − (a + bi)]z2

+ [(e + fi) − (c + di)]z − (e + fi) = z4 + [(a − 1) + bi]z3 + [(c − a) + (d − b)i]z2

+ [(e − c) + (f − d)i]z − (e + fi) So z4 − (1 + 3i)z3 + 3(i − 1)z2 + (7 + i)z − 4 − i = z4 + [(a − i) + bi]z3 + [(c − a) + (d − b)i]z2

+ [(e − c) + (f − d)i] − (e + fi) Equation coefficients gives −(1 + 3i) = (a − 1) + bi −1 − 3i = (a − 1) + bi So a − 1 = −1 b = −3 a = 0 3(i − 1) = (c − a) + (d − b)i −3 + 3i = (c − a) + (d − b)i So c − a = −3 d − b = 3 c − 0 = −3 d + 3 = 3 c = −3 d = 0 −4 − i = −(e + fi) 4 + i = e + fi So e = 4 f = 1 Therefore Q(z) = z3 + (a + bi)z2 + (c + di)z + (e + fi) = z3 + (0 − 3i)z2 + (−3 + 0i)z + (4 + i) = z3 − 3z2i − 3z + 4 + i c Q(z) = (z − a)3 + b, where a ∈ c and b ∈ R Let a = c + di, where c, d ∈ R Q(z) = (z − a)3 + b = [z − (c + di)]3 + b = z3 + 3z2(c + di) + 3z(c + di)2 − (c + di)3 + b = z3 − 3(c + di)z2 + 3(c2 + 2cdi − d 2)z

− [c3 + 3c2di + 3c(di)2 + (di) 3] + b


= z3 − 3(c + di)z2 + [(3c2 − 3d 2) + 6cdi]z − (c3 + 3c2di − 3cd 2 − d 3i) + b

= z3 − 3(c + di)z2 + [3(c2 − d 2) + 6cdi]z − [(c3 − 3cd 2 − b) + (3c2d − d 3)i] So z3 − 3iz2 − 3z + 4 + i = z3 − 3(c + di)z2 + [3(c2 − d 2) + 6cdi]z

− [(c3 − 3cd 2 − b) + (3c2d − d 3)i] Equating coefficients gives −3i = −3(c + di) 4 + i = −[(c3 − 3cd 2 − b)

− (3c2d − d 3)i] i = c + di So 4 = −(c3 − 3cd 2 − b) So c = 0 d = 1 4 = −(03 − 3(0)(1)2 − b) So a = 0 + i 4 = b a = i b = 4

18 Let P(z) = z4 + 2z3 + 8z2 + 10z + 15 Since ( 5 )z i+ is a factor of P(z), ( 5 )z i− must also be a

factor since P(z) has real coefficients Let P(z) = 2( 5 )( 5 )( )z i z i z pz q+ − + + = (z2 + 5)(z2 + pz + q) = z4 + pz3 + qz2 + 5z2 + 5pz + 5q = z4 + pz3 + (q + 5)z2 + 5pz + 5q So z4 + 2z3 + 8z2 + 10z + 15 = z4 + pz3 + (q + 5)z2

+ 5pz + 5q Equating coefficients gives p = 2 5q = 15 q = 3 So P(z) = 2( 5 )( 5 )( 2 3)z i z i z p+ − + +

= 2( 5 )( 5 )[( 2 1) 1 3]z i z i z p+ − + + − +

= 2( 5 )( 5 )[( 1) 2]z i z i z+ − + +

= 2 2( 5 )( 5 )[( 1) 2 ]z i z i z i+ − + −

= 2( 5 )( 5 )[( 1) ( 2 ) ]z i z i z i+ − + −

= ( 5 )( 5 )( 1 2 )( 1 2 )z i z i z i z i+ − + + + − 19 P(z) = 9z3 + (9i − 12)z2 + (5 − 12i)z + 5i If P(−i) = 0, (z + i) is a factor of P(z) Let P(z) = (z + i)[9z2 + (a + bi)z + (c + di)], where a, b, c, d ∈ R P(z) = (z + i)[9z2 + (a + bi)z + (c + di)] = 9z3 + (a + bi)z2 + (c + di)z + 9iz2 + i(a + bi)z

+ i(c + di) = 9z3 + (a + bi)z2 + (c +di)z + 9iz2 + (−b + ai)z

+ (−d + ci) = 9z3 + [a + (b + 9)i]z2 + [(c−b) + (d + a)i]z

+ (−d + ci) So 9z3 + (9i − 12)z2 + (5 − 12i)z + 5i = 9z3 + (a + (b + a)i)z2 + [(c−b) + (d + a)i]z

+ (−d + ci) Equating coefficients gives a + (b + a)i = 9i − 12 −d + ci = 5i a + (b + a)i = − 12 +9i −d + ci = 0 + 5i So a = −12 b + 9 = 9 So −d = 0 c = 5 b = 0 d = 0 Therefore p(z) = (z +i)[9z2 + (−12 + 0i)z + (5 + 0i)] = (z + i)(9z2 − 12z + 5)

= (z + i) 2 4 49 4 53 9

z z − + − +

= 2

2 2( ) 3 13

z i z + − +

= 2 2( ) (3 2)z i z i + − −

= (z + i)(3z − 2 + i)(3z − 2 − i)

20 a + z2 = 211a

z−

az2 + z4 = a − 11

z4 + az2 + 11 − a = 0 Let P(z) = z4 + az2 + 11 − a We want to solve for a such that P(z) = 0. If 3i− is a zero of P(z), then P ( 3 )i− = 0.

