Complex Numbers Review (solutions)
1.
If (z + 2i) is a factor then (z – 2i) is also a factor.
(z + 2i)(z – 2i) = (z2 + 4)
The other factor is (2z3 – 3z2 + 8z – 12) ÷ (z2 + 4) = (2z – 3)
The other two factors are (z – 2i) and (2z–3).
(A1)
(M1)(A1)
(C1)(C2)
[3]
2.
METHOD 1
If z  3  2i is a root, then z  3  2i is another.
P ( z )  ( z  2)  z  (3  2i)   z  (3  2i) 
(A1)
(M1)(A1)
 ( z  2)  z 2  (3  2i) z  (3  2i) z  13
 ( z  2)( z 2  6z  13)
 z 3  8z 2  25z  26
(A1)(A1)(A1)
a  8 b  25 c  26
(C2)(C2)(C2)
METHOD 2
0  8  4a  2b  c
(A1)
0  9  46i  a (5  12i)  b (3  2i)  c
(M1)
9  5a  3b  c and 46  12a  2b
(A1)
solving system of three equations
a  8 b  25 c  26
(A1)(A1)(A1)(C2)(C2)(C2)
[6]
3.
(a)
(b)
Since  is a complex number which satisfies 3 –1 = 0,   1. Hence,
13
1 +  + 2 =
= 0.
(M1)(A1)
1
(x + 2y)(2x + y) = 3x2 + 4yx + 2xy + 3y2.
Using 3 = 1 and 4 = , we get,
(x + 2y)(2x + y) = (x2 + y2) + (2 + )xy
= x2 + y2 – xy, (Since l +  + 2 = 0)
(M1)
(M1)
(M1)
(A1)
2
4
[6]
4.
(a)
z5 – 1 = (z – 1)(z4 + z3 + z2 + z + 1)
(C2)
(b)
z5 – 1 = 0
 z = cos0 + isin0 (accept z = 1).
 2π 
 2π 
 4π 
 4π 
cos  
  i sin   , cos    i sin   
 5 
 5 
 5 
 5 
2π
2π
4π
4π 

 i sin
, cos
 i sin 
 Accept cos
3
5
5
5 

(C3)
(c)
2π

 i sin
 z  cos
5

4π

 i sin
 z  cos
5

2π 
2π
 i sin
 z  cos
5 
5
4π 
4π
 i sin
 z  cos
5 
5
2
3
2π 
2π 

2
  z   2 cos  z  1 (M1)(C1)
5 
5 

4π 
4π 

2
  z   2 cos  z  1 (M1)(C1)
5 
5 

1



2π 
4π 


Thus, z4 + z3 + z2 + z + 1 =  z 2  2 cos  z  1 z 2  2 cos  z  1 (C1)
5 
5





OR



2π 
π


z4 + z3 + z2 + z + 1 =  z 2  2 cos  z  1 z 2  2 cos  z  1
5 
5





(C1)
OR
z4 + z3 + z2 + z + 1 = (z2 – 0.618z + 1)(z2 + 1.618z + 1)
(C1)
5
[10]
5.
(a)


8i = 8  cos
(b)
π
(or 1.57, 90°)
2
|8i| = 8, arg 8i =
(A1)(A1)
π
π
 i sin 
2
2
(i)
(ii)
(C2)
π
(or 0.524, 30°)
6
π
π

z = 2  cos  i sin 
6
6

|z| = 2, arg z =
(A1)(A1)
(C2)
z = 3+i (or 1.73 + i)
(A1)(A1) (C2)
[6]
2
6.
π
π 
π
π

 cos – isin   cos  isin 
4
4 
3
3
z= 
4
π
π 

– isin
 cos

24
24 

(a)
(i)
3
z = 1
(A1)
2
(ii)
 – π
π
π
 – π  
 cos
  isin 
   cos  isin 
3
3
 4  
  4 
z=
4
 – π
 – π 
 cos
  isin 
 
