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Complex Numbers
Mathematics Worksheet
This is one of a series of worksheets produced to help you increase your confidence
in handling Mathematics. This worksheet contains both theory and exercises which
cover:1. Introduction
3. The four rules
5. Roots
2. Argand Diagram
4. Modulus-Argument form
There are often different ways of doing things in Mathematics and the methods
suggested in the worksheets may not be the ones you were taught. If you are
successful and happy with the methods you use it may not be necessary for you to
change them. If you have problems or need help in any part of the work then there
are a number of ways you can get help.
For students at the University of Hull
 Ask your lecturers.
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below.
 Access more Maths Skills Guides and resources at the website below.
 Look at one of the many textbooks in the library.
Web: http://libguides.hull.ac.uk/skills
Email: [email protected]
1. What are Complex Numbers?
You should know that the solution of the quadratic equation ax 2  bx  c  0 is given
by x 
 b  b 2  4ac
. The term b 2  4ac discriminates between the three
2a
possibilities:- if b2  4ac  0 the equation has 2 distinct real roots
if b2  4ac  0 the equation has two equal real roots
if b2  4ac  0 the equation has no real roots.
Solving the equation x 2  2 x  17  0 using the formula gives
2  4  68 2   64 2  8  1
x


 1 4 1
2
2
2
So far these roots have been ignored and described as not real but by using j
to represent 1 we can write the solution to the above equation as x  1  4 j . Using
this technique means that we can solve ALL quadratic equations.
Mathematicians use i to represent 1 while many others use j , as we shall here.
Note as j   1 then j 2  1 , j 3  j 2  j  1  j   j , j 4  j 2  j 2  1  1  1
etc.
Numbers such as z  a  jb are called complex numbers with real part a and
imaginary part b . NOTE b is the imaginary part NOT jb .
2. Argand diagram
Complex numbers can be shown on co-ordinate axes - called an Argand diagram
with the real numbers on the x -axis and the imaginary numbers on the y -axis,
x  jy being shown as the point ( x, y ).
y
In the diagram P( x, y ) represents
the complex number z  x  jy
This implies that if a  jb  c  jd
then ( a, b) and (c, d ) must be the
same point which can only be true if
a  c and b  d .
P(x, y)
O
x
This leads to a very important statement:For two complex numbers to be
equal both the real and imaginary
parts must be equal.
1
3. The Four Rules
Addition, subtraction, multiplication and division of complex numbers are defined so
that they comply with normal algebra.
Addition:
( a  jb)  (c  jd )  ( a  c )  j(b  d )
Subtraction: ( a  jb)  (c  jd )  ( a  c )  j(b  d )
Multiplication:
( a  jb)(c  jd )  ac  jad  jbc  j 2 bd
 ( ac  bd )  j( ad  bc )
Note ( a  jb)( a  jb)  a 2  j 2 b 2  a 2  b 2 ie factors of a 2  b 2 are ( a  jb )( a  jb )
Also a(c  jd )  ac  jad
Conjugates
If z  a  jb then a  jb is called the conjugate of z and denoted by z * .
z  z *  a  jb   a  jb   2a and zz *  a  jb a  jb   a 2  j 2 b 2  a 2  b 2
ie both z  z * and zz * are real.
When solving quadratic equations with real coefficients the roots are always in
conjugate pairs.
For instance consider the quadratic equation 2 x 2  6 x  5  0
The solution is x 
6
 62  4  2  5
4

6 4 6 4j
3 1

  j
4
4
2 2
3 1
3 1
 j and x2    j are of the form a  jb, a  jb which are
2 2
2 2
conjugates. It can be proved that this is always the case as long as the coefficients
a, b and c are real.
The roots x1  
Note that, given equation is ax 2  bx  c  0 ,
 b  b 2  4ac
then it can be shown that the
2a
b
c
sum of the roots is  and the product .
a
a
This is true for both real and complex roots.
with solution x 
2
Division:
To divide two complex numbers multiply both the numerator and the denominator by
the conjugate of the denominator. This makes the denominator real as zz * is real
for all complex numbers z .
Divide ( 2  3 j ) by (1  2 j )
The conjugate of 1  2 j is 1  2 j hence
23j
(2  3 j )(1  2 j )
2  4 j  3 j  6 j2
27j6
4 7j

  