So P ( 3 )i− = 4 2( 3 ) ( 3 ) 11i a i a− + − + − = 0 9 − 3a + 11 − a = 0 −4a + 20 = 0 a = 5

Investigation — The exact values of

2cos5π

and

2sin5π

1 z = x +yi, z5 = 1 z5 = (x + yi)5 = x5 + 5x4(yi) + 10x3(yi)2 + 10x2 + (yi)3 + 5x(yi)4 + (yi)5 = x5 + 5x4yi − 10x3y2 − 10x2y3i + 5xy4 + y5i So x5 + 5x4yi − 10x3y2 − 10x2y3i + 5xy4 + y5i = 1 2 x5 + 5x4yi − 10x3y2 − 10x2y3i + 5xy4 + y5i = 1 (x5 − 10x3y2 + 5xy4) + (5x4y − 10x2y3 + y5)i = 1 Equating the real parts gives x5 − 10x3y2 + 5xy4 = 1

3 |z| = 2 2x y+ z5 = 1

So z = 15 21 cis

5kπ

= 21cis5kπ

= | z | cis(θ)

Therefore |z| = 1

So 2 2x y+ = 1 x2 + y2 = 1 y2 = 1 − x2 4 x5 − 10x3y2 + 5xy4 = 1 x5 − 10x3y2 + 5x(y2)2 = 1 x5 − 10x3(1 − x2) + 5x(1 − x2)2 = 1 x5 − 10x3 + 10x5 + 5x(1 − 2x2 + x4) = 1 x5 − 10x3 + 10x5 + 5x − 10x3 + 5x5 = 1 6x5 − 20x3 + 5x = 1 6x5 − 20x3 + 5x − 1 = 0

5 The two exact positive solutions for x are 1 and 5 14−

6 z = 2cis5kπ

, k ∈ N

x + yi = 2 2cos sin5 5k kiπ π +

k = 1

x + yi = 2 2cos sin5 5

iπ π +

So x = 2cos5π

= 5 1,4− from part 5

We reject the x = 1 solution because this would be when k = 0, giving x = cos(0) = 1.

7 Pythagoras’ theorem gives

2 22 2cos sin5 5π π +

= 1

2

25 1 2sin4 5

π − + = 1

25 2 5 1 2sin16 5

π− + +

= 1


26 2 5 2sin16 5

π− +

= 1

2 2sin5π

= 16 6 2 516

− +

2 2sin5π

= 10 2 516+

2sin5π

= 10 2 54+

2sin5π

= 2 5 54

+

The positive square root was taken because the angle 25π

is in the first quadrant, so 2sin5π

must be positive.

Exercise 1C — Solving equations in C 1 a x2 + 2x + 5 = 0

(x2 + 2x + 1) −1 + 5 = 0 (x + 1)2 + 4 = 0 (x + 1)2 = −4 x + 1 = 4± − x = −1 ± 2i b x2 − 8x + 25 = 0 (x2 − 8x + 16) − 16 + 25 = 0 (x − 4)2 + 9 = 0 (x − 4)2 = −9 x − 4 = 9± − x = 4 ± 3i c x2 − 14x + 149 = 0 (x2 − 14x + 49) − 49 + 149 = 0 (x − 7)2 + 100 = 0 (x − 7)2 = −100 x − 7 = 100± − x = 7 ± 10i d 4x2 − 12x + 13 = 0

4 2 934

x x − +

− 9 + 13 = 0

423

2x −

+ 4 = 0

234

2x −

= −4

23

2x −

= −1

x − 32

= 1± −

x = 32

± i

e 4x2 − 32x + 4 = 0 4x2 − 16 2x× + 4 = 0 4x2 − 4 2x + 4 = 0 x2 − 2x + 1 = 0

2 122

x x − +

− 12

+ 1 = 0

2

22

x

−

+ 1 = 0

2

22

x

−

= 12

−

22

x − = 12

−±

x = 2 12 2

i±

x = 2 22 2

i±

2 a z3 − z2 − z + 10 = 0 Let P(z) = z3 − z2 − z + 10 = 0 P(−2) = (−2)3 − (−2)2 − (−2) + 10 = −8 − 4 + 2 + 10 = 0 So z = −2 is a solution So (z + 2) is a factor Let P(z) = (z + 2)(z2 + pz + q) = z3 + pz2 + q2 + 2z2 + 2pz + 2q = z3 + (p + 2)z2 + (2p + q)z + 2q So z3 − z2 − z + 10 = z3 + (p + 2)z2 + (2p + q) + 2q Equating coefficients gives p + 2 = −1 2q = 10 p = −3 q = 5 So P(z) = (z + 2)(z2 − 3z + 5) = 0 So z + 2 = 0 z2 − 3z + 5 = 0 z = −2 a = 1 b = −3 c = 5

z = 2( 3) ( 3) 4 1 5

2 1− − ± − − × ×

×

z = 3 9 202

± −

z = 3 112

± −

z = 3 112 2

i±

b z3 − 2z2 + 3z − 2 = 0 Let P(z) = z3 − 2z2 + 3z − 2 = 0 P(1) = (1)3 − 2(1)2 + 3(1) − 2 = 1 − 2 + 3 − 2 = 0 So z = 1 is a solution So (z − 1) is a factor Let P(z) = (z − 1)(z2 + pz + q) = z3 + pz2 + qz − z2 − pz − q = z3 + (p − 1)z2 + (q − p)z − q So z3 − 2z2 + 3z − 2 = z3 + (p − 1)z2 + (q − p)z − q Equating coefficients gives p − 1 = − 2 −q = −2 p = −1 q = 2 So P(z) = (z − 1)(z2 − z + 2) = 0 So z − 1 = 0 z2 − z + 2 = 0 z = 1 a = 1 b = −1 c = 2

