1

De Moivre’s Theorem - ALL

1. Let x and y be real numbers, and be one of the complex solutions of the equation z3 = 1. Evaluate:

(a) 1 + + 2;

(b) ( x + 2y)(2

x + y). [6]

1. (a) Since is a complex number which satisfies 3 –1 = 0, 1. Hence,

1 + + 2 = = 0. (M1)(A1)

(b) (x + 2y)(2

x + y) = 3x

2 + 4

yx + 2xy + 3

y2. (M1)

Using 3 = 1 and 4

= , we get, (M1)

(x + 2y)(2

x + y) = (x2 + y

2) + (2

+ )xy (M1)

= x2 + y

2 – xy, (Since l + + 2

= 0) (A1)

2. (a) Express z5 – 1 as a product of two factors, one of which is linear.

(b) Find the zeros of z5 – 1, giving your answers in the form

r(cos θ + i sin θ) where r > 0 and –π < θ π.

(c) Express z4 + z

3 + z

2 + z + 1 as a product of two real quadratic factors. [10]

2. (a) z5 – 1 = (z – 1)(z

4 + z

3 + z

2 + z + 1) (C2)

(b) z5 – 1 = 0

z = cos 0 + i sin 0 (accept z = 1).

cos (C3)

(c) (M1)(C1)

(M1)(C1)

Thus, z4 + z

3 + z

2 + z + 1 = (C1)

OR

z4 + z

3 + z

2 + z + 1 = (C1)

OR

z4 + z

3 + z

2 + z + 1 = (z

2 – 0.618z + 1)(z

2 + 1.618z + 1) (C1)

1

1 3

5

4πsin i

5

π4cos,

5

2πsin i

5

π2

5

4πsin i

5

π4cos,

5

2πsin i

3

2πcosAccept

15

π2cos2

5

2πsin i

5

π2cos

5

2πsin i

5

π2cos 2

zzzz

15

π4cos2

5

4πsin i

5

π4cos

5

4πsin i

5

π4cos 2

zzzz

1

5

4πcos21

5

π2cos2 22 zzzz

1

5

πcos21

5

π2cos2 22 zzzz

2

3. (a) Express the complex number 8i in polar form.

(b) The cube root of 8i which lies in the first quadrant is denoted by z. Express z

(i) in polar form;

(ii) in cartesian form. [6]

3. (a) |8i| = 8, arg 8i = (or 1.57, 90°) (A1)(A1)

8i = 8 (C2)

(b) (i) |z| = 2, arg z = (or 0.524, 30°) (A1)(A1)

z = 2 (C2)

(ii) z = 3+i (or 1.73 + i) (A1)(A1) (C2) [6]

4. Consider the complex number z = .

(a) (i) Find the modulus of z.

(ii) Find the argument of z, giving your answer in radians.

(b) Using De Moivre’s theorem, show that z is a cube root of one, ie z = .

(c) Simplify (l + 2z)(2 + z2), expressing your answer in the form a + bi, where a and b are exact real

numbers. [11]

4. z =

(a) (i) z = 1 (A1)

(ii) z = (M1)

arg z = 2 (M1)

=

= (A1)

2

π

2

πsin i

2

πcos

6

π

6

πsin i

6

πcos

4

32

24

πsini–

24

πcos

3

πsin i

3

πcos

4

πsini–

4

πcos

3 1

4

32

24

πisin–

24

πcos

3

πisin

3

πcos

4

πisin–

4

πcos

4

32

24

π–isin

24

π–cos

3

πisin

3

πcos

4

π–isin

4

π–cos

24

π–4–

3

π3

4

π–

6

ππ

2

π–

3

π2

3

OR

arg z = or 2.09 radians (G3) 4

(b) z3

= (M1)

= cos 2π + i sin 2π (M1)

= l + 0i

= 1 (AG) 2

(c) (1 + 2z) (2 + z2) = 2 + z

2 + 4z + 2z

3

= 2 + z2 + 4z + 2 (since z

3 = 1) (Ml)

= 4 + z2 + 4z

OR

= 4 + z2 + 4z

4 + cos (M1)(A1)

(A1)(A1)

= 1 + z + z2 + 3 + 3z

= 3 + 3z (since 1 + z + z2 = 0) (M1)

= 3 + 3 (M1)

= 3 + 3 (A1)

= (A1) 5

5. (a) Prove, using mathematical induction, that for a positive integer n,

(cos + i sin)n = cos n + i sin n where i

2 = –1.

