# 7. Powers and Roots of Complex Numbers

by M. Bourne

Consider the following example, which follows from basic algebra:

(5

e^{3j})^{2}= 25e^{6j}

We can generalise this example as follows:

(

re^{jθ})^{n}=r^{n}e^{jnθ}

The above expression, written in polar form, leads us to DeMoivre's Theorem.

## DeMoivre's Theorem

[

r(cosθ+jsinθ)]^{n}=r^{n}(cosnθ+jsinnθ)

where `j=sqrt(-1)`. Equivalently,

(

r∠θ)^{n}=r^{n}∠nθ

### Challenge

I'm going to challenge you here...

I have never been able to find an electronics or electrical engineer that's even **heard** of DeMoivre's Theorem.
Certainly, any engineers I've asked don't know how it is applied in 'real life'.

I've always felt that while this is a nice piece of mathematics, it is rather useless.. :-)

Here are some responses I've had to my challenge:

Some possibilities

I received this reply to my challenge from user Richard Reddy:

Much of what you're doing with complex exponentials is an extension of DeMoivre's Theorem. In general, the theorem is of practical value in transforming equations so they can be worked more easily. Often, what you see in EE are the solutions to problems in physics. There was a time, before computers, when it might take 6 months to do a tensor problem by hand. DeMoivre's theorem is a time-saving identity, easier to apply than equivalent trigonometric identities.

I like your site.

And this came in from Russell Davies:

I'm an electronics engineer. In terms of practical application, I've seen DeMoivre's Theorem used in digital signal processing and the design of quadrature modulators/demodulators.

Reader Andy added:

DeMoivre's Theorem can be used to find the secondary coefficient

Z_{0}(impedance in ohms) of a transmission line, given the initial primary constantsR, L, CandG. (resistance, inductance, capacitance and conductance) using the equation`Z_0=sqrt((R+jomegaL)/(G+jomegaC))`,

where '`omega`' is the angular frequency of the supply in radians per second. In this case, the power '

n' is a half because of the square root and the terms inside the square root can be simplified to a complex number in polar form.

Reader David from IEEE responded with:

De Moivre's theorem is fundamental to digital signal processing and also finds indirect use in compensating non-linearity in analog-to-digital and digital-to-analog conversion.

Get the Daily Math Tweet!

IntMath on Twitter

After those responses, I'm becoming more convinced it's worth it for electrical engineers to learn deMoivre's Theorem. Please let me know if there are any other applications.

### Example 1

Find (1 − 2*j*)^{6}

Answer

First, we express `1 - 2j` in polar form:

`sqrt(1^2 + (-2)^2)=sqrt(5)`, and

`arctan((-2)/1) =296.6^"o"`

So

`1-2j=sqrt5\ /_ \ 296.6^text(o)`

Then

`(1-2j)^6=(sqrt5)^6/_ \ [6xx296.6^text(o)]`

`=125\ /_ \ [1779.3903^text(o)]`

`=125\ /_ \ [339.39^text(o)]`

(The last line is true because `360° × 4 = 1440°`, and we substract this from `1779.39°`.)

In rectangular form,

x= 125 cos 339.39^{o}= 117

y= 125 sin 339.39^{o}= -44

So (1 - 2*j*)^{6}= 117 - 44*j*

## Complex Roots

If *a*^{n} =* x *+* yj* then we expect
*n* complex roots for *a*.

### Example 2

If *a*^{5} = 7 + 5*j*, then we
expect `5` complex roots for *a*.

### Spacing of *n*-th roots

In general, if we are looking for the *n*-th roots of an
equation involving complex numbers, the roots will be `360^"o"/n` apart. That is,

2 roots will be `180°` apart

3 roots will be `120°` apart

4 roots will be `90°` apart

5 roots will be `72°` apart etc.

### Example 3

Find the two square roots of `-5 +
12j`*.*

Answer

For the first root, we need to find `sqrt(-5+12j`.

This is the same as (-5 + 12*j*)^{1/2}.

We express −5 + 12*j* in polar form:

`r=sqrt((-5)^2+12^2=13`

For the angle:

`alpha=tan^-1(y/x)`

`=tan^-1(12/5)~~67.38^text(o)`

The complex number −5 + 12*j* is in the **second
quadrant**, so

θ= 180° − 67.38 = 112.62°

So

`−5 + 12j = 13 ∠ 112.62°`

Using DeMoivre's Theorem:

`(r ∠ θ)^n=(r^n∠ nθ)`,

we have:

`(-5+12j)^(1"/"2)`

`=13^(1"/"2)/_(1/2xx112.62^@)`

`=3.61/_56.31^@`

This is the **first** square root. In rectangular form,

x= 3.61 cos56.31° = 2

y= 3.61 sin56.31° = 3

So the **first root** is 2 + 3*j*.