z = 2( 1) ( 1) 4 1 2

2 1− − ± − − × ×

×

z = 1 1 82

± −

z = 1 72

± −

z = 1 72 2

i±

c 2z3 − 7z2 + 10z − 8 = 0 Let P(z) = 2z3 − 7z2 + 10z − 8 = 0 P(2) = 2(2)3 − 7(2)2 + 10(2) − 8 = 16 − 28 + 20 − 8 = 0 So z = 2 is a solution So (z − 2) is a factor Let P(z) = (z − 2)(2z2 + pz + q) = 2z3 + pz2 + qz − 4z2 − 2pz − 2q = 2z3 + (p − 4)z2 + (q − 2p)z − 2q


So 2z3 − 7z2 + 10z − 8 = 2z3 + (p − 4)z2 + (q − 2p)z − 2q

Equating coefficients gives p − 4 = −7 −2q = −8 p = −3 q = 4 So P(z) = (z − 2)(2z2 − 3z + 4) = 0 So z − 2 = 0 2z2 −3z + 4 = 0 z = 2 a = 2, b = −3, c = 4

z = 2( 3) ( 3) 4 2 4

2 2− − ± − − × ×

×

z = 3 9 324

± −

z = 3 234

± −

z = 3 234 4

i±

d 3z3 − 13z2 + 5z − 4 = 0 Let P(z) = 3z3 − 13z2 + 5z − 4 = 0 P(4) = 3(4)3 − 13(4)2 + 5(4) − 4 = 192 − 208 + 20 − 4 = 0 So z = 4 is a solution So (z − 4) is a factor Let P(z) = (z − 4)(3z2 + pz + q) = 3z3 + pz2 + qz − 12z2 − 4p2 − 4q = 3z3 + (p − 12)z2 + (q − 4p)z − 4q So 3z3 − 13z2 + 5z − 4 = 3z3 + (p − 12)z2 + (q − 4p)z − 4q Equating coefficients gives p − 12 = −13 −4q = −4 p = −1 q = 1 So P(z) = (z − 4)(3z2 − z + 1) = 0 So z − 4 = 0 3z2 − z + 1 = 0 z = 4 a = 3, b = −1, c = 1

z = 2( 1) ( 1) 4 3 1

2 3− − ± − − × ×

×

z = 1 1 126

± −

z = 1 116

± −

z = 1 116 6

i±

e 4z3 − 20z2 + 34z − 20 = 0 Let P(z) = 4z3 − 20z2 + 34z − 20 = 0 P(2) = 4(2)3 − 20(2)2 + 34(2) − 20 = 32 − 80 + 68 − 20 = 0 So z = 2 is a solution So (z − 2) is a factor Let P(z) = (z − 2)(4z2 + pz + q) = 4z3 + pz2 + qz − 8z2 − 2pz − 2q = 4z3 + (p − 8)z2 + (q− 2p)z − 2q So 4z3 − 20z2 + 34z − 20 = 4z3 + (p − 8)z2 + (q − 2p)z − 2q Equating coefficients gives p − 8 = −20 −2q = −20 p = −12 q = 10 So P(z) = (z − 2)(4z2 − 12z + 10)

= (z − 2) 2 94 3 9 104

z z − + − +

= (z − 2)2

2 32 12

z − +

= (z − 2)[(2z − 3)2 − i2] = (z − 2)(2z − 3 + i)(2z − 3 − i) = 0 So z − 2 = 0 2z − 3 + i = 0 2z − 3 − i = 0 z = 2 2z = 3 − i 2z = 3 + i

z = 3 12 2

i− z = 3 12 2

i+

3 f(z) = z − 4, g(z) = z2 − z + 1, h(z) = z3 − 5z2 + 5z − 4 Show that f(z) × g(z) = h(z) f(z) × g(z) = (z − 4)(z2 − z + 1) = z3 − z2 + z − 4z2 + 4z − 4 = z3 − 5z2 + 5z − 4 = h(z), as required h(z) = 0 So f(z) × g(z) = 0 (z − 4)(z2 − z + 1) = 0 So z − 4 = 0 z2 − z + 1 = 0 z = 4 a = 1 b = −1 c = 1

z = 2( 1) ( 1) 4 1 1

2 1− − ± − − × ×

×

z = 1 1 42

± −

z = 1 32

± −

z = 1 32 2

i±

4 a x4 + 25x2 + 144 = 0 y2 + 25y + 144 = 0, with y = x2 (y + 16)(y + 9) = 0 (x2 + 16)(x2 + 9) = 0 x2 + 16 = 0 x2 + 9 = 0 x2 = −16 x = −9 x = 16± − x = 9± − x = ±4i x = ± 3i b z4 − 3z2 − 4 = 0 y2 − 3y − 4 = 0, with y = z2 (y − 4)(y + 1) = 0 (z2 − 4)(z2 + 1) = 0 z2 − 4 = 0 z2 + 1 = 0 z2 = 4 z2 = −1 z = 4± z = 1± − z = ± 2 z = ± i c 9z4 + 35z2 − 4 = 0 9y2 + 35y − 4 = 0, with y = z2 (9y − 1)(y + 4) = 0 (9z2 − 1)(z2 + 4) = 0 9z2 − 1 = 0 z2 + 4 = 0

z2 = 19

z2 = −4

z = 19

± z = 4± −

z = 13

± z = ± 2i

d 4x4 + 12x2 + 9 = 0 4y2 + 12y + 9 = 0, with y = x2 (2y + 3)(2y + 3) = 0 (2y + 3)2 = 0 (2x2 + 3)2 = 0 2x2 + 3 = 0

x2 = 32

−

x = 32

−±


x = 3 22 2i± ×

x = 62

i±

5 (z − 3)2 + 4 = 0 (z − 3)2 = −4 z − 3 = 4± − z − 3 = ± 2i z = 3 ± 2i ∴ B 6 a 1 + 3i

