Complex Number Exponentiation

Given: $z$ a complex number

Q1: What is $z^n$? Assume $n$ is an integer.

A1: Since exponentiation is repeated multiplication, it's convenient to use the complex polar form of $z$:

$z=\left|z\right|e^{j\theta}$

$z^n={\left|z\right|}^ne^{jn\theta}$

This result is known as De Moivre's formula.

Q2: Suppose you don't have $z$ raised to an integer power. Instead, you have $z^{\frac{1}{n}}$. How do you find the $n$ roots of this expression?
Q3: To answer Q2, let's look at the specific case where $n$ is 3. What are the 3 roots of $z^{\frac{1}{3}}$?
A3: Following the logic from A1,
Root 1: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
But this is only 1 root. We know we need to find 2 more. Note that $2\pi$ can be added to the angle $\theta$ without changing the result. We can write 2 more cases:
Root 2: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+2\pi)}{3}}$.
Root 3: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+4\pi)}{3}}$.
Q4: This gives us 3 roots - just what we want. But why not continue adding 2$\pi$ to the angle again? Couldn't this produce a 4th root?
A4: Suppose you added another 2$\pi$. Then you'd get the fourth case:
Root 4: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+6\pi)}{3}}$.
But note that this reduces to  $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}+2\pi}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
which is the same as Root 1. So there's no need to continue adding 2$\pi$ past the Root 3. Any other roots found by adding 2$\pi$ past this point are redundant.
A2: To answer the general case of Q2, we substitute $n$ for 3 to find that the $n$ roots of $z^{\frac{1}{n}}$ are:
 $z^{\frac{1}{n}}={\left|z\right|}^{\frac{1}{n}}e^{\frac{j(\theta+2k\pi)}{n}}$ where $k$ is an integer in the range 0, 1, ... , $n-1$