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Mathematics
Worksheet
Complex Numbers
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
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1. What are Complex Numbers?
You should know that the solution of the quadratic equation ax 2  bx  c  0 is given
 b  b2  4ac
. The term b 2  4ac discriminates between the three
2a
possibilities:- if b2  4ac  0 the equation has 2 distinct real roots
by x 
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.
( a  jb)  (c  jd )  ( a  c )  j(b  d )
Addition:
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) 3z  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
1 j
(1  j )(1  5 j )
1 6 j  5 j2
46j
2
3
c)




  
j

1 5 j
(1  5 j )(1  5 j )
1  25
26
13 13
z
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 
( 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)
z
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  b2
5  a 2  b2
(i)
12  2ab
(ii)
3
From (ii) a 
6
sub into (i)
b
2
36
6
5     b2 
 b2
2
b
 
b
Multiply through by b 2
5b 2  36  b 4
Rearrange & factorise
b4  5b2  36  0
 (b2  9)(b2  4)  0
 b 2  9 or 4 giving b  3 j or  2
but b is real (stated above)
using a  6 gives
hence b  2
a  3 when b  2 & a  3 when b  2
b
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 z2 2   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

O
x
with the definition of x for real numbers.
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
This gives
cos  =
and sin  =
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  r2e 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
For example if 1   2 
then, in the range, we would write arg( z1z2 ) =
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)

, (iii)  
z
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
N
angle PON= 
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
 cos  = 1 , sin  = giving   3
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
z
12

4 cos 11
12
 j sin 11
12
(ii)

z


z

x



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 sin12
(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 in terms of
z
2
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 z3  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
1 2 j 3  ( j 3 )
2
 

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
, z3
2


1 j 3
2
1 j 3
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:
Let cos  j sin be a cube root of 1 then
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
cos 3  1  3  2m    2m 
3
   2k (where m, n & k are integers).

3
sin 3  0  3  n    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

j 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
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



x
3
2
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.
1
2
1
4
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 14n  j sin 14n
8
8
giving ω  ω 7  cos 2 n  j sin 2n  cos 14n  j sin 14n
8
8
8
8
10




But cos 14 n  cos 2n  2n  cos  2n  cos 2n
8
and
sin 14 n
8

 sin 2n 
2n
8
8
  sin
2n
8

8
8
 sin 2n
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 z5  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 z 4 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  
cos n 
j sin n  cos   j sin 
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  
1 3
3 4

r  2 and  
5 3
,
12 12
3

 2k 
,
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
1
2
r (cos  j sin ) is a fifth root of -1
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 z 4 , z2 and z3 .
z5
z4
The roots above are solutions of the equation z5  1
z5  1  z5  1  0  ( z  1)( z 4  z3  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  21
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
(iv) j
(ix) 0
(v) 1  2 j
(x)
11  2
5
5
j
(vi) 6  2 j  2 (3  j )
5
5
5
(xii)  j
(xi) 6  12 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 
12 
3. cos 5  1 3
(ii ) 1 cos 5  j sin 5
2
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


1
a  jb
1
2
5. z 2  a 2  b 2  2abj  z 2  a 2  b 2  z and 
 
z
2
a b
2
z
6. (i ) z1  3  j, z 2   3  j (ii ) z1  2, arg( z1 )  
6


12
1
a 2  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

2;
8 8
 , 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 29th November 2004
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14