Let z2 = 1 + 3i . Take a = 1 and b = 3

x2 = 2 21 1 ( 3)

2+ +

x2 = 1 42

+

x2 = 1 22+

x2 = 32

x = 32

±

x = 3 22 2

± ×

x = 62

±

y2 = 3 12

−

y2 = 12

y = 12

±

y = 1 22 2

± ×

y = 22

±

Therefore the two roots z1, z2 are z1 = 6 22 2

i+ and

z2 = 6 22 2

i− −

So 1 3i+ is 1 ( 6 2 )2

i+ or 1 ( 6 2 )2

i− +

b 11 + 60i Let z2 = 11 + 60i. Take a = 11 and b = 60

x2 = 2 211 11 60

2+ +

x2 = 11 37212

+

x2 = 11 612+

x2 = 722

x2 = 36 x = ± 6 y2 = 36 − 11 y2 = 25

y = ± 5 Therefore the two roots z1, z2 are z1 = 6 + 5i and

z2 = −6 − 5i So 11 60i+ is 6 + 5i or −6 − 5i c 16 + 63i Let z2 = 16 + 63i. Take a = 16 and b = 63

x2 = 2 216 16 63

2+ +

x2 = 16 42252

+

x2 = 16 652+

x2 = 812

x = 812

±

x = 9 22 2

± ×

x = 9 22

±

y2 = 81 162

−

y2 = 81 322−

y2 = 492

y = 492

±

y = 7 22 2

± ×

y = 7 22

±

Therefore the two roots z1 and z2 are z1 = 9 2 7 22 2

i+

and z2 = 9 2 7 22 2

i− −

So 16 63i+ is 2 (9 7 )2

i+ or 2 (9 7 )2

i− +

7 Let z = i z2 = i. Take a = 0 b = 1

x2 = 20 0 1

2+ +

x2 = 12

x = 12

±

x = 1 22 2

± ×

x = 22

±

y2 = 1 02

−

y2 = 12

y = ± 12


y = 1 222

± ×

y = 22

±

Therefore the two roots z1 and z2 are z1 = 2 22 2

i+ and

z2 = 2 22 2

i− −

So i is 2 (1 )2

i+ and 2 (1 )2

i− +

8 One root of a (1 + i) is cis8

a π

Let z1 = cis8

a π

and z2 be the other root

The root z2 is obtained by rotating z1 through an angle of π c

So z2 = cis8

a π π −

= 7cis8

a π−

∴ C

(Note that we subtract π c rather than adding it to give an angle such that −π < Arg(z) ≤ π)

9 a z2 = 3 i− If w = 3 i− = r cis(θ)

r = 2 2( 3) ( 1)+ −

= 4 = 2

tan(θ) = 13

−

= 1 1tan3

− −

(4th quadrant)

= 6π−

So w = 2cis 26

kπ π− +

Let z2 = 2cis 26

kπ π− +

z =

12

2cis 26

kπ π − +

= 2 cis12

kπ π − +

k = 0 z = 2 cis12π−

k = 1 z = 2 cis12π π− +

= 112 cis12

π

The two square roots of 3 i− in polar form are

2 cis12π−

and 112 cis12

π

b z2 = 4 + 4i If w = 4 + 4i = r cis(θ)

r = 2 24 4+ = 32

tan(θ) = 44

tan θ = 1 θ = tan−1(1) (1st quadrant)

= 4π

So w = 32 cis 24

kπ π +

Let z2 = 32 cis 24

kπ π +

z =

12

32 cis 24

kπ π +

= 14(32) cis

8kπ π +

= 542 cis

8kπ π +

k = 0 z = 542 cis

8π

k = 1 z = 542 cis

8π π +

= 54 92 cis

8π

= 54 92 cis 2

8π π −

= 54 72 cis

8π−

The two square roots of 4 + 4i in polar form are 542 cis

8π

and 54 72 cis

8π−

c z3 = 4 4 3i− + If w = 4 4 3i− + = r cis(θ)

r = 2 2( 4) (4 3)− +

= 16 48+ = 64 = 8

tan(θ) = 4 34−

tan(θ) = 3− θ = 1tan ( 3)π −+ − (2nd quadrant)

= 3ππ −

= 23π

So w = 28cis 23

kπ π +

Let z3 = 28cis 23

kπ π +

z =

1328cis 2

3kπ π +

= 2 22cis9 3

kπ π +

k = 0 z = 22cis9π

k = 1 z = 2 22cis9 3π π +


= 2 62 cis9

π π+

= 82cis9π

k = 2 z = 2 42 cis9 3π π +

= 2 122 cis9

π π+

= 142cis9π

= 142 cis 29π π −

= 42 cis9π−

The three cube roots of 4 4 3i− + are 22 cis9π

, 82 cis9π

and 42 cis9π−

d z3 = i If w = i = r cis(θ)

r = 2 20 1+ = 1 sin(θ) = 1, cos(θ) = 0

θ = 2π

So w = cis 22

kπ π +

Let z3 = cis 22

kπ π +

z =

13

cis 22

kπ π +

= 2cis6 3

kπ π +

k = 0 z = cis6π

k = 1 z = 2cis6 3π π +

= 4cis6

π π+

= 5cis6π

k = 2 z = 4cis6 3π π +

= 8cis6

π π+

= 9cis6π

= 3cis2π

= 3cis 22π π −

= cis2π−

e z3 = −1 −i If w = −1 −i = r cis(θ)

r = 2 2( 1) ( 1)− + −

= 2

tan(θ) = 11

−−

tan(θ) = 1 θ = −π + tan−1(1) (3rd quadrant)

