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DeMoivre’s Theorem
- You may remember working with complex numbers from your algebra days or even from
precalculus
- In this tutorial, we will review the basics of complex numbers.
- We will discuss converting a complex number into polar form.
- We will cover the multiplication and division theorems.
- We will end with introducing the powerful tool of DeMoivre’s Theorem.
Complex Numbers
- Complex numbers have the standard form 𝑧 = 𝑥 + 𝑖𝑦 where x is the real part and y is the
imaginary part.
- The imaginary unit 𝑖 is the square root of - 1, in other words 𝑖 = √− 1.
- This means that the imaginary unit is the only number that when squared will become
negative, that is to say, 𝑖
2
- Complex numbers (𝑥 + 𝑖𝑦) come in conjugate pairs (𝑥 − 𝑖𝑦).
Complex Multiplication
- With complex multiplication you simply FOIL but keep in mind that 𝑖
2
- Perform the following complex multiplications
a.) ( 2 + 3 𝑖)( 1 − 𝑖) = 2 − 2 𝑖 + 3 𝑖 − 3 𝑖
2
b.)
2
c.)
2
Complex Division
- With complex division, you simply multiply both the numerator and denominator by the
conjugate of the denominator (like rationalizing)
- Your final answer must be written with the real and imaginary parts separate.
- Perform the following complex divisions.
a.)
2 −𝑖
3 + 2 𝑖
2 −𝑖
3 + 2 𝑖
3 − 2 𝑖
3 − 2 𝑖
6 − 4 𝑖− 3 𝑖+ 2 𝑖
2
9 − 6 𝑖+ 6 𝑖− 4 𝑖
2
6 − 7 𝑖− 2
9 + 4
4 − 7 𝑖
13
4
13
7
13
b.)
3
1 −𝑖
3
1 −𝑖
1 +𝑖
1 +𝑖
3 + 3 𝑖
1 +𝑖−𝑖−𝑖
2
3 + 3 𝑖
1 −𝑖
2
3 + 3 𝑖
1 + 1
3 + 3 𝑖
2
3
2
3
2
Absolute Value of Complex Numbers-The absolute value of a complex number is called the
modulus.
- If 𝑧 = 𝑥 + 𝑖𝑦 then the modulus of z is |𝑧| = √𝑥
2
2
- Find the modulus of 𝑧 = 2 − 3 𝑖.
2
2
- Imagine if you had to simplify an expression like the one below.
4
2
3
- Using the methods that we just reviewed, a problem like this would involve a lot of
arithmetic.
- We are going to learn a method that will make this process much simpler using
trigonometry.
Polar Form of Complex Numbers
- You have seen complex numbers in what is called rectangular form (𝑧 = 𝑥 + 𝑖𝑦)
- This can be plotted in rectangular coordinates by calling the y axis the imaginary axis.
Imaginary Axis
| 𝑧
| 𝑦
x-axis 𝜃
𝑥
Deriving Polar Form:
- Based on the above diagram, you should notice that the modulus |𝑧| is also the radius 𝑟.
- Remember that the polar coordinate system is a system where every point is located
based in its distance from the origin
and its angle with the positive x-axis
. Thus, an
ordered pair is
- In order to put a complex number in polar form we need to take 𝑧 = 𝑥 + 𝑖𝑦 and rewrite it
in terms of r and theta.
- Seeing the above diagram, we know that we need to find r and 𝜃.
- We also note very importantly that our complex number plots in the second quadrant.
- To find r, we will simply use Pythagorean theorem below.
2
2
2
2
2
- Now that we know r, we can update our diagram.
Now that we know 𝑟 = √ 2 , we can use inverse trig to find the value of 𝜃.
Based on the above diagram, we know that sin 𝜃 =
1
√
2
, if we rationalize the denominator,
we can say that sin 𝜃 =
√ 2
2
. Using inverse trigonometry, we can say that 𝜃 = sin
− 1
√ 2
2
). If
we take this to the Unit Circle, we will see that there is only one value in the second
quadrant that satisfies this, and it is 𝜃 = 135
𝑜
Now that we know that 𝑟 = √
2 and that 𝜃 = 135
𝑜
we can state the polar form of z below.
cos
𝑜
𝑜
Example 2: Convert 𝑧 = 2 (cos( 60
𝑜
) + 𝑖 sin( 60
𝑜
)) into standard form.
