1.
METHOD 1
Using factor theorem
Substituting z = 1 i into P(z)
 (6 + n) + (2m  2  n)i = 0
Equating both real and imaginary parts to zero
Hence m = 2 and n = 6
METHOD 2
Using Conjugate root theorem
Multiply (z + 1  i)(z + 1 + i) = z2 + 2z + 2
Let P(z) = (z2 + 2z + 2)(z  a)
2a = 8
a=4
Hence m = 2 and n = 6
(M1)
M1
A1
M1
A1A1
N2
M1
M1
(M1)
A1
A1A1
N2
[6]
2.
METHOD 1
r  2, θ  

 1 i 3

(A1)(A1)
3

3
2
3

 
  
 cos     i sin    
 3
 3 

3
1
cos   i sin  
8
1

8

METHOD 2
(1  3i )(1 
(M1)
A1
3i ) = 1  2 3i  3 (= 2  2 3i )
( 2  2 3i )(1  3i ) = 8
1
1


3
8
1  3i
METHOD 3
Attempt at Binomial expansion
(1  3i )3 = 1 + 3( 3i ) + 3 ( 3i )2 + ( 3i )3



1
1 3i 
3
M1
= 1  3 3i  9 + 3 3i
= 8
1

8
(M1)A1
(M1)(A1)
A1
M1
(A1)
(A1)
A1
M1
[5]
3.
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
x
Im w 
=
2
2
 y  1  2 xyi
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 1  4 x 2 y 2
2

y 1 x 2  y 2
2

2
(A1)

 y 2 1  4 x 2 y 2

M1
A1
(A1)
A1
1
Im w = 0  1  x2  y2 = 0 ie z = 1 as y  0
METHOD 2
w (z2 + 1) = z
w(x2  y2 + 1 + 2ixy) = x + yi
Equating real and imaginary parts
w (x2  y2 + 1) = x and 2wx = 1, y  0
1
x y2 1
Substituting w 
to give 

x
2 2x 2x
2x
1 2
x
or equivalent

y 1 
2x
2
x2 + y2 = 1, ie z = 1 as y  0
R1AG
(sin + i (1  cos))2 = sin2  (1  cos)2 + i 2 sin (1  cos)
Let  be the required argument.
M1A1

N0
(A1)
A1
M1A1
A1

(A1)
R1AG
[7]
4.
2 sin θ 1 cos θ 
tan =
M1
sin 2 θ  1 cos θ 
2 sin θ 1  cos θ 
=
2
1 cos θ  1 2 cos θ  cos 2 θ 
2 sin θ 1  cos θ 
=
2 cos θ 1  cos θ 
= tan
=
2
(M1)
A1
A1
A1
[7]
5.
(a)
EITHER
2
2 

w   cos
 i sin

5
5 

= cos 2 + i sin 2
=1
Hence w is a root of z5  1 = 0
5
5
(b)
(c)
(M1)
A1
A1
AG
OR
Solving z5 = 1
2
2
z = cos
n  i sin
n , n  0 ,1, 2 , 3 , 4.
5
5
2
2
n = 1 gives cos
which is w
 i sin
5
5
(w  1)(1 + w + w2 + w3 + w4) = w + w2 + w3 + w4 + w5  1
w w2  w3  w4
= w5  1
Since w5  1 = 0 and w  1, w4 + w3 + w2 + w + 1 = 0.
1 + w + w2 + w3 + w4 =
2
2
2 
2
2 
1  cos
 i sin
  cos
 i sin
 
5
5 
5
5 
2
2  
2
2 

 i sin
 i sin
 cos
   cos

5
5  
5
5 

2
2
4
4
1  cos
 i sin
 cos
 i sin

5
5
5
5
3
(M1)
A1
A1
M1
A1
R1
4
(M1)
2
6
6
8
8
M1
 i sin
 cos  i sin
5
5
5
5
2
2
4
4
1  cos
 i sin
 cos
 i sin

5
5
5
5
4
4
2
2
M1A1A1
cos
 i sin
 cos
 i sin
5
5
5
5
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.
4
2
A1
1  2 cos
 2 cos
0
5
5
Note: Correct methods involving equating real
parts, use of conjugates or reciprocals are
also accepted.
2
4
1
AG
cos
 cos

5
5
2
Note: Use of cis notation is acceptable throughout
this question.
cos
[12]
3