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Further complex numbers
Multiplication and division in polar
form
If s and t are complex numbers with s  p, arg s  
and t  q, arg t   , then
st  p (cos   i sin  )  q(cos   i sin  )
 pq(cos  cos   i sin  cos   i cos  sin   i 2 sin  sin  )
 pq   cos  cos   sin  sin    i  sin  cos   cos  sin   
 pq  cos(   )  i sin(   )  .
So st  p q and it may be true that arg( pq)  arg p  arg q.
Polar form
It is readily seen the x  r cos  and y  r sin  .
Hence, z  x  yi
 r cos    r sin   i
 r  cos   i sin   .
The non-zero complex number z can be written in modulus - argument form,
or polar form, as z  r  cos   i sin   , where r  z  0 is the modulus
and   arg z, with       , is the argument.
Examples
Write in modulus-argument form z  i, z  2, z  2  i and z  1  i.
z  i  r  1,  

2
z  2  r  2,   



 z  1 cos  i sin  .
2
2

 z  2  cos   i sin   .
z  2  i  r  5,   2.677  z  5  cos 2.677  i sin 2.677  .
z  1  i  r  2,   
3
4

 3
 z  2  cos  
 4


 3

i
sin



 4

 .

Multiplication and division in polar
form
If s and t are complex numbers with s  p, arg s  
and t  q, arg t   , then
st  p (cos   i sin  )  q(cos   i sin  )
 pq(cos  cos   i sin  cos   i cos  sin   i 2 sin  sin  )
 pq   cos  cos   sin  sin    i  sin  cos   cos  sin   
 pq  cos(   )  i sin(   )  .
So st  p q and it may be true that arg( pq)  arg p  arg q.
Multiplication and division in polar
form
The rules for multiplication and division
in modulus-argument from are
st  s t ,
arg( st )  arg( s)  arg(t )  k (2 ),
s
arg    arg s  arg t  k (2 ),
t
where in each case the number k ( 1, 0,1)
is chosen to ensure that the argument lies
in the interval       .
s
s
 ,
t
t
De Moivre’s Theorem
Note that
 cos   i sin  
2
  cos   i sin   cos   i sin  
 cos 2  i sin 2 .
De Moivre's Theorem:
n  , z  r  cos   i sin    z n  r n  cos n  i sin n  .
When n is not an integer, then r n  cos n  i sin n  is only
one of the possible values.
At A-Level, you are expected to be able to prove this statement
for positive integers. Use induction!
Examples
(i) Find the value of 1  i  .
10
(ii) Find the value of
1
 4  4i 
3
.
(iii) Find an expression for cos 6 and sin 6 .
Do Exercise 15A, p.334
Do Q1-Q15, pp.352-354
Some useful identities
If z  cos   i sin  then
and
so that
z n  cos n  i sin n
1
 cos n  i sin n
n
z
1
z  n  2 cos n
z
1
z n  n  2i sin n
z
n
As a special case, when n  1
we have
1
z
 2 cos 
z
1
z   2i sin 
z
Example : Show cos 4  
Show sin 6  
1
 cos 4  4 cos 2  3 .
8
1
10  15cos 2  6 cos 4  cos 6  .
32
Exponential form of a complex
number
ei
 i 
 1  i 
2!
2
 i 

3

3!
 
3 5
  i    
3! 5!
 
 2 4
 1   
2! 4!

 cos   i sin  .



So rei  r  cos  i sin  .
This is called the exponential form of a complex number.
De Moivre's Theorem now becomes
n  ,

rei

n
 r n e i n .
Note that ei   1  0
What is i i ?
Complex roots
z n  r  cos   i sin   has the n distinct roots

   2 k 
   2k  
z  n r  cos 

i
sin


 ,
n 
n 



where k  0,1, 2, n  1.
In exponential form:
z  re
n
i
has the n distinct roots
z  n re
i
Example 1 : Solve z 3  1.
  2 k
n
,
where k  0,1, 2,
n  1.
Example 2 : Solve z 4  1.
Solutions of complex equations
Solve the following:
Factorize the following:
16 z 4  ( z  1) 4
p( z )  z 4  1
z  4 3  4i  0
p( z )  z 5  1
( z  1)5  z 5
 2
Find the exact value of cos 
 5
3
Do Exercise 15B, pp.342-344

.

Useful identities in exponential form
If z  cos   i sin  then
and
so that
z n  cos n  i sin n
1
 cos n  i sin n
n
z
1
z  n  2 cos n
z
1
z n  n  2i sin n
z
n
As a special case, when n  1
we have
1
z
 2 cos 
z
1
z   2i sin 
z
ei n  ei n   2cos n
ei n  ei n   2i sin n
ei  e i  2 cos 
ei  e i   2i sin 
Summation of trigonometric series
2
sin
(10 )
k 1
Show that, provided cos   0,  (1) cos(2k  1) 
.
cos 
k 1
10
Show that
a1  a2 cos   a3 cos 3  a4 cos 4
1
,

  cos k 
40  32 cos 
k 0  2 
where a1 , a2 , a3 and a4 are to be determined.
3
k