**CHECK:** (2 + 3*j*)^{2} = 4 + 12*j* - 9
= -5 + 12*j * [Checks OK]

To obtain the other square root, we apply the fact that if we
need to find *n* roots they will be `360^text(o)/n` apart.

In this case, `n = 2`, so our roots are `180°` apart.

Adding `180°` to our first root, we have:

*x* = 3.61 cos(56.31° + 180°) = 3.61
cos(236.31°) = -2

*y* = 3.61 sin(56.31° + 180°) = 3.61
sin(236.31°) = -3

So our **second root** is `-2 - 3j`.

So the two square roots of `-5 - 12j` are `2 + 3j` and `-2 - 3j`.

### Exercises:

**1. **Evaluate
`(2 ∠ 135^@)^8`

Answer

`(2 ∠ 135^@)^8= (2)^8 ∠ (135^@ × 8)`

`= 256 ∠ 1080^@`

`= 256`

Get the Daily Math Tweet!

IntMath on Twitter

**2.** Find: (−2 + 3*j*)^{5}

Answer

(−2 + 3*j*)^{5}

= (3.60555^{} ∠ 123.69007°)^{5} (converting to polar form)

= (3.60555)^{5} ∠ (123.69007° × 5) (applying deMoivre's Theorem)

= 609.33709 ∠ 618.45035°

= −121.99966 − 596.99897*j* (converting back to rectangular form)

= −122.0 − 597.0*j* (correct to 1 decimal place)

For comparison, the exact answer (from multiplying out the brackets in the original question) is

−122 − 597

j

[**Note:** In the above answer I have kept the full number of decimal places in the calculator throughout to ensure best accuracy, but I'm only displaying the numbers correct to 5 decimal places until the last line.]

**3.** (i) Find the first 2 fourth roots
of 81(cos 60^{o} + *j* sin 60^{o})

(ii) Then sketch all fourth roots
of 81(cos 60^{o} + *j* sin 60^{o}) showing relevant values of *r* and *θ*.

Answer

**Part (i)**

There are 4 roots, so they will be `θ = 90^@` apart.

I **First root:**

`81^(1"/"4)[cos\ ( 60^text(o))/4+j\ sin\ (60^text(o))/4]`

`= 3(cos\ 15^"o" + j\ sin\ 15^"o")`

` = 2.90 + 0.78j`

II **Second root:**

Add `90°` to the first root:

3(cos(15

^{o}+ 90^{o}) +jsin(15^{o}+ 90^{o}))= 3(cos105

^{o}+jsin 105^{o})= -0.78 + 2.90

j

So the first 2 fourth roots of 81(cos 60^{o} +
*j* sin 60^{o}) are:

`2.90 + 0.78j` and `-0.78 + 2.90j`

**Part (ii)**

In each case, *r* = 81^{1/4} = 3

Easy to understand math videos:

MathTutorDVD.com

**4.** At the beginning of this section, we
expected 3 roots for

`x^3= 8`.

Find the roots and sketch them.

Answer

In polar form, 8 = 8(cos 0^{o} + *j* sin 0^{o}).

There are 3 roots, so they will be `θ = 120°` apart.

Using DeMoivre's Theorem:

a^{n}=r^{n}(cosnθ+jsinnθ),

**First root:**

`8^(1"/"3)=8^(1"/"3)(cos\ 0^text(o)/3+j\ sin\ 0^text(o)/3)`

`=2(cos\ 0^text(o)+j\ sin0^text(o))`

`=2`

**Second root:**

Add `120°` to the first root:

8

^{1/3}(cos 120^{o}+jsin 120^{o}) = −1 + 1.732j

**Third root:**

Add `120°` to the second root:

8

^{1/3}(cos 240^{o}+jsin 240^{o}) = −1 − 1.732j

So the 3 cube roots of `8` are:

To see if the roots are correct, raise each one to power `3` and multiply them out.

Easy to understand math videos:

MathTutorDVD.com

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Algebra Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand algebra lessons on DVD. See samples before you commit.

More info: Algebra videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!