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Complex Numbers
We have already looked at sets of number, ending with
, the real numbers. However if
we have x  1  0, there is no solution for x in the real numbers. The number set
2
was
created, the complex numbers, by defining i  1.
2
Thus a complex number takes the form
z  x  iy
where x is the real part Re( z )
y is the imaginary part Im( z ).
So z1  3  2i, z2  4  6i, z3  4i are all complex numbers, furthermore z3 is purely
imaginary.
From i 2  1, we have i  1 ,  z 2  1, z  i.
Now we can solve any quadratic equation by using the quadratic formula.
a) x 2  2 x  5  0
b) x 2  6 x  25  0
2  4  20
2
2  16

2
2  16  (1)

2
6  36  100
2
6  64

2
6  64  (1)

2
x
x
2  16i 2
2
2  4i

2
 1  2i
6  64i 2
2
6  8i

2
 3  4i


c) x 2  2 x  6  0
2  4  24
2
2  20

2
2  20  ( 1)

2
x
2  20i 2
2
2  i 2 5

2
 1  i 5

The Arithmetic of Complex Numbers
Addition/Subtraction
To add or subtract complex numbers we simply perform the required calculation with the
real part & with the imaginary part.
Examples
If z1  4  3i and z2  1  4i evaluate
① z1  z2
 (4  (1))  (3  (4)i
 3  7i
② z1  z2
③ z2  z1
 (4  (1))  (3  (4)i
 5i
1
 (1  4)  (4  (3)i
 5  i
Questions
Find the sum of
(a) z1  3  i & z2  2  5i
(b) z1  3  2i & z2  4  4i
(c) z1  2i & z2 
1 1
 i
2 3
Find z1  z2 when
1 3
 i
2 2
(d) z1  10  2i & z2  4  3i
(e) z1  1  2i & z2 
(f) z1  4  6i & z2  6  2i
(g) z1  4 3  i 8 & z2  3  i 2
Multiplication
To multiply two complex numbers we use the same method as we would use for multiplying
out ( x  1)(2 x  4) . We must remember here that i 2  1.
 1 1  3

 i   4i 
 2 4  2

① (4  3i )(2  i )
② (4  i )(3  2i )
 8  4i  6i  3i 2
 8  2i  3(1)
 12  8i  3i  2i 2
 11  2i
③
3
3
 2i  i  i 2
4
8
3 19
  i 1
4 8
1 19
  i
4 8

 12  11i  2
 10  11i
Questions
(a) (4  i )(3  i )
(b) (2  3i )(1  i )
(c) (5  6i)(3  2i)
(d) (2  i )(3  5i)
(e) (8  2i)(1  9i)
(f) (4  5i )(4  5i)
(g) (3  2i)(3  2i)
(h) (2  4i)(2  4i)
(g) & (h) show us important results. Here
z1  3  2i
z2  3  2i
&
z1  2  4i
z2  2  4i
.
z2 is said to be the complex conjugate of z1 and is denoted by z  if z  x  iy ,
z  x  iy.
This definition is required for division.
2
Division
To perform division it is best to consider it as a fraction. We find the conjugate of the
denominator and multiply the numerator and denominator by this. It is a little like the
method for rationalising a denominator.
Find z1  z2 when z1  2  3i and z2  1  i .
z2  1  i
z1 2  3i