2
1 2 j
(1  2 j )(1  2 j )
1 4
5
5
1 4 j
Examples
1. Given that z  1  j ,   1  5 j
(i) Express the following in the form a  jb a) 3 z  2 * ; b) z 2 ; c)
(ii) If ( m  jn ) z   find the values of m and n
z

; d)  3
(i) a) 3 z  2 *  3(1  j )  2(1  5 j )  3  3 j  2  10 j  1  7 j
b) z 2  (1  j )2  1  2 j  j 2  1  2 j  1  2 j
c)
z


46j
1 j
(1  j )(1  5 j )
1 6 j  5 j2
2
3



  
j
1 5 j
(1  5 j )(1  5 j )
1  25
26
13 13
d)  3   15 j  3   15 j 15 j  2  (1  5 j )(1  10 j  25 j 2 )
 (1  5 j )(24  10 j )   24  10 j  120 j  50 j 2   74  110 j
or, using the binomial theorem,
 3  1  5 j 3  1  35 j   35 j 2  5 j 3  1  15 j  75 j 2  125 j 3  74  110 j
(ii) ( m  jn ) z  
( m  jn ) z   
Equate real parts
Equate imaginary parts
Solving the equations together gives
Alternative method write m  jn 

z
( m  jn)(1  j )  1  5 j
( m  n )  j( n  m )  1  5 j
m  n 1
nm 5
m  2, n  3
and use division method as in (i) (c)
2. Find the square root of 5  12 j
Let the square root of
Squaring
Equating real parts
Equating imaginary parts
5  12 j be ( a  jb) with a and b real
5  12 j  a 2  2 jab  b 2
5  a 2  b 2 (i)
12  2ab
(ii)
3
From (ii) a 
6
sub into (i)
b
2
36
6
 b2
5     b2 
2
b
 
b
Multiply through by b 2
5b2  36  b4
Rearrange & factorise
b 4  5b2  36  0
 (b 2  9)(b 2  4)  0
 b 2  9 or 4 giving b  3 j or  2
but b is real (stated above)
using a  b6 gives
hence b  2
a  3 when b  2 & a  3 when b  2
so the square roots of 5  12 j are (3  2 j ) and (3  2 j )
Exercise 1
1. Given z1  1  j, z2  2  j, z3  3 j, z4  1  j find
i  3z1  z2
vi 
( z1*)  z4
z2
x 
z2 2
z1  z3
ii 
z3  3z2
vii 
xi 
iii  2 z4  z3  z2
z33
z1( z2 *) z3 z4
viii  z2 z3 2
xii 
iv 
ix 
z1
z4
v 
3( z2 *)
z3
z12 z22   z1z2 2
m and n where m  jn z1  z4
2. Find the square roots of (i) 3  4 j , (ii) 21  20 j
4. Modulus - Argument form
a) If we represent the complex number
z  x  jy by the point ( x, y ) in the
Argand diagram, then the length OP is
called the modulus of z and is written
2
as r  z  x  y
P(x, y)
y
r
2
Note by putting y  0 , this is consistent
with the definition of x for real numbers.

O
x
The angle between the positive direction of the x -axis and OP (  radians) is called
the argument of z . The angle should always be given such that       and is
written as arg z or arg ( z ).
y
x
and sin  =
This gives
cos =
r
r
y
 x  r cos , y  r sin and tan 
x
4
hence z  x  jy  r cos  jr sin  r (cos  j sin ) which is called the modulus
argument form. [Note you can also have the Euler form z  re j . ]
Multiplication and division are a lot easier if the Complex numbers are in mod-arg
form but addition and subtraction are easier when they are in Cartesian form.
b) Multiplication & division using mod-arg form
a) Multiplication
Given z1  r1(cos1  j sin1 ),
z2  r2 (cos 2  j sin 2 )
then z1z2  r1(cos1  j sin1 )  r2 (cos 2  j sin 2 )
multiplying out gives
z1z2  r1r2 (cos1 cos 2  j 2 sin1 sin 2  j sin1 cos 2  j cos1 sin 2 )
 r1r2  cos1 cos 2  sin1 sin 2   j(sin1 cos 2  cos1 sin 2 )
 r1r2 cos 1   2   j sin 1   2  
This uses the trig identities for cos 1   2  and sin 1   2  that you should know.
In the Euler form z1  r1e j1 , z 2  r2 e j 2 , giving
z1 z 2  r1e j1  r2 e j 2  r1 r2 e j1 e j 2  r1 r2 e j 1  2 
From z1z2  r1r2 cos 1   2   j sin 1   2   or from z1z2  r1 r2e j 1  2 
we have
and
| z1z2 |  r1r2  | z1 |  | z2 |
arg( z1z2 )  1   2  arg z1  arg z2
but you do need to check that the angle is in the correct range    1   2   !
7
3
then, in the range, we would write arg( z1z2 ) =
For example if 1   2 
2
2
Division:
In the same way we can show that
z1
z2