 24  
  24 
– π π – π
arg z = 2 
  3  – 4

 4   3   24 
–π
π
π
=
2
6
2π
=
3
3
(M1)
(M1)
(A1)
OR
arg z =
(b)
2π
or 2.09 radians
3

2π
2 π 
 i sin
z = 1 cos

3
3 

3
= cos 2π + i sin 2π
= l + 0i
=1
(G3)
4
3
(M1)
(M1)
(AG)
2
2
(c)
(1 + 2z) (2 + z2) = 2 + z2 + 4z + 2z3
= 2 + z2 + 4z + 2 (since z3 = 1)
= 4 + z2 + 4z
OR
= 4 + z2 + 4z
 4π 
 4π 
 2π 
 2π 
 i sin    4 cos   4i sin  


4 + cos  3 
 3 
 3 
 3 
3 3 3i

(A1)(A1)
2
2
(Ml)
(M1)(A1)
= 1 + z + z2 + 3 + 3z
= 3 + 3z (since 1 + z + z2 = 0)
(M1)
2π
2π 

cos
 isin

= 3 + 3 
3
3 
 1 i 3
– 


= 3 + 3 2
2


3
= 2

3 3
2
(M1)
(A1)
(A1)
5
[11]
7.
(a)
(b)
The result is true for n = 1, since
LHS = cos θ + i sin θ
and RHS = cos θ + isin θ
Let the proposition be true for n = k.
Consider (cos θ + isin θ)k + 1 = (cos kθ + isin kθ)(cos θ + isin θ)
= cos kθ cos θ – sin kθ sin θ + i(sin kθ cos θ
+ cos kθ sin θ)
= cos(k + 1)θ + i sin(k + 1)θ
Therefore, true for n = k = > true for n = k + 1 and the proposition is
proved by induction.
(i)
cos – i sin  
1
1


z cos  i sin   cos – i sin  
= (cos θ + isin θ)
= cos(–θ) + isin(–θ)
(R1)
(M1)
(M1)
(A1)
(R1)
5
(M1)
(A1)
(AG)
OR
1
= z–1
z
(M1)
z–1 = cos(–θ) + isin(–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)
(c)
z–n = (z–1)n = cos(–nθ) + isin(–nθ)
zn +z–n = cos nθ + i sin nθ + cos(–nθ) + i sin(–nθ)
= cos nθ + i sin nθ + cos(nθ) – i sin(nθ)
= 2cosnθ
(A1)
(M1)
(A1)
(AG)
(i)
(z + z–1)5 = z5 + 5z3 + 10z + 10z–1 + 5z–3 + z–5
(M1)(A1)
(ii)
(2cosθ)5 = 2cos5θ + 10cos3θ + 20cosθ
(M1)(A1)
5
1
giving a = 1, b = 5 and c = 10(or cos5 θ =
16
(cos5θ + 5cos3θ + 10cosθ))
(A1)
5
3
[15]
8.
(a)
(b)
(cosθ + isinθ)n = cos(nθ) + isin(nθ), n 
Let n = 1  cos + isin = cos which is true.
Assume true for n = k  (cos + i sin)k = cos(k) + i sin(kθ)
Now show n = k true implies n = k + 1 also true.
(cos + i sin)k+1 = (cos + i sin)k(cos + isin)
= (cos(k) + i sin(k)(cos + i sin)
= cos(k) cos – sin(k)sin +i(sin(k) cos + cos(k) sinθ)
= cos(k + ) + i sin(kθ + θ)
= cos(k + 1) + 1 sin(k + 1)  n = k + 1 is true.
Therefore by mathematical induction statement is true for n ≥ 1.
(i)
(A1)
(A1)
(R1)
7
  2π 
 2π  
z1 = 2  cos   isin   
 5 
  5 
 z15 = 25(cos2 + 1sin2)
= 32
Therefore z1 is a root of z5 –32 = 0
(ii)
(A1)
(M1)
(M1)
(M1)
(M1)
(AG)
4π
4π 

z12  4 cos
 isin

5
5 

6π
6π 

z12  8 cos
 isin

5
5 

8π
8π 

z14  16 cos  isin

5
5

z15 = 32(cos2 + isin2)(= 32(cos0 + isin0) = 32)
(A2)
Note: Award (A2) for all 4 correct, (A1) for 3 correct, (A0)
otherwise.
(iii)
Im
z 21
6
4
2
–8 –6 –4 –2
–2
–4
z 31
–6
–8
–10
–12
–14
–16
z1
z 51
Re
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32
z 41
(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)
4
a rotation (anti-clockwise) of
2π
(72°), centre (0, 0)
5
8π