= 4ππ− +

= 34π−

So w = 32 cis 24

kπ π− +

Let z3 = 32 cis 24

kπ π− +

z =

1332 cis 2

4kπ π − +

= 16 22 cis

4 3kπ π− +

k = 0 z = 6 2 cis4π−

k = 1 z = 16 22 cis

4 3π π− +

= 16 3 82 cis

12π π− +

= 6 52 cis12π

k = 2 z = 16 42 cis

4 3π π− +

= 16 3 162 cis

12π π− +

= 16 132 cis

12π

= 16 132 cis 2

12π π −

= 6 112 cis12

π−

The three cube roots of −1 − i are 6 2 cis4π−

,

6 52 cis12π

and 6 112 cis12

π−

10 Let z = 13( 125 )i−

So z3 = −125i Let z3 = w w = −125i = r cis(θ)

r = 2 20 ( 125)+ − = 125

sin(θ) = 125 ,125

− cos(θ) = 0

= 1

θ = 2π−

So w = 125cis 22

kπ π− +


So z3 = 125cis 22

kπ π− +

z = 3

125cis 22

kπ π1

− +

= 25cis6 3

kπ π− +

k = 0 z = 5cis6π−

= 5 cos sin6 6

iπ π − − +

= 3 152 2

i

−

= 5 3 52 2

i−

k = 1 z = 25 cis6 3π π− +

= 35cis6π

= 5cis2π

= 5 cos sin2 2

iπ π +

= 5i

k = 2 z = 45 cis6 3π π − +

= 75cis6π

= 7 75 cos sin6 6

iπ π +

= 5 cos sin6 6

iπ π − −

= 3 152 2

i

− −

= 5 3 52 2

i− −

Sum = 5 3 5 5 3 552 2 2 2

i i i− + − −

= 5 32

5 5 352 2

i i− + − 52

i−

= −5i + 5i = 0

So 13( 125 )i− is 5i, 5 3 5

2 2i− or 5 3 5

2 2i− − and they sum

to zero. 11 a Let z3 = 64

If w = 64 = r cis(θ) r = 64 sin(θ) = 0 cos(θ) = 1 θ = 0 So w = 64 cis(2kπ) Let z3 = 64 cis(2kπ)

z = [ ]1364cis(2 )kπ

= 24cis3kπ

k = 0 z = 4 cis(0) = 4 (cos(0) + i sin(0)) = 4

k = 1 z = 24cis3π

= 2 24 cos sin3 3

iπ π +

= 4 cos sin3 3

iπ π − +

= 1 342 2

i

− +

= 2 2 3i− +

k = 2 z = 44cis3π

= 4 44 cos sin3 3

iπ π +

= 4 cos sin3 3

iπ π − −

= 1 342 2

i

− −

= 2 2 3i− − The three cube roots of 64 are 4, 2 2 3i− + and

2 2 3i− − b

12 a z4 = 16 z4 = 16 cis(2kπ)

z = 14(16cis(2 ))kπ

= 14 216 cis

4kπ

= 2cis2

kπ

k = 0 z = 2 cis(0) = 2

k = 1 z = 2 cis2π

= 2i k = 2 z = 2 cis(π) = −2

k = 3 z = 32 cis2π

= −2i The four solutions in Cartesian form are 2, 2i, −2 and −2i b z4 = 25 z4 = 25 cis(2kπ)

z = 14(25cis (2 ))kπ

= 14 225 cis

4kπ

= 5 cis2

kπ


k = 0 z = 5 cis(0)

= 5

k = 1 z = 5 cis2π

= 5i k =2 z = 5 cis( )π

= 5−

k = 3 z = 35 cis2π

= 5i− The four solutions in Cartesian form are 5, 5 ,i

5− and 5i− c z6 = 64 z6 = 64 cis(2kπ)

z = 16(64cis(2 ))kπ

= 16 264 cis

6kπ

= 2cis3

kπ

k = 0 z = 2 cis(0) = 2

k = 1 z = 2cis3π

z = 2 cos sin3 3

iπ π +

= 1 322 2

i

+

= 1 + 3i

k = 2 z = 22cis3π

= 2 22 cos sin3 3

iπ π +

= 2 cos sin3 3

iπ π − +

= 1 322 2

i

− +

= −1 + 3i k = 3 z = 2 cis(π) = −2

k = 4 z = 42cis3π

= 4 42 cos sin3 3

iπ π +

= 2 cos sin3 3

iπ π − −

= 1 322 2

i

− −

= 1 3i− −

k = 5 z = 52cis3π

= 5 52 cos sin3 3

iπ π +

= 2 cos sin3 3

iπ π −

= 1 322 2

i

−

= 1 3i− The six solutions in Cartesian form are 2, 1 3 ,i+

1 3 ,i− + −2, 1 3i− − and 1 3 .i− d z6 = 27 z6 = 27 cis(2kπ)

z = 16(27cis(2 ))kπ

= 16 227 cis

6kπ

= 3 cis3

kπ

k = 0 z = 3 cis(0)