z2 1  i
2  3i 1  i

1 i 1 i
2  2i  3i  3i 2

1  i2
1  5i

2
1 5
  i
2 2

Further Examples
①

3i
2  2i
(3  i )(2  2i )
(2  2i )(2  2i )
6  6i  2i  2i 2
4  4i 2
8  4i

8
1
 1 i
2

②
3  5i
3  i

③
(3  5i )(3  i)
(3  i )(3  i)
9  3i  15i  5i 2
9  i2
14  12i

10
7 6
  i
5 5


2  6i
4i
(2  6i )(4  i )
(4  i )(4  i )
8  2i  24i  6i 2
16  i 2
14  22i

17
14 22
  i
17 17

Questions
Find z1  z2 when
(a) z1  2  4i & z2  7  i
(b) z1  4  2i & z2  3  i
(c) z1  4  i & z2  3  2i
1
(d) z1  2  i & z2  6  4i
2
3
Square Roots
By equating real and imaginary parts we can achieve the square root of a complex number.
Clearly, as with the real numbers, there will be two solutions.
Examples
①
5  12i
Then x 2  y 2  5
36
 y2  5
2
y
36  y 4  5 y 2
If x  iy  5  12i
Then ( x  iy)2  5  12i
x2  2ixy  i 2 y 2  5  12i
( x2  y 2 )  (2 xy)i  5  12i
2 xy  12
6
x
y
y 4  5 y 2  36  0
( y 2  9)( y 2  4)  0
y2  4
y  2
y 2  9
so no solns
y
6
 3  3  2i
2
6
 3  3  2i
When y  2, x 
2
When y  2, x 
② 3  4i
Then x 2  y 2  3
4
 y 2  3
2
y
If x  iy  3  4i
Then ( x  iy)2  3  4i
x 2  2ixy  i 2 y 2  3  4i
2 xy  4
2
x
y
4  y 4  3 y 2
y4  3y2  4  0
( y 2  1)( y 2  4)  0
( x 2  y 2 )  (2 xy)i  3  4i
y2  4
y  2
y 2  1
so no solns
y
2
 1  1  2i
2
2
 1  1  2i
When y  2, x 
2
When y  2, x 
Questions
Calculate
(a) 15  8i
(b)
4
24  10i
(c) 9  40i
Argand Diagrams
An argand diagram is used to geometrically represent a complex number. If z  x  iy we
can plot the point P ( x, y ) in the complex plane.
OP the position vector represents z.
Im( z )
x
y
The length of OP is called the modulus
of z and is denoted z .
P
The angle is called the argument of z
Arg ( z )    2 n, n .
y

x
O
Re( z )
The principle Arg ( z ),     
In order to help us find the principle argument the following diagram may help.
 

  

Therefore each time we are dealing with a complex number
question, an Argand diagram will help us pick the correct
principle argument, so it should always be drawn.
Once we know the modulus and principle argument we can
change the number from Cartesian form ( x  iy ) to another
form known as polar form.
Modulus
Argument
From Pythagoras:
From Trigonometry:
z
y
y
tan  
y
x

x
x
r  z  x2  y 2
Arg z      
The polar form, mentioned before, is then created as follows
x
y
cos  
sin  
y
r
r
r
y  r sin 
x  r cos

x
Thus z  x  iy ;
z  r cos  ir sin  , i.e.
5
z  r (cos   i sin  )
POLAR FORM
Examples
For each of the following complex numbers, find the modulus, argument write in polar form.
tan  
r  z  12  12
① 1 i

4

Arg z 
4

r 2
Im( z )
1

1
1
1
Re( z )
Polar Form z  r (cos   i sin  )



 2  cos  i sin 
4
4

tan  
r  z  12  12
② 1 i

4

Arg z 
4

r 2
Im( z )
1

1
1
Re( z )
Polar Form z  r (cos   i sin  )
1

  
 
 2  cos 
  i sin 
 4 
 4




 2  cos  i sin 
4
4




(Recall cos( )  cos  ,sin( )   sin( ). )
r  z  02  32
③ 3i

3
0

Read from
2

Arg z 
Argand Diag.
2

r 3
Im( z )
3
tan  
Re( z )
Polar Form z  r (cos   i sin  )



 3  cos  i sin 
2
2

6
④ 2 3  2i
2 3


6
r4
Im( z )
2

2 3
2
tan  
r  z  (2 3)2  22
Arg z    
Re( z )
 