(A1)
 or clockwise (288) 
5


Note: Do not penalize if centre of enlargement or rotation not
given.
9
[16]
9.
(a)
6 i 2
2
z1 =
=
z1 =
2
6 2

4 4
(A1)
 1 
π
  
arg z1 = arctan 
(A1)
6
 3
  π
 π 
Therefore, z1 = 2  cos    i sin     (C1)
 6 
  4
z2 = 1 – i
z2 = 1  1  2
π
arg z2 = arctan(–1) = –
(A1)
4
  π
 π 
z2 = 2  cos    i sin  –  
(C1)
 4 
  4
(b)
  

2  cos    i sin  –
z1

  6

z2
  π

2  cos    i sin  –
6


 
6
π 

6  
π 

4  
  π π
 π π 
= 1  cos     i sin  –   
 6 4 
  6 4
π
π
= cos
+ i sin
12
12
(c)
(M1)
z1  6  i 2  1  1  i 


 1  i  1  i 
z 2 
2

(M2)
(AG)
2
(M1)
6  2  i( 6  2 )
.
(A1)
4
π
6 2
Therefore, a = cos 
(A1)
2
4
π
6 2
b = sin 
(A1)
12
4
Note: Some students may use the half-angle formulas. Answers will only differ in form.
=
cos
π

12
2 3
2
sin
π

12
4
2 3
2
[12]
10. Let z = x + iy , x, y 
Then, z + 162 = 16z + 12
.
 (x + 16)2 + y2 = 16{(x + 1)2 + y2}
 x2 + 32x + 256 + y2 = 16x2 + 32x + 16 + 16y2
 15x2 + 15y2 = 240
(M1)
5
 x2 + y2 = 16
Therefore, z = 4.
(a + bi)(2 – i) = (5 – i)
(5  i)
(a + bi) =
(2 – i)
11.
11 3
 i (using a graphic display calculator)
5 5
11
3
Therefore a =
,b=
5
5
=
(5  i) (2  i)

(2 – i) (2  i)
10  5i – 2i  1 11  3i

=
4 1
5
11
3
Therefore a =
,b=
5
5
OR a + bi =
OR (a + bi)(2 – i) = (5 – i)
(2a + b) + (2b – a)i = (5 – i)
2a + b = 5
–a + 2b = –1
11
3
therefore a =
,b=
5
5
(A1)
(A1)(C3)
[3]
(M1)
(M1)
(A1)
(A1)
(M1)
(A1)
(A1)
(M1)
(A1)
(A1)
[3]
12.
z=1+
i(i  3 )
(i – 3 )(i  3 )
i(i  3 )
–4
 5  i 3  5  i 3 

=
 4 
4 
=1+
=
(M1)(A1)
(A1)
(A2)
5 i 3 
5
3 

Accept
a

(

1
.
25
),
b


(A1)(C3)(C3)
4
4 
4
4 
Note: Do not award the last (A1) for b = –0.433. Award (C0)
for b = –0.433 with no working.
[6]
1.
METHOD 1
z = (2 – i)(z + 2)
= 2z + 4 – iz – 2i
z(1 – i) = – 4 + 2i
 4  2i
z=
1 i
 4  2i 1  i