(b) The complex number z is defined by z = cos + i sin.

(i) Show that = cos (–) + i sin (–).

(ii) Deduce that zn + z

–n = 2 cos nθ.

(c) (i) Find the binomial expansion of (z + z–l

)5.

(ii) Hence show that cos5 = (a cos 5 + b cos 3 + c cos ),

where a, b, c are positive integers to be found. [15]

5. (a) The result is true for n = 1, since

LHS = cos θ + i sin θ

and RHS = cos θ + i sin θ (R1)

Let the proposition be true for n = k. (M1)

Consider (cos θ + i sin θ)k + 1

= (cos kθ + i sin kθ)(cos θ + i sin θ) (M1)

= cos kθ cos θ – sin kθ sin θ + i(sin kθ cos θ

+ cos kθ sin θ)

3

π2

3

3

π2sini

3

π2cos1

3

π2sini4

3

π2cos4

3

π4sini

3

π4

2

i33

2

3

3

π2sini

3

π2cos

2

3i

2

1–

i2

33

2

3

z

1

16

1

4

= cos (k + 1)θ + i sin (k + 1)θ (A1)

Therefore, true for n = k true for n = k + 1 and the proposition is

proved by induction. (R1) 5

(b) (i) (M1)

= (cos θ + i sin θ) (A1)

= cos (–θ) + i sin (–θ) (AG)

OR

= z–1

(M1)

z–1

= cos (–θ) + i sin (–θ), by de Moivre’s theorem

(accept the cis notation). (A1)(AG)

Note: Award (M0)(A0) to candidates who use the result of part (a) with

no consideration that in this part, n < 0.

(ii) z–n

= (z–1

)n = cos (–nθ) + i sin (–nθ) (A1)

zn +z

–n = cos nθ + i sin nθ + cos (–nθ) + i sin (–nθ) (M1)

= cos nθ + i sin nθ + cos (nθ) – i sin (nθ) (A1)

= 2 cos nθ (AG) 5

(c) (i) (z + z–1

)5 = z

5 + 5z

3 + 10z + 10z

–1 + 5z

–3 + z

–5 (M1)(A1)

(ii) (2 cos θ)5

= 2 cos 5θ + 10 cos 3θ + 20 cos θ (M1)(A1)

giving a = 1, b = 5 and c = 10(or cos5 θ =

(cos 5θ + 5 cos 3θ + 10 cos θ)) (A1) 5

6. (a) Use mathematical induction to prove De Moivre’s theorem

(cos + i sin)n = cos (n) + i sin (n), n

+.

(b) Consider z5 – 32 = 0.

(i) Show that z1 = 2 is one of the complex roots of this equation.

(ii) Find z12, z1

3, z1

4, z1

5, giving your answer in the modulus argument form.

(iii) Plot the points that represent z1, z12, z1

3, z1

4 and z1

5, in the complex plane.

(iv) The point z1n is mapped to z1

n+1 by a composition of two linear transformations, where n = 1,

2, 3, 4. Give a full geometric description of the two transformations. [16]

6. (a) (cos θ + i sin θ)n = cos (nθ) + i sin (nθ), n

+

Let n = 1 cos + i sin = cos which is true. (A1)

Assume true for n = k (cos + i sin)k = cos (k) + i sin (kθ) (M1)

Now show n = k true implies n = k + 1 also true. (M1)

(cos + i sin)k+1

= (cos + i sin)k(cos + i sin ) (M1)

= (cos (k) + i sin (k)(cos + i sin)

= cos (k) cos – sin (k)sin +i(sin (k) cos + cos (k) sinθ) (A1)

= cos (k + ) + i sin (kθ + θ)

sini–cos

sini–cos

sinicos

11

z

z

1

16

1

5

2πsin i

5

2πcos

5

= cos (k + 1) + 1 sin (k + 1) n = k + 1 is true. (A1)

Therefore by mathematical induction statement is true for n ≥ 1. (R1) 7

(b) (i) z1 = 2

= 25(cos 2 + 1 sin 2) (M1)

= 32

Therefore z1 is a root of z5 –32 = 0 (AG)

(ii)

= 32(cos 2 + i sin 2)(= 32(cos 0 + i sin 0) = 32) (A2)

Note: Award (A2) for all 4 correct, (A1) for 3 correct, (A0) otherwise.