= 3

k = 1 z = 3 cis3π

= 3 cos sin3 3

iπ π +

= 1 332 2

i

+

= 3 32 2

i+

k = 2 z = 23 cis3π

z = 2 23 cos sin3 3

iπ π +

= 3 cos sin3 3

iπ π − +

= 1 332 2

i − +

= 3 32 2

i− +

k = 3 z = 3 cis ( )π

= 3−

k = 4 z = 43 cis3π

= 4 43 cos sin3 3

iπ π +

= 3 cos sin3 3

iπ π − −

= 1 332 2

i

− −

= 3 32 2

i− −

k = 5 z = 53 cis3π

= 5 53 cos sin3 3

iπ π +


= 3 cos sin3 3

iπ π −

= 1 332 2

i

−

= 3 32 2

i−

The six solutions in Cartesian form are 3, 3 3 ,2 2

i+ 3 3 ,2 2

i− + 3,− 3 32 2

i− − and 3 32 2

i−

13 z5 = 1 z5 = cis(2kπ)

z = 15(cis (2 ))kπ

= 2cis5kπ

k = 0 z = cis(0)

k = 1 z = 2cis5π

k = 2 z = 4cis5π

k = 3 z = 6cis5π

= 6cis 25π π −

= 4cis5π−

k = 4 z = 8cis5π

= 8cis 25π π −

= 2cis5π−

The five solutions in polar form are cis(0), 2cis5π

, 4cis5π

, 4cis5π−

and 2cis5π−

Investigation — Current and voltage

1 V = 1Re i tci L R I e

i Cωω

ω + +

I = 5 cos(130t) A = I0 cos(ω t) A So I0 = s and ω = 130 R = 80 Ω L = 0.15 H C = 3 × 10−6 F

V = 1306

1Re 130 0.15 80 5130 3 10

i ti ei

×−

× × + + × × × × ×

= 1304

1Re 19.5 80 5(3.9 10 )

iti ei−

+ + ×

×

= 41Re 19.5 80 5(cos(130 ) sin(130 ))

(3.9 10 )ii t i tii−

+ + × × +

×

= 4

1Re 80 19.5 5(cos(130 ) sin(130 ))(3.9 10 )

i t i t−

+ − × + ×

= 24

180 (5cos(130 )) 19.5 sin(130 )(3.9 10 )

t i t−

× + − ×

×

= 4

1400 cos(130 ) 19.5 sin(130 )(3.9 10 )

t t−

− − ×

×

≈ 400 cos(130t) + 12720 sin(130t)


Exercise 1D — Graphs of complex relations: rays, lines and circles 1 a

b

c

d

e

f

2 a

b

c

d

e

f

g

h

i

j

3 a C

b E 4 B 5 D

6 a

b

c

d

e

f

g

h

7 a

b

c

d

e

f

g

h

8 a

b


c

d

e

f

g

h

9 a

b

c

d

e

f

10 B 11 D 12 a

b

c

d

e

f

g

h

Chapter review 1 a 2u

= 2(4 − 3i) = 8 − 6i

b u + v = (4 − 3i) + (2 + 5i) = 6 + 2i c u × v = (4 − 3i) × (2 + 5i) = 8 + 20i − 6i − 15i2 = 23 + 14i

d uv

= 4 3 2 52 5 2 5

i ii i

− −×+ −

= 2

28 20 6 15

4 25i i i

i− − +

−

= 7 269

i− −

= 7 269 9

i− −

e 3u + 2v = 3(4 − 3i) + 2(2+5i) = 12 − 9i + 4 + 10i = 16 + i f |3u + 2v| = |16 + i|, from part (e)

= 2 216 1+ = 256 1+ = 257 ≈ 16.03

2 a |z|

= 2 23 4+ = 25 = 5 b |w| = 2 2( 1) 2− +

= 5 c z = 3 − 4i d |z + w| = |(3 + 4i) + (−1 + 2i)| = |2 + 6i|

= 2 22 6+ = 4 36+ = 40 = 2 10 ≈ 6.32 e zw = (3 + 4i)(−1 − 2i) = −3 − 6i − 4i − 8i2 = 5 − 10i f w−1 = (−1 + 2i)−1

= 1 1 21 2 1 2

ii i

− −×− + − −

= 21 2

1 4i

i− −

−

= 1 25

i− −

= 1 25 5

i− −

3 z = 2 + 2i

r = 2 22 2+


= 8 = 2 2

tan(θ) = 22

tan(θ) = 1 θ = tan−1(1) (1st quadrant)

= 4π

z = 2 2 cis4π

4 3cis6π

= 3 cos sin6 6π π +

= 3 132 2

i

+

= 3 3 32 2

i+

5 a u × v

= 33cis 2cis4 3π π

= 33 2cis4 3π π × +

= 9 46cis12

π π+

= 136cis12

π

= 136cis 212

π π −

= 116cis12

π−

b uv

=

33cis4

2cis3

π

π

= 3 3cis2 4 3

π π −

= 3 9 4cis2 12

π π−

= 3 5cis2 12

π

c v3

= 3

2cis3π

= 32 cis 33π ×

= 8 cis(π)

d u

= 33cis4π

=

1233cis

4π

= 12 1 33 cis

2 4π ×

= 33 cis8π

6 a Let z = 1 32 2

i+ and w = 6

1 32 2

i

+

z = 1 32 2

i+

r = 221 3

2 2 +

= 1 34+

= 1 = 1

tan(θ) =

3212

= 3 θ = 1tan ( 3)− (1st quadrant)