z=
1 i
1 i
=–3–i
M1
A1
M1
A1
METHOD 2
let z = a + ib
a  ib
=2–i
a  ib  2
M1
6
a + ib = (2 – i)((a + 2) + ib)
a + ib = 2(a + 2) + 2bi – i(a + 2) + b
a + ib = 2a + b + 4 + (2b – a – 2)i
attempt to equate real and imaginary parts
a = 2a + b + 4(  a + b + 4 = 0)
and b = 2b – a – 2(  – a + b – 2 = 0)
M1
A1
Note: Award A1 for two correct equations.
b = –1;a = –3
z=–3–i
A1
[4]
2.
(a)
12  ( 2  3 ) 2
AB =
=
88  4 3
= 2 2 3
(b)
M1
A1
A1
METHOD 1
arg z1  
Note: Allow
π
π
, arg z 2  
4
3
A1A1
π
π
and .
4
3
Note: Allow degrees at this stage.
AÔB 
π π

3 4
=
π
π
(accept  )
12
12
A1
Note: Allow FT for final A1.
METHOD 2
attempt to use scalar product or cosine rule
cos AÔB 
1 3
M1
A1
2 2
AÔB 
π
12
A1
[6]
3.
(a)
using de Moivre’s theorem
1
zn + n = cos nθ + i sin nθ + cos nθ – i sin nθ (= 2 cos nθ), imaginary
z
part of which is 0
M1A1
1
 n

so Im  z  n  = 0
AG
z 

z  1 cos   i sin   1

z  1 cos   i sin   1
(cos   1  i sin  )(cos  1  i sin  )
=
(cos   1  i sin  )(cos  1  i sin  )
(b)
M1A1
Note: Award M1 for an attempt to multiply numerator and denominator
by the complex conjugate of their denominator.
7
2
 z  1  (cos   1)(cos   1)  sin 
 Re

real denominato r
 z 1
M1A1
Note: Award M1 for multiplying out the numerator.
cos 2   sin 2   1
real denominato r
=0
A1
AG
[7]
4.
(a)
METHOD 1
z i
=i
z2
z + i = iz + 2i
(1 – i)z = i
i
z=
1 i
M1
A1
A1
EITHER
z=
π
cis  
2
 3π 
2cis  
 4 
z=
2  3π 
cis  
2
 4 
M1

1
 3π  
 or
cis   
2  4 

A1A1
OR
z=
z=
1  i
2
1 1 

   i
2 2 

2  3π 
1
 3π  
cis  or
cis  
2
 4 
2  4 
M1
A1A1
METHOD 2
x  i( y  1)
x  2  iy
x + i(y + 1) = –y + i(x + 2)
x = –y; x + 2 = y + 1
1
1
solving, x =  ; y 
2
2
1 1
z=   i
2 2
2  3π  
1
 3π  
z=
cis    or
cis   
2
 4 
2  4 
i=
M1
A1
A1
A1
A1A1
Note: Award A1 fort the correct modulus and A1 for the correct argument,
but the final answer must be in the form r cis θ.
Accept 135° for the argument.
(b)
x  ( y  1)i
( x  2)  yi
use of (x + 2) – yi to rationalize the denominator
substituting z = x + iy to obtain w =
(A1)
M1
8
ω=
=
(c)
x( x  2)  y ( y  1)  i(  xy  ( y  1)( x  2))
( x  2) 2  y 2
( x 2  2 x  y 2  y)  i( x  2 y  2)
( x  2) 2  y 2
Re ω =
x 2  2x  y 2  y
( x  2) 2  y 2
=1
 x2 + 2x + y2 + y = x2 + 4x + 4 + y2
 y = 2x + 4
which has gradient m = 2
(d)
A1
AG
M1
A1
A1
A1
EITHER
π
 x = y (and x, y > 0)
4
i(3 x  2)
2 x 2  3x
ω=