(iii)

(A1)(A3)

Note: Award (A1) for graph of reasonable size, scale, axes marked, (A3)

for all 5 points correctly plotted, (A2) for 4 points correctly plotted. (A1)

for 3 points correctly plotted.

(iv) Composite transformation is a combination of (in any order)

an enlargement scale factor 2, centre (0, 0); (A1)

a rotation (anti-clockwise) of (72°), centre (0, 0)

(A1) 9

Note: Do not penalize if centre of enlargement or rotation not given. [16]

7. Given that z , solve the equation z3 – 8i = 0, giving your answers in the form z = r (cos + i sin). [6]

5

2πisin

5

π2cos

51z

5

4πisin

5

π4cos42

1z

5

6πisin

5

π6cos82

1z

5

8πisin

5

π8cos164

1z

51z

–8

–8

–6

–6

–4

–4

–2–2

2

2

4

4

6

6

8 10 12 14 16 18 20 22 24 26 28 30 32

–10

–12

–14

–16

zz z

z

z

Re

Im

121

51

41

31

5

π2

)288(

5

8π clockwiseor

6

7. (A1)

where (A1)(A1)

(A1)

(A1)

(A1) (C6)

8. Consider the complex number z = cos + i sin.

(a) Using De Moivre’s theorem show that

zn + = 2 cos n.

(b) By expanding show that

cos4 = (cos 4 + 4 cos 2 + 3).

(c) Let g (a) = .

(i) Find g (a).

(ii) Solve g (a) = 1 [11]

8. (a) zn = cos n + i sin n

= cos (–n ) + i sin (–n) (M1)

= cos n – i sin (n ) (A1)

Therefore (AG) 2

(b) (M1)

(M1)

(2 cos )4 = 2 cos 4θ + 8 cos 2θ + 6 (A1)

(A1)

2sini

2cos8i8 33

zz

)sini(cos rz

)2(2

3and83

nr

6sini

6cos21

z

6

5sini

6

5cos22

z

2–sini

2–cos2or

2

3sini

2

3cos2 33

zz

nz

1

41

zz

8

1

a

0

4 dcos

nz

1

nzn

n cos22

1

432

234

411

41

)(61

41

zzz

zz

zzz

zz

61

41

2

2

4

4

zz

zz

)62cos84cos2(16

1cos4

7

(AG) 4

(c) (i) (M1)

(A1)

(A1)

(ii)

a = 2.96 (A1)

Since cos4 θ 0 then g (a) is an increasing function so

there is only one root. (R1) 5

9. Let z = cos + i sin , for – .

(a) (i) Find z3 using the binomial theorem.

(ii) Use de Moivre’s theorem to show that

cos 3 = 4 cos3 – 3 cos and sin 3 = 3 sin – 4 sin

3.

(b) Hence prove that = tan.

(c) Given that sin = , find the exact value of tan 3. [21]

9. (a) (i) (cos + i sin)3 = cos

3 + 3 cos2 i sin + 3 cos i

2 sin

2

+ i3 sin

3 A1A1A1A1

(= cos3 + 3 cos

2 sin i 3 cos

sin2 i sin

3)

(= cos3 3 cos sin

2 + (3 cos2

sin sin3) i)

Note: Award A1 for each term in the expansion.

(ii) (cos + i sin)3 = cos 3 + i sin 3 (A1)

equating real and imaginary parts (M1)

cos 3 = cos3 3 cos sin

2 A1

= cos3 3 cos (1 cos

2) A1

= 4 cos3 3 cos AG N0

and sin 3 = 3 cos2 sin sin

3 A1

= 3 (1 sin2 ) sin sin

3 A1

= 3 sin 4 sin3 AG N0

)32cos44(cos8

1

a a

0 0

4 d)32cos44(cos8

1dcos

a

0

32sin24sin4

1

8

1

aaaag 32sin24sin

4

1

8

1)(

aaa 32sin24sin

4

1

8

11

4

π

4

π

θθ

θθ

cos3cos

sin3sin

3

1

8

(b) A1A1

= A1A1

Using 1 2 sin2 = 2 cos

2 1 = cos 2 M1

A1

= tan AG

(c) METHOD 1

A1

M1

= (A1)

M1

=

A1 N0

METHOD 2

M1

= (A1)

M1A1

= A1 N0

10. Let y = cos + i sin.

(a) Show that = iy.

