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FP3: Complex Numbers
De Moivre’s theorem
Lesson objectives.:
* to understand de Moivre’s theorem for positive and negative powers
* to find the square roots of a complex number using this theorem.
Suppose z is a complex number with modulus r and argument θ, i.e. z  r (cos   i sin  ) OR, using
shorthand notation, z = [r, θ].
Then
z2 
z3 
z4 
This suggests that z n 
De Moivre’s theorem states that if
z  r (cos   i sin  )
then
zn 
This result applies for any positive or negative power n.
Note: We will be able to prove this result more formally next lesson.
Example: z  3  i .
a) Find z 2 and z3 .
b) Find z8 and z 3
Solution:
First write z in modulus-argument form:
z=[
,
(Note: we should write both of these in exact form if we can)
So z 2  [
,
] and z3  [
,
]
]
The Cartesian versions of this complex numbers are:
z2 
and
z3 
We can show these complex numbers on an Argand diagram:
Im
8
6
4
2
-8
-4
0
-2
-2
-4
-8
b) Likewise: z8  [
]
,
z 3  [
,
]
Example 2: w  0.8  0.6i . Find w9 .
Solution: In mod-arg form: w = [
Therefore, w9  [
So w9 
,
,
].
].
2
4
6
8
Real
The Argand diagram in this case is as follows:
Im
1
0.5
-1
0
-0.5
Real
1
0.5
-0.5
-1
Note: Accuracy is important in this work – try to use exact values for the modulus and argument
where possible.
Finding the square root of a complex number
In FP1, we considered one way to find the square roots of a complex number. There is an alternative
way to find the roots using de Moivre’s theorem.
Example: Find the square roots of 5 – 12i, giving your answers in the form a + bi.
Solution:
Write 5 – 12i in modulus-argument form: 5 – 12i = [13, -1.1760052]
Using de Moivre’s theorem:
Therefore
5  12i 

(we want to keep accuracy)

5  12i  (5  12i)1/ 2  131/ 2 , 12  1.1760052
13, 0.5880026


So:
5  12i  13(cos(0.5880026)  i sin(0.5880026))
= 13(0.83205  0.55470 i)
= 3  2i
But… there is a second square root. To find this other root we find an equivalent form for 5 – 12i by
adding 2π to the argument (this gives an equivalent angle).
So we also have: 5 – 12i = [13, -1.1760052 + 2π] = [13, 5.107180]
Using the above method:
Therefore:

 
5  12i  131/ 2 , 12  5.107180 
13, 2.553590
5  12i  13(cos 2.553590  i sin 2.553590)  3  2i

Note: If one square root is a + bi, the other will be –a – bi. But we will need to extend the above
method in order to deal with cube roots, fourth roots etc.
Example 2: Find the square roots of 1 + i.
Solution: 1 + i = [
So: (1  i)1/ 2  [
,
,
] OR [
]=[
,
,
]=[
] OR [
,
].
,
]=[
Therefore (1  i)1/ 2 
OR (1  i)1/ 2 
Past examination question: Use de Moivre’s theorem to find the value of (1  2i)10 .
Solution:
Let z = 1 – 2i.
Corrected version:
Converting z to mod-arg form gives
z = [2.24, -1.11].
Therefore
z10   2.2410 , (1.11)10 


 3180.4, 2.84
So (1  2i)20 = 3180.4(sin2.84 + icos2.84)
= 3180.4(0.297 + i(-0.955))
= 944.6 – 3037.3i
,
].