z1
z2
z 
arg 1   arg( z1 )  arg( z 2 )
 z2 
again it is important to check that the angle is in the correct range.
And
5
b) Value of z n
z  r(cos  j sin )
Given
Then, from above,
z 2  r 2 (cos 2  j sin 2 )
And
z 3  r(cos  j sin )  r 2 (cos 2  j sin 2 )
 r 3 (cos 3  j sin 3 )
De Moivre developed this further and proved that
z n  r n cos n  j sin n  for all values of n .
This is fairly obvious for positive integer values of n , but it can be proved for negative
values (see Example 2(a) below) and, with care, for fractional values.
Examples
1. Given z   2  j 2   1  j 3
(a) Express z and  in modulus-argument form
and then
(b) find (in modulus-argument form) (i)  z (ii)

z
, (iii)  
and hence
(c) find the values of sin 11 , cos 11 in surd form
12
12




1 (a) There are at least two methods for this
Draw a diagram:
Method 1 – the safest & best!
y
z 2 j 2
P
From the diagram, using
 2 , 2 
triangle PON, by Pythagoras,
OP = 2 and, by trigonometry
O
angle PON= 
N
x
4
hence angle POx = 3
4
this gives
similarly
Method 2

Let
then
cos =
4
  1  j 3  2 cos 3  i sin 3
equate real parts
and imaginary parts
Squaring and adding gives
hence

z   2  j 2  2 cos 3  j sin 3

4

z  r(cos  j sin )
 2  j 2  r(cos  j sin )
 2  r cos
2  r sin 
r2
2
, sin  = - 2
2
2
6
1
giving   3
 cos  = 1 , sin  = 2
2

4

so z   2  j 2  2 cos 3  j sin 3 as before
4
4
y
x 2  y 2 and tan  but you really
x
need the diagram to be sure you have the correct value for  .
y
(b) (i)  z   z  2  2  4
Note from section 4 we could have said r  z
arg( z )  arg  arg z  3    13 .
4
3


12
But the argument of a complex number
cannot be greater than  hence from the
diagram arg( z )  11

12
11


z  4 cos
 j sin 11
12
12

x



(ii)

z


z

2
1
2
 3  5
 
arg   arg  arg z  

3
4
12
z

 5
 5 

 1 cos
 j sin
hence

z
12
12 

(iii)
 
 2  2 |  | 4; arg  2  2 arg 
2
2 

 j sin
hence  2  4 cos

3
3 

(c)
but
2
3
 11
 11 

 j sin
from part (i)  z  4 cos

12
12 


z  2 j 2
 1  j 3   
=  2 (1  3 )  j 2 (1  3 )
comparing (A) & (B)

2 2 3 j
(A)

2 2 3

(B)

11  j sin 11 =  2 (1  3 )  j 2 (1  3 )
 z  4 cos 12
12
equating real and imaginary parts:
 11
real: 4 cos 
 12
  11  
cos

 12 

 11
   2 (1  3 ) imaginary: 4 sin  12


2 (1  3 )
2 (1  3 )
  11 
; sin

4
4
 12 

  2 (1  3 )

7
Example 2. Use De Moivre’s Theorem to, simplify
cos 5  j sin 5
1
(a)
(b) cos 3  j sin 3 4 (c)
6
cos   j sin 
cos   j sin 

3
3

(a)
1
cos   j sin 
cos   j sin 


cos   j sin   cos   j sin   cos   j sin  
cos2   j 2 sin2 


cos   j sin 
cos2   sin2  


cos   j sin 
 cos   j sin 
1
If De Moivre’s theorem works for negative indices we’d get
1
 cos    j sin    cos   j sin 
cos   j sin 
But this is the same as above, so De Moivre’s theorem works for negative indices.
(b) By De Moivre’s Theorem cos 3  j sin 3 4  cos 12  j sin 12
(c) Using De Moivre’s theorem
cos 5  j sin 5
cos 5  j sin 5 cos 5  j sin 5