6
5
6
Polar Form z  r (cos   i sin  )
5
5 

 4  cos
 i sin

6
6 

Questions
Express the following complex numbers in polar form.
(a) z   2  i 2
(c) z  i
(b) z  1  i 3
(e) z  4
(d) z  1  i 3
Geometric Interpretations in the Complex Plane (Loci)
If we place a restriction on the modulus of a complex number we create a locus of the point
which moves on the complex plane.
Example
① Given z  x  iy draw the locus of the point which moves on the complex plane so that
(a) z  4
(b) z  4 .
 x2  y 2  4
z 4
x 2  y 2  42
x  iy  4
Equation of a circle, centre origin and radius 4.
y
a)
4
z 4
0
4
b)
4
x
7
y
z 4
4
x
② If z  x  iy , find the equation of the locus z  2  3 & draw on an Argand diagram.
 ( x  2) 2  y 2  3
z2 3
( x  2)2  y 2  32 A circle, centre (2,0) and radius 3.
x  iy  2  3
( x  2)  iy  3
y
z2 3
1
2
5
x
③ Find the equation of the locus when z 1  z  2i .
z  1  z  2i
x  iy 1  x  iy  2i
( x 1)  iy  x  ( y  2)i
( x  1)  y  x  ( y  2)
2
2
2
Im( z )
2
( x  1)2  y 2  x 2  ( y  2)2
x2  2 x  1  y 2  x2  y 2  4 y  4
1 2x  4  4 y
4 y  2x  3
1
3
y  x
2
4
3
4
3
2
Re( z )
Questions
a) Given that z  x  iy find the equation of each of the following loci and draw a
representation of them on an Argand Diagram.
i) z  5
ii) z  3
iii) z  1  4
iv) z  2i  3
v) z  2  z  i
vi) z  3  z  2i
8
The Fundamental Theorem of Algebra
The Fundamental Theorem of algebra stares that if P( z )  z n  an 1 z n 1  ...  a2 z 2  a1 z  a0
is a polynomial of degree n (with real or complex coeffiecients) then P ( z ) has n solutions
1 ,  2 , 3 ,... n in the complex numbers & P( z )  ( z  1 )( z   2 )...( z   n ) .
i.e. a cubic equation has 3 roots, a quartic equation has 4 roots…etc.
Also, if z  x  iy is a solution of P ( z ) , z will also be a solution (conjugate pairs are roots).
Examples
① Find all solution of the equation z 3  z 2  z  2  0.
2 1 1 1 2
0 2 2 2
1 1 1 0
 ( z  2) is a factor, z  2 is a solution.
f ( z)  ( z  2)( z 2  z  1)
1  12  4  1 1
2
1
3
z   i
2
2
z
1
3
1
3
,z   i
.
 Solutions are z  2, z    i
2
2
2
2
② f ( z )  2 z 3  5z 2  8z  20  0. If z  2i is a factor, find all roots of the equation.
z  2i is a factor  z  2i is also a factor (the conjugate)
( z  2i )( z  2i )
 f ( z )  ( z  2i)( z  2i)(
 z 2  4i 2
 ( z  4)(
2
 z2  4
)
)
Long Division will enable us to find the remaining factor.
z 2  4 2z3

2z3

5 z 2
5z 2
5z 2
2z
5
8 z 20
8z
20
20
0
If the remainder is not 0 you have made a mistake.
 f ( z )  ( z  2i)( z  2i )(2 z  5)
5
 z  2i, z  2i, z  .
2
9
③ f ( z )  z 3  z 2  z  15. If for f ( z )  0, z  3 is a root, find the two remaining roots.
3 1 1 1 15
0
1
3
2
6
5
 ( z  3) is a factor, z  3 is a solution.
15
0
2  22  4 1 5
2
z  1  2i
z
f ( z )  ( z  3)( z  2 z  5)
2
 Solutions are z  3, z  1  2i, z  1  2i.
④ Past Paper 2002.
Verify i is a solution of z 4  4 z 3  3z 2  4 z  2  0. Hence find all solutions.
To verify either do synthetic division, or sub in solution and check you get 0 out.
 z 4  4 z 3  3z 2  4 z  2
 i 4  4i 3  3i 2  4i  2
 1  4(i)  3(1)  4(i)  2
0
( z  i )( z  i )
 i is a solution, i is also a solution (conjugate)
 f ( z )  ( z  i )( z  i )(
 z2  i2
 ( z  1)(
2
 z2 1
z2 1 z4