2
2
( x  2)  x
( x  2) 2  x 2
3x  2
if arg(ω) = θ  tan   2
2 x  3x
3x  2
1
2 x 2  3x
arg (z) =
(A1)
(M1)
M1A1
OR
π
 x = y (and x, y > 0)
4
π
 x2 + 2x + y2 + y = x + 2y + 2
arg (w) =
4
solve simultaneously
x2 + 2x + x2 + x = x + 2x + 2 (or equivalent)
arg (z) =
A1
M1
M1
A1
THEN
x2 = 1
x = 1 (as x > 0)
A1
Note: Award A0 for x = ±1.
│z│ =
2
A1
Note: A1low FT from incorrect values of x.
[19]
5.
(a)
│z│ = 5 and │w│=
│w│ = 2│z│
4  a2
4  a2  2 5
attempt to solve equation
M1
Note: Award M0 if modulus is not used.
a = ±4
(b)
zw = (2 – 2a) + (4 + a)i
forming equation 2 – 2a = 2 (4 + a)
3
a= 
2
A1A1
N0
A1
M1
A1
N0
[6]
9
6.
(a) │z│ = z, arg(z) = 0
so L(z) = ln z
(b)
(c)
A1A1
AG
N0
A1A1
N2
A1A1
N2
A1
N1
for comparing the product of two of the above results with the third
M1
for stating the result –1 + i = –1 × (1 – i) and L (–1 + i) ≠ L (–1) + L (1 – i)R1
hence, the property L(z1z2) = L(z1) + L(z2)
does not hold for all values of z1 and z2
AG
N0
(i)
L(–1) = ln 1 + iπ = iπ
(ii)
L(1 – i) = ln 2  i
(iii)
L(–1 + i) = ln 2  i
7π
4
3π
4
[9]
7.
(sin
+ i (1  cos))2 = sin2  (1  cos)2 + i 2 sin (1  cos)
M1A1
Let  be the required argument.
tan =
2 sin θ 1  cos θ 
sin 2 θ  1  cos θ 
M1
2
2 sin θ 1  cos θ 
=
1  cos θ  1  2 cos θ  cos θ 
=
2 sin θ 1  cos θ 
2 cos θ 1  cos θ 
2
2
(M1)
A1
= tan
A1
=
A1
[7]
8.
METHOD 1
Substituting z = x + iy to obtain w 
w
x  yi
x  yi2 1
x  yi
x  y 2  1  2 xyi
A1
2
Use of (x2  y2 + 1  2xyi) to make the denominator real.
=
x  yi x 2  y 2 1 2 xyi 
x
Im w 
=
2

2
 y 2 1  4 x 2 y 2


y x 2  y 2 1  2 x 2 y
x
x
2

2
 y 2 1  4 x 2 y 2

y 1 x 2  y 2
2

(A1)

2
 y 2 1  4 x 2 y 2
Im w = 0  1  x2  y2 = 0 i.e. z = 1 as y ≠ 0
M1
A1
(A1)
A1
R1AG
N0
METHOD 2
10
w (z2 + 1) = z
(A1)
w(x2  y2 + 1 + 2ixy) = x + yi
A1
Equating real and imaginary parts
w (x2  y2 + 1) = x and 2wx = 1, y ≠ 0
Substituting w 

M1A1
1
x y2 1

x
to give 
2 2x 2x
2x

A1
1 2
x
y 1 
or equivalent
2x
2
(A1)
x2 + y2 = 1, i.e. │z│ = 1 as y ≠ 0
R1AG

[7]
1
4
9.
(a) z = (1  i)
Let 1 – i = r(cos θ + i sin θ)
 r  2 A1
π
θ= 
A1
4
1
   π
 π  4
z =  2  cos    isin     
 4 
   4
M1
1
   π

 π
 4
=  2  cos   2nπ   i sin    2nπ   

 4

   4
=
1
8
2
  π nπ 
 π nπ  
 cos     i sin     
 16 2  
  16 2 
=
1
28
  π
 π 
 cos    i sin    
 16  
  16 
M1
Note: Award M1 above for this line if the candidate has forgotten to
add 2π and no other solution given.
=
1
28
  7π 
 7π  
 cos   i sin   
 16  
  16 
=
1
28
  15π 
 15π  
 cos
  i sin 
 
