## What are $$n$$th roots?

Let $$n \in \mathbb{N}$$. We say that $$b$$ is an $$n$$th root of $$a$$ if $$b^n = a$$. When $$n = 2$$, $$b$$ is called a square root of $$a$$. When $$n = 3$$, $$b$$ is called a cube root of $$a$$.

It is well known that every positive real number has two square roots. For example, the square roots of $$4$$ are $$2$$ and $$-2$$ because $$2^2 = 4$$ and $$(-2)^2 = 4$$. We now turn to finding $$n$$th roots for complex numbers.

## Motivating example

Consider the complex numbers $$z = 2\operatorname{cis}\frac{\pi}{6}$$ and $$w = 2\operatorname{cis}\frac{5\pi}{6}$$. By De Moivre's Formula, we get $z^3 = 2^3 \operatorname{cis}\left(3\cdot \frac{\pi}{6}\right) = 8 \operatorname{cis}\frac{\pi}{2}$ and $w^3 = 2^3 \operatorname{cis}\left(3\cdot \frac{5\pi}{6}\right) = 8 \operatorname{cis}\frac{5\pi}{2} = 8 \operatorname{cis}\left(2\pi + \frac{\pi}{2}\right) = 8 \operatorname{cis}\frac{\pi}{2}.$ Even though $$z\neq w$$, we have $$z^3 = w^3 = 8 \operatorname{cis} \frac{\pi}{2}$$. Hence, $$z$$ and $$w$$ are distinct cube roots of $$8 \operatorname{cis} \frac{\pi}{2}$$. (Check that $$2\operatorname{cis}\frac{3\pi}{2}$$ is also a cube root of $$8 \operatorname{cis} \frac{\pi}{2}$$.)

Observe that the argument of $$w$$, which is $$\frac{5\pi}{6}$$, is $$\frac{2\pi}{3}$$ more than the argument of $$z$$, which is $$\frac{\pi}{6}$$. Even if we did not know the actual arguments of $$z$$ and $$w$$, we could tell from this observation that the argument of $$w^3$$ is $$2\pi$$ more than the argument of $$z^3$$ since three times $$\frac{2\pi}{3}$$ is $$2\pi$$. As complex numbers in polar form having the same modulus but with arguments differing by integer multiplies of $$2\pi$$ are the same, we see that the cubes of $$z$$ and $$w$$ are the same complex number.

## $$n$$th roots

In general, with $$n \in \mathbb{N}$$ and $$r > 0$$, the $$n$$th power of each of the following $$n$$ complex numbers is equal to $$r \operatorname{cis} \theta$$: $r^{\frac{1}{n}}\operatorname{cis}\frac{\theta}{n}, r^{\frac{1}{n}}\operatorname{cis}\left(\frac{2\pi}{n}+\frac{\theta}{n}\right), r^{\frac{1}{n}}\operatorname{cis}\left(\frac{4\pi}{n}+\frac{\theta}{n}\right), \ldots, r^{\frac{1}{n}}\operatorname{cis}\left(\frac{2(n-1)\pi}{n}+\frac{\theta}{n}\right).$ These are all the $$n$$th roots of $$r \operatorname{cis} \theta$$.

### Example

Find all cube roots of $$-8$$.

The polar form of $$-8$$ is $$8 \operatorname{cis} \pi$$. Hence, applying the generic list above with $$\theta = \pi$$ and $$n = 3$$, we obtain the cube roots $2\operatorname{cis}\frac{\pi}{3},~ 2\operatorname{cis}\left(\frac{2\pi}{3}+\frac{\pi}{3}\right),~ 2\operatorname{cis}\left(\frac{4\pi}{3} + \frac{\pi}{3}\right).$ Simplifying gives $2\operatorname{cis}\frac{\pi}{3},~ 2\operatorname{cis}\pi,~ 2\operatorname{cis}\frac{5\pi}{3}.$ Converting to rectangular form gives $1+\sqrt{3}i,~ -2,~ 1-\sqrt{3}i$

## Exercises

1. Find all square roots of $$17-14i$$. Express the answers in rectangular form with real and imaginary parts rounded to 4 decimal places.

2. Find all 4th roots of $$-1$$. Express the answers in rectangular form with real and imaginary parts rounded to 4 decimal places.