6
6  j sin 6
cos 2  j sin 2


cos
cos  j sin
3
3

3
3

 cos 5  j sin 5 cos 2  j sin 2 1
 cos 5  j sin 5 cos  2   j sin  2  
 cos 3  j sin 3
Exercise 2
1. Express each of the following in modulus  argument form giving
your answers as fractions of  except in part (v) :
i  1 
j ii   1  j 3
iii   1 iv 
12  2 j
v  3  4 j vi  
j
2. Using the answers from question 1, express the following

in mod  arg form i  1  j   1  j 3

ii   1 
j 3
12  2 j
3. Using your answers from question 2, express cos
  as a fraction
5
12
4. Use De Moivre' s Theorem to simplify the following
i  (cos 2 
j sin 2 )(cos 5  j sin 5 )
(ii ) (cos 3  j sin 3 ) 4
(iv )
(iii ) (cos 2  j sin 2 )(cos 3  j sin 3 ) 2
(cos 5  j sin 5 )(cos 2  j sin 2
(cos 3  j sin 3 ) 2
(v )
1
cos 3  j sin 3
8
vi   cos 3 
6

3

j sin 
3
2


cos   j sin  
4
3
3
( vii ) cos   j sin 
4
5. Given the complex number a  jb , where a and b are real numbers, find z 2 and
1
z
2
in terms of a and b . Verify that z 2  z and 1  1 .
z
z
6. (i) By writing z  a  jb , (with a and b real), solve the equation z 2  2  2 3 j
(ii) Express z and 2  2 3 j in modulus-argument form and verify that
arg( 2  2 3 j )  2  arg z , 2  2 3 j  z 2
5. Roots of Unity and Roots of Complex Numbers
If z 3  1 then
z3  1  0
factorising
( z  1)( z 2  z  1)  0
hence
z  1 or z 2  z  1  0
1 1 4
2
Solving z 2  z  1  0 leads to
z
The 3 cube roots of unity are
z1  1, z2 
But z1 2  

1 j 3 
2
1 j
and z 2 2  
 2
2
 

3
2
1 2 j 3 ( j 3 )
 

2
4
1 2 j 3 ( j 3 )
4

 2 2 j 3
4
2


2 2 j 3
4
1 j 3
2

1 3
2

1 j 3
2
1 j 3
1 j 3
, z3 
2
2
 z2
1 j 3
2

 z1
The roots are often written as 1,  and 2 where 3  1
And as z   is a solution of z 2  z  1  0 then 2    1  0
also z  2 is a solution so 1  2  4  0  1  2    0 as 3  1
In Modulus-Argument form:
cos  
j sin  3  1
by De Moivre' s Theorem this gives
cos 3 
j sin 3   1
equating real and imaginary parts gives
cos 3  1, sin 3  0
Let cos  j sin be a cube root of 1 then
cos 3  1  3  2m    2m 
3
2 k
    3 (where m, n & k are integers).
n

sin 3  0  3  n   

3 
giving
 
2
3
, 0, 2 in the range -     
3
9
The values, outside the range       , are repeats of the values you already
have. For example   4 is equivalent to   2 etc.
3
3
So the 3 cube roots of unity are
 
 
 
 
cos 2  j sin 2    1 
3
3   2

cos  0   j sin  0    1,
cos 2  j sin 2     1 
3
3   2

which is the same as before (mathematics is consistent)!
j 3
j 3
2



y

If the 3 roots are put on an Argand Diagram the
lines from the origin to the 3 points representing
the complex roots are at angles of 2 or 120o


2
3
x
to each other. Points representing  and  are
reflections of each other in the x -axis as they
are conjugates.
Note the labels  and 2 can be interchanged.
2
4
1
2
1
4
If z  1 then z  1  0 which factorises into
( z 2  1)( z 2  1)  0 giving ( z  1)( z  1)( z  j )( z  j )  0
so the 4 fourth roots of 1 are 1, -1, j ,  j and on the
Argand diagram we have the points representing the
4 fourth roots at (1, 0), (-1, 0), (0, 1) and (0, -1)
and the lines from the origin to the points are at
angles of 90o to each other.
x
-1
1
y
-1
Examples
1. If  is a complex cube root of 1, simplify (1  2 )(1  )
If  is a complex cube root of 1 then 1    2  0 and 3  1
Hence 1 2   and 1   2
giving (1  2 )(1  )  ( )  ( 2 )  3  1
Or: multiply out: (1  2 )(1  )  1    2  3
 (1    2 )  3  0  1  1.
2. If  is any complex eighth root of 1 show that   7 is real
In mod-arg form 1  cos0  j sin0 or cos(2n )  j sin(2n ) ( n an integer)
If  is an eigth root of unity then 8 = 1 = cos(2n )  j sin(2n )
by De Moivre’s Theorem
  cos 28n  j sin 28n for n  0 to 7
Or, to get the argument correct, take n as –3 to 4, but for all values of n :
ω  cos 2 n  j sin 2 n
8
8