[You may assume that for the purposes of differentiation and integration, i may be treated in the

θθθ

θθθ

θθ

θθ

coscos3cos4

sinsin4sin3

cos3cos

sin3sin3

3

1cos2cos2

sin21sin22

2

θθ

θθ

θ

θ

θθ

θθ

cos2

sin2

cos3cos

sin3sin

23

2cos

3

1sin θθ

27

14

3

133sin θ

27

23

23

232

3

243cos

3

θ

227

10

2

20

23

210

233tan θ

27

14

3

133sin

3

1sin θθ

27

23

27

210

27

2313cos

2

2

θ

27

210

27

23

3tan θ

2

20

23

210

23

θ

y

d

d

9

same way as a real constant.]

(b) Hence show, using integration, that y = ei

.

(c) Use this result to deduce de Moivre’s theorem.

(d) (i) Given that = a cos5 + b cos

3 + c cos, where sin 0, use de Moivre’s theorem

with n = 6 to find the values of the constants a, b and c.

(ii) Hence deduce the value of . [20]

10. (a) A1

EITHER

A1

= i (cos + i sin ) A1

= i y AG N0

OR

i y = i(cos + i sin) (= i cos + i2 sin) A1

= i cos sin A1

= AG N0

(b) M1A1

ln y = i + c A1

Substituting (0, 1) 0 = 0 + c c = 0 A1

ln y = i A1

y = ei

AG N0

(c) cos n + i sin n = ein

M1

= (ei

)n A1

= (cos + i sin )n AG N0

Note: Accept this proof in reverse.

θ

θ

sin

6sin

θ

θ

sin

6sinlim

0

θθx

ycosisin

d

d

θθθ

ycosisini

d

d 2

θ

y

d

d

θy

ydi

d

10

(d) (i) cos 6 + i sin 6 = (cos + i sin)6 M1

Expanding rhs using the binomial theorem M1A1

= cos6 + 6 cos

5 i sin + 15 cos4 (i sin)

2 + 20 cos

3 (i sin)3

+ 15 cos2 (i sin)

4 + 6 cos (i sin)

5 + (i sin)

6

Equating imaginary parts (M1)

sin 6 = 6 cos5 sin 20 cos

3 sin3 + 6 cos sin

5 A1

6 cos5 20 cos

3 (1 cos2) + 6 cos (1 cos

2)2 A1

= 32 cos5 32 cos

3 + 6 cos (a = 32, b = 32, c = 6) A2 N0

(ii) M1

= 32 32 + 6

= 6 A1 N0

11. Prove by induction that 12n + 2(5

n−1) is a multiple of 7 for n

+. [10]

11. Let P(n) be the proposition 12n + 2(5

n 1) is a multiple of 7

When n = 1 121 + 2(5

0) = 14 (which is a multiple of 7) R1

P(1) is true

Assume P(k) is true,

ie 12k + 2(5

k 1) = 7m, m

+ M1

For n = k + 1 12k + 1

+ 2(5k) = 12(12

k) + 2(5

k) M1

= 12(7m 2 5k 1

) + 2 5k A1

= 12(7m 2 5k 1

) + 10 5k 1

A1

= 12(7m) 24 (5k 1

) + 10(5k 1

) A1

= 7(12m 2 5k 1

) A1

12k + 1

+ 2(5k) is a multiple of 7 R1

So P(k) is true P(k + 1) is true, and P(1) is true. R1

Hence by induction P(n) is true. R1 N0

Note: Award the final R1R1R1 only

if the previous reasoning is correct. [10]

12. Prove that is real, where n +. [6]

12. EITHER

A1

A1

θ

θ

sin

6sin

θθθθ

θ

θθcos6cos32cos32lim

sin

6sinlim 35

00

nn

i3i3

6sini

6cos2i3

6sini

6cos2i3

11

Hence using De Moivre’s Theorem

M1A1

= (A1)

