PURE MATHEMATICS UNIT 2 Bossi Inc

SOLUTION ARCHIVES

W507 COMPLEX NUMIBERS

Sol Ica Given ZZ 102 31 0 then the quadraticdiscriminant is

A BZ Tac

C 10 4 c 31100 12424

0 i roots are complex

so Z b JJ29

Glo t 1242440 I 14 6 i

40 I 2560

5 I 566

i The roots of the equation in rectangular

form are

Z 5 t Tbi and

Zz 2ft5 Foi

I cont'd

Now in polar form

Z 1211Ldl

152 4651 tan

13T 126.100

a Zz z t

13T 26.100

Sol I

Given 22 82 17 0 then its roots are

Z Zz C 8 I IC 8 2 4G iz2 l

8 I J 64 68

2

8 I 542

8 I Zi

4 ti and 4 i

Icb cont'd

Now 1271 14 tilIt

TH

9 agedtan Te

14.040

O 2450 rad

Therefore in trigonometric form

Z 1211 Gsd ishida

517 cos14.040 t isin 14.040

Zz 2ftJIT cos14040 isin 14047

And in exponential form

Z JTF0.2450i

and Zz Ff eO 24506

Sol 2cal

U2 60 t 320

4 15 80 Quadrant II

4 J 15782 1800 tan Es

68 151.930

So by taking the square root of both sides

U I 68 151.9302

12 151.930

I 168 1

I 568 960

I 568 cos 75.960 t i sin 75.960

I t si

so the required values of u are

2 86 and 2 86

2Cb Given the quadratic equation

22

3 zi z t 5 Ji O l

it discriminant is

A b Tac

f 3 Zi2

4G 5 si

9 12 i 4 20 t 20615 t 8 i

U24

So by applying the quadratic formula to Equus

Z3 Zi JJ

Z c

I 3 Zi JIE3 Zi Uz

3 Zi I 2280

I 3 Zi i ki

2 ti or l 3J

Sol 3 aIf Z 3 Zi is a factor of f z then

p 3 Zi should be a roof of fCzsuch that fCp 0

Verification

f p p 8ps t 25ps 26

3 zip 8 3 zi t 25 3 20 26

33 3 3 4 Zi t 3 3 ti t C zip

8 9 126 1462 125 50 i 26

27 54 i 36 t Bi 40 t 96I t 99 506

27 36 40 99 t 54 t 8 t 96 50 i

O t Ou

By the factor theorem z 3 26 is indeed

a factor of ftz QED

Sole 3 b

since the polynomial Hz has real coefficients thenit complex roots occur in conjugate pairs

So with p 3 zi then the other root isB't 3 Zi

Now the quadratic factor of Hz is

z p z p't z2 ptp z pp'tE 2 ReB z IpZZ 2 3 2 321 22

22

GZ t 13

Sol 3Cc

Using the previous result let

Az B ZZ Gz 13 23 82

2t 25 Z 26

By comparison of terms

A23 23 A I

13 B 26 B 2

So the 3rd factor of Hz is z 2

So all three roots of fee o are

2 3 t Zi 3 Zi 2

So II 4

Using De Moines TheoremI 464 I

sin 70 TIM cos 70 is ind I 5 w ws ii b is ro is b i

TIM cos 0 t islandt

c a 21 35 35217

Im is7

where C cosd and

S since

Ilm Ct t 7 is t 21C Eis 35C cis

35C is t 21 chis 7C is

Cist

Im f I t ti c's 21C's 35 i c'st 35C's 21 i c S5 7CSb

7 is 7

7Cbs 35C't s't 21 255 757

7 I 5 s 35 G 5 25321 I 5 S5 75

75 l 35 1 354 S6 3553C 25 542155 2157 757

75 2153 2155 7St 3553 7055 35t 2155 28 St

sin78 7 S 5653 t 11255 70 S

7 sind 56 sin'd 112 sin50 70 sin78

Sol 5

First I 3 t i 531 JC 3 2 435

19 t 3

JT2

and

avg 3 it 3 1800 tan Eg

1800 300

1500

i Using DeMoines Theorem

C 3 t c B 7 T2 14500t

fit 112 1500127 1242

123512 1 104 5

1728 JI 100

1728512 t Oi

Sol 6 a

Given z 3il 2 2 21 let z octiysuch that

Ixtiy sit 21 at icy 21Grouping real and imaginary terms yield

Ix t y 3 il 2 Getz tiyApplying Pythagoras Theorem gives

1 2t y 3

22 I Get25Ty

squaring both sides we get

set Cy 35 4 Getz y

sit y by 6 4 x't 4 4 t y422 t 162C t 16 t 4g

Hence

3 2 t 3g t 16K by t 10 0

XZ t y t 1632C t Zy 103

By completing the quadratic squares we get

Get Eg t y t 1 Eg Eg t 45

t 6q4 t I

30 t 64 t 99

419

c The locus of 2 mapped to the cartesian

plane for the given relationship is a

circle centered at f Eg 1 and withradius r Etz

Sol 6lb

avg Z Z 3D Ez

avg Z 2 35 Ez

Therefore the locus is a K line with fixed pointat 2,3in and angle Eg radians

Reader Please produce an

accurate sketch ofthe locus for 6Caand 6Cb

C NEIN Oi