ω 7  cos 14 n  j sin 14 n
8
8
giving ω  ω 7  cos 2 n  j sin 2 n  cos 14 n  j sin 14 n
8
8
8
8
10




But cos 14 n  cos 2n  2n  cos  2n  cos 2n
8
and
sin 14 n
8

 sin 2n 
2 n
8
8
  sin
2 n
8
8
   sin
8
2 n
8
hence   7  cos 28n  j sin 28n  cos 28n  j sin 28n  2cos 28n which is real.
3. Find the 5 fifth roots of unity and put them on an Argand diagram.
Let z  cos  j sin be a fifth root of 1 then z 5  1
or cos   j sin 5  1  cos 5  j sin 5  1
 cos 5  1 and sin 5  0
 5  2n
where n is an integer
Solving for  in the range - <    to
give distinct values of cos + j sin
gives   45 , 25 , 0 , 25 , 45
y
z1
z2
z5
and the 5 fifth roots of 1 are
z1  cos 2  j sin 2 ,
x
 5

5
z 2  cos 4  j sin 4 ,
5
5
4


z 3  cos
 j sin 4 
5
5
2



z 4  cos
 j sin 2 
5
5
z3
z4
z 5  1,
Note that z1 and z4 are conjugates as
are z2 and z3
6. The nth roots of  cosα  jsinα 
Assume that cos   j sin   is an n th root of cos   j sin  
Then
cos  
j sin n  cos   j sin  
cos n  j sin n   cos   j sin  
cos n  j sin n  cos   j sin   cos  2 p   j sin  2 p 
 
giving
  2 p
n
where p is an integer
taking p  1, 2, 3, 4, ... n gives n roots (though you need to check that      )
Hence there are n distinct n th roots of  cos   j sin  , with successive arguments
differing by
2
n
where the one root has argument

n
.
It can further be shown that the n distinct n th roots of s cos   j sin   have
modulus n s and arguments as above.
11
Examples
1. Find the cube roots of  2  j 2 and mark the point and its cube roots on an
Argand diagram.
On page 6 example 1 showed that

 2  j 2 = 2 cos 3  j sin 3
4
4
let
r(cos  +j sin ) be a cube root of  2  j 2
then
r3(cos  +j sin )3 = r3(cos 3 +j sin 3) = 2 cos 3  j sin 3
 4

4
= 2 cos 3  2k  j sin 3  2k


r  3 2 and  
3
r  2 and  

1 3
3 4

 2k 
5 3
,
12
12
,
11
12


4
4


for integer values of k
3  8k 
12
for      
So there are 3 distinct cube roots of
 2  j 2 as shown on the diagram. A is the
point  2  j 2 .
A
z2
z3
  12   12 
z 2  3 2 cos   j sin 
4
4
z 3  3 2 cos11   j sin11 
12
12
z1  3 2 cos 5  j sin 5 ,
O
z1
The points lie on the circumference of a circle
of radius 3 2
and the angle between the lines z1, z2 etc. are
8
12

2
3
or 120o.
Note as  2  j 2 is not real the roots are not in conjugate pairs.
2. Find the 5 fifth roots of –1 and hence prove that cos 35  cos 5 
If
then
r (cos  j sin ) is a fifth root of -1
1
2
r 5 (cos  j sin )5  r 5 (cos5  j sin5 )  1[cos   j sin  ]
 1[cos(  2k )  j sin(  2k )]
for integer values of k

r 5  1 and cos5  j sin5  [cos(  2k )  j sin(  2k )]

r  1 and  

r  1 and
  2k  
5
1
1 2k 
3   3
,
, ,
,

5
5 5 5
5
for      
12
The 5 fifth roots of -1 are
z1  cos   j sin  , z 2  cos 3  j sin 3 ,
 5