z4


4 z 3
4z3
4z3
z2
3 z 2
 z2
2 z 2
4 z 2
4 z 2
2z2
2z2
2
2
4 z 2
4 z
)(
)(
)
)
 f ( z )  ( z  i)( z  i)( z 2  4 z  2)
0
4  42  4 1 2
2
z  2  2
z
 Solutions are z  i, z  i, z  2  2, z  2  2.
10
Multiplying & Dividing Numbers in Polar Form
If we are multiplying two numbers in polar form we multiply the moduli & add arguments.
If we are dividing two numbers in polar form we divide the moduli & subtract arguments.
(Remember to always bring the argument back into the principle argument in the range
     )
Examples






① z  3  cos  i sin  ; w  4  cos  i sin 
3
3
2
2



  
   
zw  3  4  cos     i sin    
3 2
 3 2 

5
5 

 12  cos
 i sin

6
6 

z 3
  
   
  cos     i sin    
w 4
3 2
 3 2 
3
  
 
  cos 
  i sin 
4
 6 
 6
3


  cos  i sin 
4
6
6









② z  2  cos  i sin  ; w  5  cos  i sin 
4
4
6
6



  
   
zw  2  5  cos     i sin    
4 6
 4 6 

5
5 

 10  cos
 i sin

12
12 

z 2
  
   
  cos     i sin    
w 5
4 6
 4 6 
2

 
  cos  i sin 
5
12
12 
Questions
Find zw &
z
if
w



z  2  cos  i sin 
3
3

(a)



w  4  cos  i sin 
3
3




z  4  cos  i sin 
2
2

(b)



w  2  cos  i sin 
3
3

11



z  5  cos  i sin 
4
4

(c)



w  2  cos  i sin 
8
8

De Moivre’s Theorem
By extending the results previous, we obtain the result commonly known as De Moivre’s
Theorem.
z n  r n (cos n  i sin n )
If z  r (cos   i sin  )
Again we must remember to ensure we return our answer to having the principle argument.
Examples



① If z  3  cos  i sin  find z 3 .
2
2

3
3 

z 3  33  cos
 i sin 
2
2 


  
   
 27  cos 
  i sin 

 2 
 2 




 27  cos  i sin 
2
2

Subtract 2 to return to the principle arg.
3
3 

4
 i sin
②If z  2  cos
 find z .
4
4 

z 4  24  cos3  i sin 3 
 16(cos   i sin  )
If we are given a complex number in Cartesian form, we can still use De Moivre’s Theorem
by converting it into polar form (Using an Argand diagram to help.)
Examples
① Find (1  i) 4 .
r  z  12  (1)2

r 2
Im( z )
tan  
1
1
1

4
Arg z   

Re( z )

1
12
3
4

4
Polar Form z  r (cos   i sin  )
z4 

 3 
 3  
 2  cos 
  i sin 

 4 
 4 

3
3 

 2  cos
 i sin

4
4 


② Calculate  3  i

3
4
 4(cos 3  i sin 3 )
 4(cos   i sin  )
by using De Moivre’s Theorem.
 3
r z 
r2
Im( z )
1

 3
 2   cos 124  i sin 124 
2
 12
1
3


6
tan  
Arg z    
Re( z )
 

5
5 

 2  cos
 i sin

6
6 

3i

(1  i )3
4
.
z1  3  i
r z 
 3
2
 12
tan  
1
3

6

Arg z 
6

r2
Im( z )
1
5
6
15
15 

 z  23  cos
 i sin

6
6 

5
5 

 8  cos
 i sin

2
2 




 8  cos  i sin 
2
2

Polar Form z  r (cos   i sin  )

③ Find

6

3 Re( z )



Polar Form z1  2  cos  i sin 
6
6

2
2 

z14  16  cos
 i sin

3
3 

13
z2  1  i
tan  
r  z  12  (1)2

4

Arg z 
4

r 2
Im( z )
1

1
1
Re( z )
1

 3
z23  2 2  cos 
 4

Polar Form z  r (cos   i sin  )