 16  
  16 
=
1
8
2
  9π 
 9π  
 cos    i sin    
 16  
  16 
A2
Note: Award A1 for 2 correct answers. Accept any equivalent form.
(b)
11
A2
Note: Award A1 for roots being shown equidistant from the origin
and one in each quadrant.
A1 for correct angular positions. It is not necessary to
see written evidence of angle, but must agree with the diagram.
1
(c)
z2

z1

15 π 
 15π  
2 8   cos
  i sin 
 
16 
 16  

1
28

7π 
 7π  
  cos   i sin   
16 
 16  

π
π
= cos  i sin
2
2
=i
(  a = 0, b = 1)
M1A1
(A1)
A1
N2
[12]
10.
5zz* + 10 = (6 – 18i)z*
Let z = a + ib
5 × 10 + 10 = (6 – 18i)(a – bi) (= 6a – 6bi – 18ai – 18b)
Equate real and imaginary parts
 6a – 18b = 60 and 6b + 18a = 0
 a = 1 and b = –3
z = 1 – 3i
M1
M1A1
(M1)
A1A1
A1
[7]
1
3
iz1 + 2z2 = 3  z2 =  iz1 
2
2
z1 + (1 – i)z2 = 4
3
 1
 z 1 + (1 – i)   iz1   = 4
2
 2
1
3 1 2
3
 z1  iz1   i z1  i = 4
2
2 2
2
1
1
5 3
z1  iz1   i

2
2
2 2
 z1 – iz1 = 5 + 3i
11.
M1A1
A1
12
EITHER
Let z1 = x + iy
 x + iy – ix – i2y = 5 + 3i
Equate real and imaginary parts
 x+y=5
–x + y = 3
2y = 8
y = 4  x = 1 i.e. z1 = 1 + 4i
1
3
z2 =  i(1  4i) 
2
2
1
3
2
z2 =  i  2i 
2
2
7 1
z2 =  i
2 2
(M1)
M1
A1A1
M1
A1
OR
5  3i
1 i
(5  3i)(1  i)  5  8i  3 
z1 =


(1  i)(1  i) 
2

z1 = 1 + 4i
1
3
z2 = – i(1 + 4i) +
2
2
1
3
2
z2 =  i  2i 
2
2
7 1
z2 =  i
2 2
z1 =
M1
M1A1
A1
M1
A1
[9]
12.
METHOD 1
20 + 10bi = (1 – bi)(–7 + 9i)
20 + 10bi = (–7 + 9b) + (9 + 7b)i
Equate real and imaginary parts
(M1)
A1A1
(M1)
EITHER
–7 + 9b = 20
b=3
(M1)A1
OR
10b = 9 + 7b
3b = 9
b=3
(M1)A1
METHOD 2
=
(2  bi)(1  bi)  7  9i

(1  bi)(1  bi)
10
2  b 2  3bi
 7  9i
10
1 b
Equate real and imaginary parts
2  b2
7

Equation A
2
10
1 b
2

(M1)
A1
(M1)
13
3b
1 b
2

9
Equation B
10
From equation A
20 – 10b2 = –7 – 7b2
3b2 = 27
b = ±3
From equation B
30b = 9 + 9b2
3b2 – 10b + 3 = 0
By factorisation or using the quadratic formula
1
b=
or 3
3
Since 3 is the common solution to both equations b = 3
A1
A1
R1
[6]
13.
METHOD 1
since b > 0
 arg(b + i) = 30°
1
= tan 30°
b
b=
3
(M1)
A1
M1A1
A2
N2
METHOD 2
arg(b + i)2 = 60°  arg(b2 – 1 + 2bi) = 60°
2b
= tan 60° = 3
2
(b  1)
3b 2  2b  3 = 0
( 3b  1)(b  3 ) = 0
since b > 0
b= 3
M1
M1A1
A1
(M1)
A1
N2
[6]
14.
(a)
(b)
(c)
(d)
1 2i
e
z = 2 i
e
1 i
e
z=
2
(M1)
A1
1
2
│z│< 1
│z│ 
Using S∞ =
N2
A2
AG
a
1 r
S∞ =
e i
1
1  e i
2
(i)
S∞ =
e i
cis 