 5
 z3  1,
5
5
z 4  cos 3  j sin 3 , z 5  cos   j sin  
5
5
5
5
y
z2
z1
z3
x
Note that as -1 is real then the roots are in conjugate
pairs z1 and z4 , z2 and z3 .
z5
z4
The roots above are solutions of the equation z 5  1
z 5  1  z 5  1  0  ( z  1)( z 4  z 3  z 2  z  1)  0
The complex roots are the solutions of z 4  z 3  z 2  z  1  0
Substitute the root z1  cos
   j sin 


5
5
cos 5  j sin 5 4  cos 5  j sin 5 3  cos 5  j sin 5 2  cos 5  j sin 5   1  0
Hence by De Moivre’s Theorem
cos 4  j sin 4  cos 3  j sin 3  cos 2  j sin 2  cos   j sin   1  0
5
5

5
5

5

5
Using cos   cos(   ) and sin  sin(   ) gives


5

5


 cos   j sin   cos 3  j sin 3  cos 3  j sin 3  cos   j sin   1  0
5
5
5
5
5
5
5
5
3
3
3
3







 cos  j sin  cos
 j sin
 cos
 j sin
 cos  j sin   1  0
5
5
5
5
5
5
5
5
3


 2 cos  2 cos  1  0
5
5
cos 5  cos 35 
1
2
Exercise 3
1. Use De Moivre’s Theorem to find the (i) square roots, (ii) the cube roots of
(a) 2 (1  j )
(b) j
(c) 32( 3  j )
2. If  is a complex cube root of unity simplify (1  6)(1  62 )
3. If  is a complex fifth root of unity simplify (1   4 )(1   2 )(1   )
4. If  is one of the complex cube roots of –1 show that the roots can denoted by –
1,  and 2 . Prove that   2  1.
5. Find the 7 seventh roots of –1 and prove that cos   cos 3  cos 5
7
7
7

1
2
13
ANSWERS
Exercise 1
1. (i) 5  2 j (ii) 6 (iii) 4  6 j
(vii) 27 j
(viii) 9( 3  4 j )
2. (i) 1  2 j and 1  2 j
(v) 1  2 j
(iv) j
(ix) 0
(x)
11  2 j
5
5
(vi) 6  2 j  2 (3  j )
5
5
5
(xi) 6  12 j
(xii)  j
(ii) 5  2 j and 5  2 j
Exercise 2
1. (i ) 2 cos   j sin 
  4   4  (ii) 2cos23   j sin23  (iii) 1cos   j sin 
(iv ) 4cos   j sin  (v ) 5cos   j sin   where   tan 1 4  (vi ) cos   j sin 
6
6
3
2
2
 12 
12 
2. (i ) 2 2 cos 5  j sin 5
 6 
 6 
(ii ) 1 cos 5  j sin 5
2
12 
3. cos 5  1 3
2 2
4. (i ) cos 7  j sin 7 (ii ) cos12  j sin12 (iii ) cos 8  j sin 8 (iv ) cos  j sin



(v ) cos( 3 )  j sin( 3 ) (vi ) cos 2  j sin 2  1 (vii ) cos 17  j sin 17


5. z 2  a 2  b 2  2abj  z 2  a 2  b 2  z
2
and
1
a  jb
1



2
2
z a b
z
6. (i ) z1  3  j, z 2   3  j (ii ) z1  2, arg( z1 )  
6


12
1
a2  b2

12

1
z
z 2  2, arg( z 2 )  5
6
2  2 3 j  4, arg 2  2 3 j  
3
Exercise 3
1. In each case the modulus and the relevant arguments are given.
Square roots: (a) 2 ;  , 7 (b) 1;  , 3 (c) 8;  , 11
Cube roots:(a)
2. 31
3

8 8
2 ;  , 3 , 7 (b)
12 4 12
4
1;

4
 , 5 , 
6 6 2
12
12
(c) 4;  , 13 , 11
18
18
18
3. 3 1   2  1    2
4. Roots are 1, 1  1 2j 3 ,  2  1 2j



3

 12
5. Roots are cos   2n  j sin   2n for n  3 to 3
7
7
We would appreciate your comments on this worksheet,
especially if you’ve found any errors, so that we can improve it for
future use. Please contact the Maths tutor by email at
[email protected]
Updated 22nd June 2012
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alternative format on request. Telephone 01482 466199
© 2009
14