 
 2  cos 
 4


 3  
  i sin 


 4 

   
  i sin 


 4 
Leave the argument as negative at this stage.
z14
16 
 2  3  
 2  3

      i sin 

 cos 
3
z2
2 2
 3  4 
 3  4
8 
17
17 

 i sin
 cos

12
12 
2

 7 
 7
 4 2  cos 
  i sin 
 12 
 12

7
7 

 4 2  cos
 i sin

12
12 







Fractional Indices
Fractional Indices may also be considered here, but we must be careful. We already know if
1
we find z 2 we have 2 solutions; solving z 3 for z should give 3 solutions, etc.
To help achieve this we add multiples of 2 to the argument.

z 2  cos
Then




 2
z   cos   2k   i sin   2k  
3

3


3
 i sin

If we have
3
1
where k  0,1
(as this will give 2 solns)
14
By De Moivre’s Theorem


1 
 3  2 k
2
z  1  cos 
2




When k  0
z0  cos

6
 i sin

6




 3  2 k  
  i sin 

2


 



When k  1
7
7
z1  cos
 i sin
6
6
5
5
 cos
 i sin
6
6
 In general
1
1

   2 k
z n  r n  cos 
n



   2k  
  i sin 
 for k  0,1, 2,..., n  1.
n  


You will need to remember this formula.
Examples
3
3

 i sin
① Considering z 3  8  cos
4
4


 3

 4  2 k
z  8  cos 
3




1
3

 3  8k
 2  cos 
 12


 , find all possible results for z.


 3

 4  2 k
  i sin 
3





 3  8k  
  i sin 
  for k  0,1, 2

 12

When
k 0



z0  2  cos  i sin 
4
4

k 1
11
11 

z1  2  cos
 i sin

12
12 

k 2



 

19
19 

z2  2  cos
 i sin

12
12 


 5 
 5  
 2  cos 
  i sin 

 12 
 12  

5
5 

 2  cos
 i sin

12
12 

15



② Solve z 3  8  cos  i sin  .
3
3




 3  2 k
z  8  cos 
3




1
3

   6k
 2  cos 
9





 3  2 k
  i sin 
3





   6k
  i sin 
9





 


  for k  0,1, 2

When
k 0



z0  2  cos  i sin 
9
9

k 1
7
7 

z1  2  cos
 i sin

9
9 

13
13 

z2  2  cos
 i sin

9
9 


 5 
 5  
 2  cos 
  i sin 

 9 
 9 

5
5 

 2  cos
 i sin

9
9 

k 2
Nth Roots of Unity
Solving the equation z n  1 is a special case which gives solutions known as the nth roots of
unity. These & the other roots previously calculated can be drawn on an Argand Diagram
and evaluated geometrically.
Examples
① Solve z 3  1 to find the cube roots of unity ( z 3  1(cos 0  i sin 0) .
z 3  1(cos 0  i sin 0)
2k
2k 

z  1  cos
 i sin
 for k  0,1, 2.
3
3 

1
3
16
When
k 0
z0  1 cos 0  i sin 0 
( 1) for plotting
k 1
z1  cos
k 2
z2  cos
2
2
 i sin
3
3

1
3 
i
   
2 2 

4
4
 i sin
3
3
 2 
 2 
 cos 
  i sin 

 3 
 3 
2
2
 cos
 i sin
3
3

1
3 
i
   
2 2 

Im( z )
z1
z0
Re( z )
z2
② Solve z 4  1 to find the fourth roots of unity.
z 4  1(cos 0  i sin 0)
0  2k
0  2k 

z  14  cos
 i sin

4
4 

k
k
 cos
 i sin
for k  0,1, 2,3.
2
2
1
When
k 0
z0  1 cos 0  i sin 0 
k 1
 i
( 1) for plotting
k 2
z1  cos
z2  cos   i sin 
k 3
  1
z3  cos
 cos
  i 
17

