Complex Numbers
Section 3: Complex numbers and trigonometry
Notes and Examples
These notes contain subsections on:
 Using de Moivre’s theorem to find trigonometric identities
 Using complex numbers to sum real series
Using de Moivre’s theorem to find trigonometric identities
De Moivre’s theorem can be used to give a multiple angle formula in terms of
powers and to express powers of sine and cosine in terms of multiple angles.
This can be very useful in integration.
z n  z n
z n  z n
and cos n 
are
2i
2
used frequently. You should be happy with these identities and know how
they arise (they are proved on page 345).
In the example below, the identities sin n 
Example 1
(i)
Express sin 6  in terms of multiple angles
(ii)
Hence find  sin 6  d .
putting n = 1
Solution
(i)
z  cos  isin   sin n 
So
 2i 
6
sin 6    z  z 1 
z n  z n
 2i sin   z  z 1 .
2i
6
.
 z 6  6 z 5 z 1  15 z 4 z 2  20 z 3 z 3  15 z 2 z 4  6 zz 5  z 6
 z 6  6 z 4  15 z 2  20  15 z 2  6 z 4  z 6
 z 6  z 6  6  z 4  z 4   15  z 2  z 2   20
 2 cos 6  12 cos 4  30 cos 2  20
Therefore:

using z  z  2cos n for
n = 6, 4 and 2 respectively
n
n
64sin 6   2cos 6  12cos 4  30cos 2  20 .
20  2 cos 6  12 cos 4  30 cos 2
sin 6  
64
10  cos 6  6 cos 4  15cos 2

32
1/5
 sin
(ii)
6
 10  cos 6  6 cos 4  15cos 2 
 d
32


1 
1
6
15

 10  sin 6  sin 4  sin 2   c
32 
6
4
2

1

 60  sin 6  9sin 4  45sin 2   c
192
 d   
Using complex numbers to sum real series.
Consider the sum
n
n
 n 
 n
1    cos     cos 2  ...  
 cos (n 1)     cos n .
1
 2
 n  1
 n
This sum is not an arithmetic series, it is not a geometric series, nor is it a
binomial series. However, in Example 2 below, complex numbers will be used
to show that it is equal to
n


 2cos  cos .
2
2

n
Similarly, it is not immediately clear that the series
sin  sin 2 sin 3 sin 4
 2  3  4  ...
2
2
2
2
can be dealt with using the techniques that we have learnt so far. It is shown
in Example 3, using complex numbers, that the sum of this series is
sin 
5  4 cos 
In both examples, the strategy is the same:

introduce some complex terms so that the given summation becomes the
real or imaginary part of an arithmetic, geometric or binomial series. Then,
using the familiar formulae for these types of sums, find an expression for
the (usually complex) value of this new summation. In both examples
n
below, De Moivre’s Theorem, that  cos  isin    cos n  isin n , or
e 
i
n
 ein , is used.

manipulate this expression to express it as a single complex number. In
both of the examples below, some of this algebraic manipulation is dealt
with in the first part of the question.

equate real or imaginary parts to give the value of the original summation.
2/5
Example 2
(i) Using the identities 1  cos 2  2cos 2  and sin 2  2cos sin  , prove that
i

1  ei  2e 2 cos .
2
n
n
n
 
 


 n
(ii) Let C  1    cos    cos 2  ...  
 cos (n  1)     cos n .
1
 2
 n  1
 n
By considering C  iS , where
n
n
 n 
 n
S    sin     sin 2  ...  
 sin (n  1)     sin n
1
 2
 n  1
 n

n

show that C   2cos  cos .
2
2

n
gives 1  cos   2cos
Solution
(i) 1  ei  1  cos   i sin 
 2 cos 2


 2e cos
2

2

2

 2i cos sin
2
2
2

 

 2  cos  i sin  cos
2
2
2

i
2
The identity 1  cos 2  2cos

sin 2  2cos  sin 


gives sin   2 cos sin
2
2
The identity
2
 n

n
 n 
n
(ii) C  iS  1    cos     cos 2  ...  
 cos (n  1)     cos n 
 2
 n  1
 n
 1

 n 

n
 n 
n
 i   sin     sin 2  ...  
 sin  (n  1)     sin n 
 2
 n  1
 n
 1 

n
n
 1     cos   i sin       cos 2  i sin 2   ...
1
 2
 n 
n

  cos  (n  1)   i sin  (n  1)      cos n  i sin n 
 n  1
n
n
n
2
 1     cos   i sin       cos   i sin    ...
1
 2
 n 
n
n 1
n

  cos   i sin       cos   i sin  
 n  1
n
2
n
n
 n  i n 1  n  i n
 1    ei     ei   ...  
 e     e 
using de Moivre’s
1
 2
 n  1
n
theorem
 1  ei 
n
This is just the standard result
1  x 
n
 n  n
 n  n1  n  n
i
 1    x    x 2  ...  
 x    x , with x  e
 1  2
 n  1
 n
3/5
Using the answer to part (i) gives
C  iS  1  ei 
n
 i

  2e 2 cos 
2

n

  

 2n  cos  i sin   cos 
2
2 
2

n
n
n
n  


 2  cos
 i sin   cos  .
2
2 
2

n
n
n


Equating the real parts of the equation above gives C   2cos  cos .
2
2

n
Example 3
(i) Show that  2  ei  2  ei   5  4cos  .
sin  sin 2 sin 3 sin 4



 .....
2
22
23
24
cos cos 2 cos3 cos 4
By considering C  iS where C  1 



 ... ,
2
22
23
24
2sin 
show that S 
.
5  4cos 
(ii) Let S 
Solution
(i)  2  ei  2  ei   4  2ei  2e i  ei e i
 5  2  ei  ei 
 5  2  cos   i sin     cos( )  i sin( )  
 5  2  cos   i sin   cos   i sin  
 5  2  (2 cos  )
 5  4 cos  .
cos 
sin  cos 2
sin2 cos 3
sin3
i

i 2 
 i 3  ...
2
3
2
2
2
2
2
2
 cos   i sin    cos 2  i sin 2   cos 3  i sin 3 
 1 


  ...
2
22
23

 
 

(ii) C  iS  1 
 cos   i sin 
 1 
2

  cos   i sin    cos   i sin  

 ...

22
23

2
3
 cos   i sin    cos   i sin    cos   i sin  
 1 

 
  ...
2
2
2

 
 

2
3
4/5
2
3
e i  e i   e i 
 1
       ...
2  2   2 
1

 e i 
1   
 2 

This is just a geometric
series with first term 1 and
common ratio 
ei
2
2
2  e i
using the standard result that the sum of an infinite
geometric progression with first term a and common ration
e i
a


r
r is
, with a = 1 and
. Note r  1 is
2
1 r
required for the series to converge, check you are happy
e i
that this is the case for r  
.)
2
Using part (i):
2
2  ei
 i
 2  2  e 

i  
 i 
 2  e  2  e  .
4  2e i

5  4 cos 
4  2 cos   2i sin 

5  4 cos 
C  iS 
Equating imaginary parts gives S 
2sin 
.
5  4cos 
5/5