SOLUTION:
This is actually very simple; we use the unit circle to find the actual values of cos
𝑜
and
of sin
𝑜
Next distribute the 2 and you are done.
Example 3: Convert 𝑧 = 2 𝑖 into polar form.
SOLUTION:
It is important to remember that our number is actually 𝑧 = 0 + 2 𝑖.
Just like before, we need to plot this number in the complex plane.
This is a fun little diagram because we can get a lot of information from it.
We can clear see that z is plotted directly on the imaginary (y) axis, this means that 𝜃 =
𝑜
We also remember that r is the distance from the origin, and we can clearly see that we are
a distance of 2 from the origin, therefore, 𝑟 = 2 ,
Therefore, the polar form of 𝑧 = 2 𝑖 is 𝑧 = 2 (cos 90
𝑜
𝑜
1
We have plotted 𝑧
1
on the complex plane and now we need to find r using the Pythagorean
theorem.
2
2
2
2
Now we can complete the above diagram below.
1
Since we know r, we can determine that sin 𝜃 =
1
√ 2
√ 2
2
and this means that 𝜃 = sin
− 1
√ 2
2
and if we take this to the unit circle, there is only one value in the 1
st
quadrant that will
work and it is 𝜃 = 45
𝑜
. Since we know r and 𝜃, we can say the 𝑧
1
2 (cos 45
𝑜
𝑜
Next, we will convert 𝑧
2
= − 1 + 𝑖 into Polar form, first, we plot 𝑧
2
on the complex plane.
2
Next, we use Pythagorean theorem to find the value of r.
2
2
2
2
So, now we can complete the diagram below.
2
We can see that sin 𝜃 =
1
√
2
√ 2
2
and with inverse trigonometry we get 𝜃 = sin
− 1
√ 2
2
According to the unit circle in quadrant 2 the solution is 𝜃 = 135
𝑜
This means that 𝑧
2
2 (cos 135
𝑜
𝑜
We now know that 𝑧
1
2 (cos 45
𝑜
𝑜
) and 𝑧
2
2 (cos 135
𝑜
𝑜
According to the multiplication theorem
1
2
cos
𝑜
𝑜
𝑜
𝑜
1
2
cos 180
𝑜
𝑜
We can see that the product is 𝑧 1
2
= 2 (cos 180
𝑜
𝑜
) in Polar form, lastly, we are
asked to convert it into standard form to see if we get the same answer that we did when
we FOILed.
1
2
= 2 (cos 180
𝑜
𝑜
1
2
1
2
1
2
We got the same answer!
Our next task is to convert these two complex numbers into polar form.
We start with 𝑧
1
3 + 𝑖. Of course, the first thing that we will do is sketch a diagram.
In order to find r, we will use Pythagorean theorem below.
2
2
2
2
2
Since r=2, we can update our diagram below.
Since we know r, we can say that sin 𝜃 =
1
2
, this means that 𝜃 = sin
− 1
1
2
If we take this to the first quadrant of the unit circle, we will see that 𝜃 = 30
𝑜
This means that 𝑧 1
cos 30
𝑜
𝑜
Next, we turn our attention to 𝑧
2
= 2 𝑖, we start with a diagram.
We can see that since this number is actually on the imaginary axis, 𝜃 must be 90
𝑜
and we
can also see that r (the distance from the origin) is 2.
Therefore, it becomes clear that 𝑧 2
cos 90
𝑜
𝑜
Since we now know that 𝑧 1
= 2 (cos 30
𝑜
𝑜
) and 𝑧
2
= 2 (cos 90
𝑜
𝑜
), we can
apply the division theorem.
1
2
cos 30
𝑜
𝑜
cos 90
𝑜
𝑜
[
cos
𝑜
𝑜
𝑜
𝑜
)]
If we simplify this, we get 1
[
cos
𝑜
𝑜
)]
Does the negative angle look strange? It very well may but it is nothing to worry about, we
can simply add 360 to both angles to get their coterminal versions.
[cos(− 60
𝑜
) + 𝑖 sin(− 60
𝑜
)] → [cos 300
𝑜
𝑜
]
Now that our polar complex number looks a little more respectable, we can take each value
to the unit circle to convert our number to standard form.