2
 i sin

2
3
3
 i sin
2
2

2
 i sin

2
Im( z )
z1
z2
z0
Re( z )
z3
De Moivres’ Theorem & Multiple Angle Trigonometric Formulae
Best shown by example.
Examples
① Using De Moivre’s and the Binomial Theorem, express cos 4 as a polynomial in cos .
By De Moivre
 cos   i sin  
4
 cos 4  i sin 4
By Binomial Theorem
 cos   i sin  
4
 cos 4   4i cos3  sin   6i 2 cos 2  sin 2   4i 3 cos  sin 3   i 4 sin 4 
 cos 4   4i cos3  sin   6 cos 2  sin 2   4i cos  sin 3   sin 4 
  cos 4   6 cos 2  sin 2   sin 4    i  4 cos3  sin   4 cos  sin 3  
Equating Real Parts gives us
cos 4  cos 4   6 cos 2  sin 2   sin 4 
(Note, equating imaginary parts would give sin 3  4 cos3  sin   4 cos  sin 3  )
cos 4  cos4   6cos2  (1  cos2  )  (1  cos 2  )2
replacing all sin 
 cos   6cos   6cos   1  2cos   cos 4 
4
2
4
 8cos 4   8cos 2   1
18
2
② Find a formula for cos3 in terms of powers of cos by using De Moivre’s Theorem.
Hence express cos3  in terms of cos & cos3 .
By De Moivre
 cos   i sin  
3
 cos 3  i sin 3
By Binomial Theorem
 cos   i sin  
3
 cos3   3i cos 2  sin   3i 2 cos  sin 2   i 3 sin 3 
 cos3   3cos sin 2   i(3cos2  sin   sin3  )
Equating Real Parts gives us
cos 3  cos3   3cos  sin 2 
 cos3   3cos  (1  cos 2  )
 4 cos3   3cos 
3
Hence cos  
③ Express
cos 3  3cos 
4
sin 5
as a polynomial in sin  .
sin 
By De Moivre
 cos   i sin  
5
 cos 5  i sin 5
By Binomial Theorem
 cos   i sin  
5
 cos5   5i cos 4  sin   10i 2 cos3  sin 2   10i 3 cos 2  sin 3   5i 4 cos  sin 4   i 5 sin 5 
 cos5   5i cos4  sin   10cos3  sin 2   10i cos 2  sin 3   5cos  sin 4   i sin 5 
 cos5   10cos3  sin 2   5cos  sin 4   i(5cos 4  sin   10cos 2  sin 3   sin 5  )
Equating Imaginary Parts
sin 5  5cos 4  sin   10 cos 2  sin 3   sin 5 
sin 5
 5cos 4   10 cos 2  sin 2   sin 4 
sin 
 5(1  sin 2  ) 2  10(1  sin 2  ) sin 2   sin 4 
 5  10sin 2   5sin 4   10sin 2   10sin 4   sin 4 
 16sin 4   20sin 2   5
19
Past Paper Questions
2001
(a) Given that 1  cos   i sin  ,      , state the value of  .
(b) Use de Moivre’s Theorem to find the non-real solutions, z1 and z 2 , of the equation
z 3  1  0.
Hence show that z12   z2 and z22   z1.
(c) Plot all the solution of z 3  1  0 on an Argand diagram and state their geometrical
significance.
(1,5,2,3 marks)
2002
Verify that i is a solution of z 4  4 z 3  3z 2  4 z  2  0.
Hence find all the solutions
(5 marks)
2003
① Identify the locus in the complex plane given by z  i  2
(3 marks)
1
 cos   i sin  .
w
Use de Moivre’s Theorem to prove wk  w k  2cos k , where k is a natural number.
② Given that w  cos   i sin  , show that
Expand  w  w1  by the binomial theorem and hence show that
4
1
1
3
cos 4   cos 4  cos 2  .
8
2
8
(1, 3, 5 marks)
2004
Given z  1  2i, express z 2 ( z  3) in the form a  ib.
Hence, or otherwise, verify that 1  2i is a root of the equation
z 3  3z 2  5 z  25  0.
Obtain the other roots of this equation.
(2, 2, 2 marks)
2005
① Given the equation z  2iz  8  7i, express z in the form a  ib.
20
(4 marks)
② Let z  cos  i sin  .
(a) Use the binomial expansion to express z 4 in the form u  iv, where u and v are
expression involving sin  and cos .
(b) Use de Moivre’s theorem to write down a second expression for z 4 .
(c) Using the results of (a) and (b), show that
cos 4


