ONLINE: MATHEMATICS EXTENSION 2
Topic 2
2.3
COMPLEX NUMBERS
Powers and Roots of Complex Numbers
DE MOIVRE’S THEOREM
This is an important and useful theorem in complex number theory and can be derived
from the product of two complex numbers in polar and exponential form.
z  R ei   R  cos  i sin  
We can very easily compute the value of the complex number zn when z is expressed in
exponential form.
z 2   R ei   R ei    R 2 e
i  2 
 R 2 cos  2    i sin  2   
Multiplying by z once more gives


z 3  z z 2   R ei   R 2 ei  2   R 3 ei 3   R 3 cos  3   i sin  3  
so we can generalize the result
zn  Rn e
i n 
 R n cos  n    i sin  n   
If we take the special where R = 1, we obtain de Moivre’s theorem
cos   i sin  n  cos  n    i sin  n  
n
real
Application to trigonometric formulae
Multiple angle formula for the sine and cosine functions can be easily derived from
de Moivre’s theorem, for example sin  3  and cos  3 
cos  3   i sin  3    cos   i sin  
3
 cos3   3cos  sin 2   3 i sin  cos2   i sin 3 
  4 cos3   3cos    i  3sin   4sin 3  
sin 2   1  cos 2 
cos 2   1  sin 2 
Equating the real and imaginary parts
cos  3   4 cos3   3cos 
sin  3   3sin   4sin 3 
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We can also use de Moivre’s theorem to obtain expressions for cosn  and sin n  .
z  cos   i sin 
z n  cos   i sin    cos  n    i sin  n  
n
z 1  cos   i sin    cos   i sin 
1
z  z 1  2 cos 
z  z 1  2 i sin 
z n  z  n  2 cos  n  
z n  z  n  2 i sin  n  
For example, let n = 3
z  z   z
1 3
 2 cos  
3
3
 z 3   3  z  z 1 
z z 1  1
 2 cos  3   3  2 cos  
8cos3   2 cos  3   6cos 
cos3  
1
3
cos  3   cos 
4
4
PROOF BY INDUCTION



Prove
In proof by induction show that the statement is correct for n = 1.
Assume the statement is true for n.
Show that the statement is true for n+1 which is a statement with n replaced by
n+1.
cos   i sin  n  cos  n    i sin  n  
by induction
n  1 LHS  cos   i sin  RHS  cos   i sin 
statement is true for n = 1
 LHS  RHS
Assume that cos   i sin    cos  n    i sin  n   is true
n
LHS for n+1
LHS  cos   i sin  
n 1
 cos   i sin   cos   i sin  
n
 cos  n    i sin  n    cos   i sin  
 cos  n   cos   sin  n   sin    i sin  n   cos   cos  n   sin  
We can use the trigonometric identities
cos( A  B )  cos A cos B  sin A sin B
sin( A  B )  sin A cos B  cos A sin B
LHS  cos   n  1    i sin   n  1  
This is the statement with n replaced by n+1 on the RHS  statement is true
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ROOTS OF COMPLEX NUMBERS
Another important application of de Moivre’s theorem is to calculate the roots of a
number.
Suppose we require the nth roots (n = 1, 2, 3, … ) of z.
z  R ei 
We can add any integer multiple of 2 to  without changing the number
z  Re
i   2 k 
k is an integer
Then
1
zn  Re
   2 k 
i

 n 
and we allow k to take the values 0, 1, 2, … , (n-1).
Example
Find the fifth roots of z  3  i
Express z  x  i y in exponential form z  R e i 
x 3
z  2e
z
1
5
y 1
 
i 
6
1
 2e
2 e
5
 y
 


i   2 k 
6

 1  

 3 6
  a tan    a tan 
x
R  x2  y 2  2
k  0,1, 2,3, 4
  12 k 
i

 30 
Since k = 0, 1, 2, 3, 4 we have 5 distinct roots
k 0
k 1
z
z
1
k 2
z
k 3
z
k 4
1
z
5
5
1
1
1
2 e
1
5
2 e
5
1
5
2 e
5
2 5e
1
5
5
1
1
2 e
5
 
i 
 30 
1 
 
  
 2 5 cos    i sin   
 30  
  30 
 13 
i

 30 
1 
 13
 2 5 cos 
  30

 13
  i sin 

 30



 25 
i

 30 
1 
 25 
 25
 2 5 cos 
  i sin 
 30
  30 



 37 
i

 30 
1 
 37
 2 5 cos 
  30

 37
  i sin 

 30



 49 
i

 30 
1 
 49
 2 5 cos 
  30
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
 49  
  i sin 


 30  
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Example
Find the square root of a complex number z  x  i y
Express the complex number in exponential form z  R e i 
zz
12
Re
   2 k 
i

 2 
k  0, 1,
since there are two roots
Express the result in rectangular form
z  3x  i
z
12
2 e
12
 1  
 y
y  1 R  x 2  y 2  2   a tan    a tan 

x
 3 6
x 3
  / 6  2 k 
i

2


z11 2  21 2 e i  /12 
2 e
12


i   k 
 12

z21 2  21 2 e
k  0, 1


i   
 12

z11 2  21 2 cos  /12   i sin  /12  
z21 2  21 2 cos  /12   i sin  /12  
z11 2  1.3360  i  0.3660 
z21 2   1.3360  i  0.3660  
Alternative procedure
z
1
2
 w2  3 x  i
w  a i b
a 2  b2  3
w2   a  i b    a 2  b 2   i  2 a b 
2
2ab  1
Using the magnitudes of w 2 and w
w2  w w
a 2  b2  2
3  1  2  a 2  b2
a 2  b2  3
 2a 2  2  3
2 3
 1.3660
2
1
1
b

 0.3660
2a 2 1.3660  a
a
Therefore, the two solutions are
w1  1.3660  i  0.3660 
w2   1.3660  i  0.3660  
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The point z and the two roots z1/2 are shown on
the Argand diagram
COMPLEX nth ROOTS OF  1
We can write the number 1 as a complex number
z  1  e i  2 k 
k  0, 1, 2, 3,
We can now find the complex numbers w corresponding to the nth roots of the number 1
1
n
wz e
 2 k 
i

 n 
k  0, 1, 2, 3,
,( n  1)
The n roots can be shown on an Argand diagram. The roots are equally spaced around
the circumference of the unit circle with centre (0,0). The angular spacing between each
root is
angular spacing between each root =
2
n
n  1, 2, 3,
The lines joining the roots forms the shape of a regular n-sided polygon.
n = 2 (k = 0 and 1) two roots
1
2
k  0 w  z  e i 0  1
1
k  1 w  z 2  e i    1
angular spacing between roots =  rad (180o)
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n = 3 (k = 0, 2, 3) three roots
1
z3  e
 2 k 
i

 3 
k  0, 1, 2
1
k  0 w  z3  e
1
k  1 w  z3  e
1
k  2 w  z3  e
i 0
1
i  2  / 3
i  4  / 3
angular spacing between roots = 2/3 rad (120o)
nth roots of 1 are equally spaced around the unit circle  
2
360o
rad 
with centre
n
n
0 and so from the vertices of a regular n-sided polygon.
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