= which is real. R1

OR

M1A1

M1A1

The terms in odd powers of i are of opposite sign in each series expansion

and hence cancel. R1

Hence + is real. R1

OR

+ has the form R2

= M1A1

= 2 Re A1

expression is real R1

13. Express in the form where a, b . [5]

13. METHOD 1

(A1)(A1)

M1

(M1)

A1

METHOD 2

(1 )(1 ) = 1 2 3 (= 2 2 ) (M1)A1

6sini

6cos2

6sini

6cos2i3i3

nnnn nnnn

6sini

6cos

6sini

6cos2

nnnnn

6cos2 1 nn

...

!3

3i21

!2

3i13i3i3

33221

nnnnn nnnnn

n

...!3

3i21

!2

3i13i3i3

33221

nnnnn nnnnn

n

ni3 ni3

ni3 ni3 nn zz

nn zz

nz

33i1

1

b

a

3,2

θr

3

33

3sini

3cos23i1

sinicos8

1

8

1

i3 i3 i3 i3

12

( 2 2 )(1 ) = 8 (M1)(A1)

A1

METHOD 3

Attempt at Binomial expansion M1

(1 )3 = 1 + 3( ) + 3 ( )

2 + ( )

3 (A1)

= 1 3 9 + 3 (A1)

= 8 A1

M1

14. Let w = cos

(a) Show that w is a root of the equation z5 − 1 = 0.

(b) Show that (w − 1) (w4 + w

3 + w

2 + w + 1) = w

5 − 1 and deduce that w

4 + w

3 + w

2 + w + 1 = 0.

(c) Hence show that cos [12]

14. (a) EITHER

(M1)

= cos 2 + i sin 2 A1

= 1 A1

Hence w is a root of z5 1 = 0 AG

OR

Solving z5 = 1 (M1)

z = cos A1

n = 1 gives cos which is w A1

(b) (w 1)(1 + w + w2 + w

3 + w

4) = w + w

2 + w

3 + w

4 + w

5 1

w w2 w

3 w

4 M1

= w5 1 A1

Since w5 1 = 0 and w 1, w

4 + w

3 + w

2 + w + 1 = 0. R1

(c) 1 + w + w2 + w

3 + w

4 =

i3 i3

8

1

i31

13

i3 i3 i3 i3

i3 i3

8

1

i31

13

.5

2sini

5

2

.2

1

5

4cos

5

2

5

5

5

2sini

5

2cos

w

.4,3,2,1,0,5

2sini

5

2 nnn

5

2sini

5

2

2

5

2sini

5

2cos

5

2sini

5

2cos1

13

(M1)

M1

M1A1A1

Notes: Award M1 for attempting to replace 6

and 8 by 4 and 2.

Award A1 for correct cosine terms and

A1 for correct sine terms.

A1

Note: Correct methods involving equating real

parts, use of conjugates or reciprocals are

also accepted.

AG

Note: Use of cis notation is acceptable throughout

this question.

15. z1 = and z2 =

(a) Find the modulus and argument of z1 and z2 in terms of m and n, respectively.

(b) Hence, find the smallest positive integers m and n such that z1 = z2. [14]

15. (a) (A1)

arg or arg (1 i) = (A1)

A1

A1

arg (z1) = m arctan A1

arg (z2) = n arctan (1) = A1 N2

(b) (M1)A1

M1A1

43

5

2sini

5

2cos

5

2sini

5

2cos

5

4sini

5

4cos

5

2sini

5

2cos1

5

8sini

5

8cos

5

6sini

5

6cos

5

4sini

5

4cos

5

2sini

5

2cos1

5

2sini

5

2cos

5

4sini

5

4cos

05

2cos2

5

4cos21

2

1

5

4cos

5

2cos

m3i1 .i1n

2i1or23i1

3

3i1

4

4

7accept

π

mz 21

n

z 22

33

m

4

n

4

7accept

πn

mnnm 222

integeran is where,243

kknm

14

(M1)

A1

The smallest value of k such that m is an integer is 5, hence

m = 12 A1

n = 24 . A1 N2 [14]

knm

243

kmm

24

23

km 26

5

km5

12