 p cos 2   q sec2   r , where     ,
2
cos 
2
2
stating the values of p , q and r.
(3, 1, 6 marks)
2006
Express the complex number z  i 
1
in the form z  x  iy, stating the values of x
1 i
and y.
Find the modulus and argument of z and plot z and z on an Argand diagram.
(3, 4 marks)
2007
① Show that z  3  3i is a root of the equation z 3  18 z  108  0 and obtain the
remaining roots of the equation.
(4 marks)
② Given that z  2  z  i , where z  x  iy, show that ax  by  c  0 for suitable values
of a, b and c.
Indicate on an Argand diagram the locus of complex numbers z which satisfy
z 2  z i .
(3, 1 marks)
2008
Given z  cos   i sin  , use de Moivre’s theorem to write down an expression for z k in
terms of  , where k is a positive integer.
1
Hence show that k  cos k  i sin k .
z
Deduce expression for cos k and sin k in terms of z.
2
Show that cos 2  sin 2   
1 2 1
z  2  .
16 
z 
Hence show that cos2  sin 2   a  b cos 4 , for suitable constants a and b.
(3, 2, 3, 2 marks)
21
2009
(1  2i) 2
in the form a  ib where a and b are real numbers.
7 1
Show z on an Argand diagram and evaluate z and arg( z ).
Express z 
(6 marks)
2010
Given z  r (cos   i sin  ), use de Moivre’s theorem to express z 3 in polar form.
2
2 

Hence obtain  cos
 i sin
 in the form a  ib.
3
3 

Hence, or otherwise, obtain the roots of the equation z 3  8 in Cartesian form.
3
Denoting the roots of z 3  8 by z1 , z 2 , z3 :
(a) state the value z1  z2  z3 ;
(b) obtain the value of z16  z26  z36 .
(1, 2, 4, 3 marks)
2011
Identify the locus in the complex plane given by
z  1  3.
Show in a diagram the region given by z  1  3.
(5 marks)
2012
① Given that ( 1  2i ) is a root of the equation
z 3  5z 2  11z  15  0,
obtain all roots.
Plot all the roots on an Argand diagram.
(4, 2 marks)
22
② (a) Prove by induction that
(cos   i sin  )n  cos n  i sin n
for all integers n  1.

 

 cos  i sin 
18
18 
(b) Show that the real part of 
is zero.
4

 

 cos  i sin 
36
36 

11
(6, 4 marks)
2013
2
① Given that z  1  3i, write down z and express z in polar form.
(4 marks)
② Describe the loci in the complex plane given by:
(a) z  i  1;
(b) z  1  z  5 .
(2, 3 marks)
2014
(a) Express 1 as a complex number in polar form and hence determine the solutions to the
equation z 4  1  0.
(b) Write down the four solutions to the equation z 4  1  0.
(c) Plot the solutions of both equations on an Argand diagram.
(d) Show that the solutions of z 4  1  0 and the solutions of z 4  1  0 are also solutions of
the equation z 8  1  0.
(e) Hence identify all the solutions to the equation
z 6  z 4  z 2  1  0.
(3, 2, 1, 2, 2 marks)
2015
By writing z in the form x  iy :
(a) solve the equation z 2  z  4 ;
2


(b) find the solutions to the equation z 2  i z  4 .
2
23
(3, 4 marks)
2016
Let z  3  i.
(a) Plot z on an Argand diagram.
(b) Let w  az where a  0, a  .
Express w in polar form.
(c) Express w8 in the form ka n ( x  iy), k , x, y  .
24
(1, 2, 3 marks)