De Moivre’s theorem – exam questions  Question 1: Jan 2007 

 Question 2: June 2007 

 Question 3: Jan 2006 

22

2 22

4 3

It is given that .

1) )Show that 2cos (2 )

1) Find a similar expression for (2 )

1 1) Hence show that 2 4cos 2cos (3 )

) Hence solve the quartic equation 2

iz e

a i z marksz

ii z marksz

iii z z marksz z

b z z z

2 1 0

giving the roots in the form . (5 )

z

a ib marks

 

 Question 4: Jan 2010 

    

Question 5: June 2009 

 Question 6: Jan 2009 

  Question 7: June 2006 

    

Question 8: Jan 2008 

 

Question 9: June 2008 

     

De Moivre’s theorem – exam questions ‐ answers  Question 1: Jan 2007 

6

2 2

6 6

6 6)

6 6 6 6

) ( sin )(1 sin ) ( ) ( )( sin )

( )

) (1 ) (1 ) (6 6 6 6

1

6

1b Cos iSin Cos iSin Cos iSin

c Cos i Cos i Cos iSin Cos iSi

Cos iSin

n Cos i

Cos iSin Cos Sin

d Cos iSin Cos iSin Cos i

6

6

6 66 6 6

sin )(1 sin ) (1 )6 6 6 6 6

( ) 1 (1 ) 1 1 (1 )6 6 6 6 6 6 6 6 6 6

0

Cos i Cos iSin

Cos iSin Cos iSin Cos iSin Cos iSin Cos iSin

Question 2: June 2007 

1515 15 0

(15 ) 0 (15 ) 1

3 315

0 02 13

Cos iSin Cos iSin i

Cos and Sin

so

 

Question 3: Jan 2006 

2 2 2 22 2

2

2

22

2

2

1 1) )

1 1)

2 2 2 2

1 1) 2 2

12

12

2 2 2

2 2 1

12

2

i

i i ii

i i ii

z e

a i z e e ez e

Cos iSin Cos iSin

ii z e e ez e

Cos iSin Cos iSin

iii z z Cos Cosz z

weknow that Co

z Co

s Cos so

z z

sz

z Cosz

z

22

4 3 2 2

2 2

2 2 22 2

2

2

12(2 1) 2 2

) 2 1 0 factorise by

( 0 )

1 1 1 1( 2 ) 0 2 0

4 2 0

2 (2 1) 0

10

2

14

2

12

2

2

Cos Cosz

b z z z z z

z is not a solution

z z z This gives z zz z z z

Cos Cos

Cos

z z Cos Cosz z

Cos

Cos or Cos

or

321 3

2

3 3

2

iiz e

o

i or

r r

z e i

o

 

 

Question 4: Jan 2010 2

7

727 27

7

4 62 37 7

8 104 57 7

1267

) )

cos 2 sin 2 1

1

2)7 2

7

2, 3,

4, 5,

6,

i

i i

i i

i i

i

a i e

so e e i

is a solutionof z

ii k k

the other non real solutions are

for k e for k e

for k e for k e

for k e

       

2 3 4 5 6

7

4 42

2 3 4

5 2 2 7 7

2 3 4 5 6

2 3 3 2 1

1

5

2 2 3

6

3

)1

is a geometric series with common ration

1 1 1

1 1

4) ) 2cos

7

)1

1

0

1

2

0

1

i

b

c i e e

ii

2 4 6cos 2cos 2cos 1

2 4 6 1cos cos cos

7 7

7

7

7 7

2

 

 Question 5: June 2009 

4

4 4312 12

4

4 4 4

4

1 3) 2 2 16 16 cos 16

3 3 2 2

8(1 3)

) let's write ,

is a solution of this equation when

16 and 4 = 23

2 2, 1,0,11 2

8

2

ii i

i i

a z e so z e e z iSin i

z i

b z re z r e

z

r k

r and k k

a

S

5 11 7

12 12 12 12: 2 , 2 , 2 ,2i i i i

olutions are e e e e

 

 Question 6: Jan 2009 

4 4 8 4 4 8 4

8 4 8

4 3

4

8 4 8 4

4 43 3

) 1 1

2cos 1 2 cos 1

1) for cos ( ) , 2 cos 1 1 0

2

We can factorise as 0

We need to solve

(

3

i i i i i

i

i

i i

a z e z e z z e z e z z e e

z z z z

b z z becomes z

z e

e

e

z

z

z

11 5 9

12 12 12

4 4 43

12

3)

1 4 23

12 2

This gives , , ,

)

i i

i

i

i

i

i i

re e r e e

r or k

k

c

z e e e e

 

    

Question 7: June 2006 6 6 6 0

6 6 6 0

6

2 203 3 3 3

2 2

) 1 1

The equation z 1 1

1 1

6 0 2 3 3

2 33

1 1 ( )) )

i i

i i i

i i

i i i i

i i

i ii i

a Let note z re then z r e and e

is equivalent to r e e

This gives r r

k k

k k

so

w e e e eb i

z e or e or e or e or e o

w e e

r e

2

2

2

2

2

6 6

1 1)

1 2 2

2 2 1)

1 2

2)

1

2

) ) ( 2 ) is equivalen

2

2

t or

2

t o

i i

i

i

e e

w iii

w iSin iSin i

i i e Cos iSiniii

w iwSin e Sin Sin S

iS

inCos Sin

iSin Sin

iiv z Cot i

in

i

so zw

i zw z

c i z i

Sin

Cot i

z i w

z

z

6

66 6 2 6

2 3 6

der 5 polynomial=0

(the term in z )

2) ( 2 ) 1 ( ) 1

( )

2 5cot 0 , cot , cot

3 3

, c

, 3

ot

, , , 3

, cot6

3

3

3

6 3

k ki i

cancel out

z iii z i z w

z

So w e question a w e

This gives

z

z i i i i i

i i i i i

 

Question 8: Jan 2008 

4

5 5 5 5

5

5 5 4

3

5

4

1 1) 4 4 4 2 4 2

2 2

) Let's write ( )

4 4

4 2

4 2 5 24

22 2, 1,0,1,2

20 515 7 9 17

2 , , , ,20 20 20 20 20

5 :

2

4 4

,

i

i i

ii

th

i

a i i e

b z re r e

z i becomes

r e e

so r and k

r and k k

r and

The roots of i are

e

7 9 17

20 20 20 202 , 2 , 2 , 2i i i i

e e e e

   

Question 9: June 2008 

2 22 2

24 2 2

22 2

2 2

2 42 4

6 2 2 42 6

6

2 4

1 1 1 1) ) 1 1

1 1 1 1 1)

1 12

1 12 2

1 1 2 22 2 4

a i z z z zz z z z

ii z z z z zz z z z z

z zz z

z zz z

z z z zz z z z

z

2 46 2 4

1 1 12 4z z

z z z  

4 24 2

24

4 2

6 26 2

1) ) cos sin cos sin 2cos

1) cos sin cos sin 2 sin

1 1 1 1)cos sin

2 2

1 1 1

64

1 1 1

64

nn

nn

b i z n i n n i n nz

ii z n i n n i n i nz

c z zz zi

z zz z

z zz z

44

4 2

12 4

12cos 6 2cos 2 4cos 4 4

64

1 1 1 1) cos sin cos 6 cos 4 cos 2

32 16

1 1 1 1cos 6 cos 4 cos 2

32 16 32 16

1 1 1 1sin 6 sin 4 sin 2

192 64 64 1

32 16

6

zz

d d d

c