θ = 3π

z = cis3π

So w = z6

= 6

cis3π

= 61 cis 63π ×

= cis(2π) = cos(2π)+ isin(2π) = 1

b Let z = 3 − 3i and w =

13 3i−

z = 3 − 3i

r = 2 23 3+ = 18

tan(θ) = 33

−

tan(θ) = –1 θ = tan−1(−1) (4th quadrant)

= 4π−

z = 18 cis4π−

So w = 1z

= 12z

−

=

12

18cis4π −

−

= 14 118 cis

2 4π− − − ×

= 41 cis

818π


= 4 41 cos sin

8 818 18iπ π +

≈ 0.45 + 0.19i 7 x2 + 6x + 1 = (x2 + 6x + 9) −9 + 1 = (x + 3)2 − 8 = 2 2( 3) ( 8)x + −

= (x + 3 + 8) (x + 3 − 8)

= ( 3 2 2)+ +x ( 3 2 2)+ −x ∴ A 8 Let P(z) = z8 − 3z2 + 4z − 12 If 2i is a zero, then −2i must also be a zero. From the multiple choice the other zero must be either 3 or

−3, so we test them P(3) = (3)3 − 3(3)2 + 4(3) − 12 = 27 − 27 + 12 − 12 = 0 P(−3) = (−3)3 − 3(−3)2 + 4(−3) − 12 = −27 − 27 − 12 − 12

= −78 ≠ 0 So the third zero must be 3. ∴ A 9 a P(z) = z2 − 16z + 89

= (z2 − 16z + 64) − 64 + 89 = (z − 8)2 + 25 = (z − 8)2 − 25i2 = (z − 8)2 − (5i)2 = (z −8 + 5i)(z − 8 − 5i) b P(z) = 2z3 + 6z2 − 7z + 4 P(−4) = 2(−4)3 + 6(−4)2 − 7(−4) + 4 = −128 + 96 + 28 + 4 = 0 So (z + 4) is a factor of p(z) Let P(z) = (z + 4)(2z2 + pz + q) = 2z3 + pz2 + qz + 8z2 + 4pz + 4q = 2z3 + (p + 8)z2 + (q + 4p)z + 4q So 2z2 + 6z2 − 72 + 4 = 2z3 + (p + 8)z2 + (q + 4p)z + 4q Equating coefficients gives p + 8 = 6 4q = 4 p = −2 q = 1 So P(z) = (z + 4)(2z2 − 2z + 1)

= 2(z + 4)(z2 − z + 12

)

= 2(z + 4) 2 1 1 14 4 2

z z − + − +

= 2(z + 4)21 1

2 4z

− +

= 2(z + 4)2

21 12 4

z i − −

= 2(z + 4)2 21 1

2 2z i

− −

= 2(z + 4) 1 1 1 12 2 2 2

z i z i − + − −

10 P(z) = z4 + z3 + az2 + 9z − 9, P(3i) = 0 P(3i) = (3i)4 + (3i)3 + a(3i)2 + 9(3i) − 9 = 81 − 27i − 9a + 27i − 9 = 81 27i− 9 27a i− + 9− = 72 − 9a = 0

a = 729

−−

a = 8

11 Let z = 5 5 3i− So z2 = 5 − 5 3i , Take a = 5 and

b = − 5 3

x2 = 2 25 5 ( 5 3)

2+ + −

x2 = 5 25 752

+ +

x2 = 5 1002

+

x2 = 5 102

+

x2 = 152

x = 152

±

= 15 22 2

± ×

= 302

±

y2 = 152

− 5

y2 = 52

y = 52

±

= 5 22 2

± ×

= 102

±

So 5 5 3i− is 30 102 2

i

± −

∴ D

12 z3 = 3 i+ If w = 3 + i = r cis(θ)

r = 2 2( 3) 1+

= 4 = 2

tan(θ) = 13

θ = 1 1tan3

−

(1st quadrant)

= 6π

So w = 2cis 26

kπ π +

Let z3 = 2cis 26

kπ π +

z = 13

2cis 26

kπ π +

= 13 22 cis

18 3kπ π +

k = 0 z = 3 2 cis18π

k = 1 z = 3 22 cis18 3π π +

= 3 122 cis18

π π+


= 3 132 cis18

π

k = 2 z = 3 42 cis18 3π π +

= 3 242 cis18

π π+

= 3 252 cis18

π

= 3 252 cis 218

π π −

= 3 112 cis18

π−

C

13 z8 = 1 will have 8 solutions which will be evenly spaced on

an Argand diagram, so they will be spaced 3608

° = 45° apart

E

14 a x2 − 2 5x + 13 = 0 a = 1, b = 2 5,− c = 13

x = 22 5 ( 2 5) 4 1 13

2 1± − − × ×

×

= 2 5 20 522

± −

= 2 5 322

± −

= 2 5 4 22

i±

= 5 2 2i±

b z2 = 8 + 15i. Take a = 8 and b = 15

x2 = 2 28 8 152

+ +

x2 = 8 64 2252

+ +

x2 = 8 2892

+

x2 = 8 172

+

x2 = 252

x = 252

±

= 5 22 2

± ×

= 5 22

±

y2 = 252

− 8

y2 = 92

y = 92

±

y = 3 22 2

± ×

= 3 22

±

Therefore z = 5 2 3 22 2

i

± +

= 2 (5 3 )2

i± +

c z3 + 8 = 0 z3 = −8 z3 = − 8 cis(2kπ)

z = (−8 cis(2kπ)13)