If we distribute, we get the following final solution.
1
2
This is the same as the previous answer we got!
DeMoivre’s Theorem
- At last, we come to the grand finale!
- DeMoivre’s Theorem, while relatively simple, has many far-reaching consequences that
often arise in advanced mathematics.
- What the Precalculus students’ needs this for is to simplify expressions where a complex
number is being raised to a power.
- If I asked you to calculate (− 1 − 𝑖 √
12
, you could do it by writing the expression out 12
times and doing quite a bit of FOIL, however DeMoivre’s Theorem shows us an easier way
to simplify the expression.
Now that we know r, we can see that sin 𝜃 =
1
2
and 𝜃 = sin
− 1
1
2
), if we go to the 2
nd
quadrant of the unit circle, we can see that 𝜃 = 150
𝑜
This means that − √
3 + 𝑖 = 2 (cos 150
𝑜
𝑜
Now, we remember that we are raising this to the 4
th
power, therefore.
[
cos 150
𝑜
𝑜
)]
4
2 - Now we can apply DeMoivre’s Theorem and simplify
[ 2 (cos 150
𝑜
𝑜
)]
4
4
[cos( 4 ∙ 150
𝑜
) + 𝑖 sin( 4 ∙ 150
𝑜
)]
This will become.
[
cos 600
𝑜
𝑜
]
However, 600 is not on the unit circle, but we can subtract 360 from it to put the
angle within the range of 0 to 360.
16 [cos 240
𝑜
𝑜
]
3 - Lastly, we will convert 16
[
cos 240
𝑜
𝑜
]
into standard form.
16 [−
)]
16 [−
]
This is the final answer! While it might seem like a lot of work, the elegance of this method
is impossible to deny.
Final Example: Convert all complex numbers into standard form and them simplify the
expression, lastly convert your answer back into standard form.
4
2
3
Note: In this example, we will not cover converting the numbers to polar form, that exercise
is performed in appendix.
SOLUTION:
4
2
3
Convert every number to
Polar
( 2 (cos 60
𝑜
𝑜
4
( 2 (cos 330
𝑜
𝑜
2
( 2 (cos 300
𝑜
𝑜
3
Apply DeMoivre’s
Theorem
4
(cos 4 ∙ 60
𝑜
𝑜
2
(cos 2 ∙ 330
𝑜
𝑜
3
cos 3 ∙ 300
𝑜
𝑜
Simplify 16
cos 240
𝑜
𝑜
cos 660
𝑜
𝑜
cos 900
𝑜
𝑜
Simplify the numerator
using the multiplication
theorem.
cos
𝑜
𝑜
𝑜
𝑜
cos 900
𝑜
𝑜
Simplify
64 (cos 900
𝑜
𝑜
8 (cos 900
𝑜
𝑜
Apply the Division
Theorem
[
cos
𝑜
𝑜
𝑜
𝑜
)]
Simplify 8 (cos 0
𝑜
𝑜
Convert to Standard
Form
Simplify 8
You must admit, its pretty cool that we can simplify a complicated expression in only a few
simple steps.
b.) Convert √
3 − 𝑖 into polar form.
We need to remember that 𝑥 = √ 3 and 𝑦 = − 1 and we will draw a sketch below.
r
We will use Pythagorean theorem to find the value of r.
2
2
2
2
2
We can now update our diagram below.
We can now see that sin 𝜃 = −
1
2
and if we take that to the 3
rd
quadrant of the unit circle,
we find that 𝜃 = 330
𝑜
This means that √ 3 − 𝑖 = 2
cos 330
𝑜
𝑜
c.) Convert 1 − 𝑖 √
3 into Polar Form.
In this case we remember that 𝑥 = 1 and 𝑦 = − √
3 when we draw our sketch below.
r
We can use Pythagorean theorem below to find the value of r.
2
2
2
2
2
Since we know the value of r, we can update our diagram below.
We can see that sin 𝜃 = −
√
3
2
and if we take this to the 4
th
quadrant of the unit circle,
we will see that 𝜃 = 300
𝑜
This means that 1 − 𝑖 √
3 = 2 (cos 300
𝑜
𝑜