1
1
1  e i 1  cis 
2
2
(M1)
A1
N2
(M1)
14
cos   i sin 
(A1)
1
1  (cos   i sin  )
2
1 2i 1 3i
Also S∞ = eiθ + e  e  ...
2
4
1
1
= cis θ + cis2  cis3  ...
2
4
1
1
1
1

 

S∞ =  cos   cos 2  cos 3  ...  i sin   sin 2  sin 3  ...
2
4
2
4

 

(ii)
(M1)
A1
Taking real parts,




1
1
cos


i
sin


cos   cos 2  cos 3  ...  Re
 1

2
4
 1  (cos   i sin  ) 
 2



1
1

1  cos   i sin  
(cos   i sin  )


2
2

= Re
1
1
 1
  1

 1  cos   i sin   1  cos   i sin   
2
2
  2

 2
1
1
cos   cos 2   sin 2 
2
2
=
2
1
 1

2
1  cos    sin 
4
 2

1

 cos   
2

=
1
1  cos   (sin 2   cos 2  )
4
(2 cos   1)  2
4(2 cos   1)

=
(4  4 cos   1)  4 2(5  4 cos  )
=
4 cos   2
5  4 cos 
A1
M1
A1
A1
A1
A1AG
N0
[25]
15.
 2  4  16
 1  i 3
z=
2
(a)
i
–1 + i 3  re  r = 2
3 2π

1
3
θ = arctan
M1
A1
A1
–1 – i 3 = reiθ  r = 2
3
2π

1
3
θ = arctan
   2e
 
i
2π
3
2π
i
2e 3
A1
A1
A1
(b)
15
A1A1
(c)
cos nθ + i sin nθ = (cos θ + i sin θ)n
Let n = 1
Left hand side = cos 1θ + i sin 1θ = cos θ + i sin θ
Right hand side = (cos θ + i sin θ)1 = cos θ + i sin θ
Hence true for n = 1
Assume true for n = k
cos kθ + i sin kθ = (cos θ + i sin θ)k
 cos(k + 1)θ + i sin(k + 1)θ = (cosθ + i sin θ)k(cos θ + i sin θ)
= (cos kθ + i sin kθ)(cos θ + i sin θ)
= cos kθ cosθ – sin kθ sin θ + i(cos kθ sinθ + sin kθ cosθ)
= cos(k + 1)θ + i sin(k + 1)θ
Hence if true for n = k, true for n = k + 1
However if it is true for n = 1
 true for n = 2 etc.
 hence proved by induction
i
3
8e i2π
3


2
e
4π
i
2
3
M1A1
M1
M1A1
A1
A1
R1
4π
(d)
A1
4e
4π
4π
 2i sin
= 2 cos
3
3
2
i 3
 1  i 3
=  2
2
2
(M1)
A1A1
(e)
a3 = 8ei2π
β3 = 8e–i2π
Since e2π and e–2π are the same α3 = β3
(f)
EITHER
α = –1 + i 3
β = –1 – i 3
α* = – 1 – i 3
β* = – 1 + i 3
A1
A1
R1
A1
αβ* = (–1 + i 3 ) (–1 + i 3 ) = 1 – 2 i 3 – 3 = 2 – 2 i 3
M1A1
βα* = (–1 – i 3 )(–1 – i 3 ) = 1 + 2 i 3 – 3 = –2 + 2 i 3
 αβ* + βα* = –4
A1
A1
OR
Since α* = β and β* = α
αβ* = 2e
i
2π
3
 2e
i
2π
3
 4e
i
4π
3
M1A1
16
i
2π
3
i
2π
i
4π
 2e 3  4e 3
4π
 i 4π

i
αβ* + βα* = 4 e 3  e 3 




4π
4π
4π
4π 

 i sin
 cos
 i sin
= 4 cos

3
3
3
3 

4π
1
 8    4
= 8 cos
3
2
βα* = 2e
i2
(g)
A1
A1
A1
πn
αn = 2 n e 3
This is real when n is a multiple of 3
i.e. n = 3N where N  +
M1A1
R1
[31]
17