= 13 28 cis

3kπ −

= 22 cis3kπ −

k = 0 z = −2cis(0) = −2

k = 1 z = 22cis3π −

= 2 22 cos sin3 3

iπ π − +

= 2 cos sin3 3

iπ π − − +

= 1 322 2

i

− − +

= 1 − 3i

k = 2 z = 42cis3π −

= 4 42 cos sin3 3

iπ π − +

= 2 cos sin3 3

iπ π − − −

= 1 322 2

i

− − −

= 1 + 3i The three cube roots of −8 are −2, 1 3i− and

1 + 3i

15 The graph is a ray with angle 5π clockwise from the

horizontal and shifted 4 in the positive direction of the Im(z) axis so

E 16 vertical line at Re(z) = −7 So B

17 Ray with angle 34π anticlockwise from the horizontal and

shifted 3 units in the negative direction of the Re(z) axis and 4 units in the positive direction of the Im(z) axis. So

C 18 Circle of radius 4 with centre at 0 + 2i. So A

19 Circle of radius 1 ( 5)2

− − − = 42

= 2 with centre at −3 − i. So

D

20 Sketch the graph of 2: Arg ( 5 3 )5

z z i π − − = −

on an

Argand diagram.


21 Sketch the graph of z: Im (z + 20 + 3i) = −1 on the complex plane.

22 Illustrate z: | z − 3 + 3i | = 1 on the complex plane.

Modelling and problem solving

1 a zw

= 2cis

3

2cis4

π

π

= 2 cis2 3 4

π π −

= 4 3cis12

π π−

= cis12π

= cos sin12 12

iπ π +

b z = 2cis3π

= 2 cos sin3 3

iπ π +

= 1 322 2

i

+

= 1 + 3i

w = 2cis4π

= 2 cos sin4 4

iπ π +

= 1 122 2

i +

= 2 2 2 22 2 2 2

i× + ×

= 2 2i+

c zw

= 1 3 2 22 2 2 2

i ii i

+ −×+ −

= 2

22 2 6 6

2 2i i i

i− + −

−

= ( 2 6) ( 6 2)4

i+ + −

d i Equating the results from parts (a) and (c) gives

cos sin12 12

iπ π +

= 2 6 6 24 4

i+ −+

Equating the real parts gives

2 6cos12 4π + =

ii Equating the imaginary parts of part (d) i gives

= sin12π

= 6 24−

iii tan12π

= sin

12

cos12

π

π

=

6 24

2 64

−

+

= 6 2 2 62 6 2 6

− −×+ −

= 2( 6 2)

2 6− −

−

= (6 2 12 2)4

− − +−

= (8 4 3)4

− −−

= 4− (2 3)

4−

−

= 2 − 3

c z = 2cis4π

, w = 2cos6π

zw = 2cis4π

2cis6π

= 2 2cis4 6π π × +

= 3 24cis12

π π+

= 54cis12π

= 5 54 cos sin12 12

iπ π +

= 5 54cos 4sin12 12

iπ π +

z = 2cis4π

= 2 cos sin4 4

iπ π +

= 1 122 2

i +

= 2 2 2 22 2 2 2

i× + ×

= 2 2i+

w = 2cis6π

= 2 cos sin6 6

iπ π +

= 3 122 2

i

+

= 3 i+ zw = ( 2 2 )i+ ( 3 )i+

= 26 2 6 2i i i+ + + = ( 6 2) ( 6 2)i− + +

Therefore 5 54cos 4sin12 12

iπ π +

= ( 6 2) ( 6 2)− + + i


Equating the real and imaginary paths gives

54cos12π

= 6 2 − 54sin12π

= 6 2+

5cos12π

= 6 24− 5sin

12π

= 6 24+

5tan12π

=

5sin125cos12

π

π

=

6 24

6 24

+

−

= 6 2 6 26 2 6 2

+ +×− +

= 2( 6 2)

6 2+−

= 6 2 12 24

+ +

= 8 4 34

+

= 2 + 3, as required

2 a z2 + 64 = z2 − 64i2 = z2 − (8i)2 = (z + 8i)( z − 8i) b z4 + 64 = y2 − 64i2 = y2 − (8i)2, with y = z2 = (y + 8i)(y − 8i) = (z2 + 8i)(z2 − 8i) c i Show that (2 + 2i)2 = 8i (2 + 2i)2 = 4 + 2 × 2 × 2i + 4i2 = 4 + 8i − 4 = 8i, as required ii Show that (2 − 2i)2 = −8i (2 − 2i)2 = 4 + 2 × 2 × − 2i + 4i2 = 4 − 8i − 4 = −8i, as required d z4 + 64 = (z2 + 8i)(z2 − 8i) = [z2 − (2 − 2i)2][z2 − (2 + 2i)2] = (z + 2 − 2i)(z − 2 + 2i)(z + 2 + 2i)(z − 2 − 2i) e z4 + 64 = (z + 2 − 2i)(z − 2 + 2i)(z + 2 + 2i)(z − 2 − 2i) = [(z + 2 − 2i)(z + 2 + 2i)][(z − 2 + 2i)(z − 2 − 2i)] = [z2 + (2 + 2i) z + (2 − 2i) z + (2 − 2i)(2 + 2i)]

× [z2 + (−2 − 2i) z + (−2 + 2i) z + (−2 + 2i)(−2 − 2i)] = [z2 + (2 + 2i + 2 − 2i)z + 4 + 4i − 4i − 4i2] ×

[z2 + (− 2 − 2i − 2 + 2i) z + 4 + 4i − 4i − 4i2] = (z2 + 4z + 8)(z2 − 4z + 8)

chapter 1 — complex numbers -...

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