# 7. Powers and Roots of Complex Numbers

by M. Bourne

Consider the following example, which follows from basic algebra:

`(5e^(3j))^2=25e^(6j)`

We can generalise this example as follows:

`(re^(jtheta))^n=r^(n)e^(jntheta)`

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

## DeMoivre's Theorem

`[r(cos theta+j sin theta)]^n` `=r^n(cos ntheta+j sin ntheta)`

or equivalently,

`(r/_theta)^n=r^n/_ ntheta`

### 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.. :-)

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.

Please let me know if there are any other applications.

### Example 1

Find (1 - 2*j*)^{6}

Continues below ⇩

## 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`*.*

### Exercises:

**1. **Evaluate
`(2 ∠ 135^@)^8`

**2.** Find: (−2 + 3*j*)^{5}

**3.** (i) Find the first 2 fourth roots
of 81(cos 60° + *j* sin 60°)

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

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

`x^3= 8`.

Find the roots and sketch them.

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