# nth Roots of Complex Numbers

Recall from the De Moivre's Formula for the Polar Representation of Powers of Complex Numbers page that if $z \in \mathbb{C}$, $r = \mid z \mid$, and $\theta = \arg (z)$ then for all $n \in \mathbb{N}$ we have that:

(1)This important formula is known as De Moivre's formula. Using this formula, we will prove that for all nonzero complex numbers $z \in \mathbb{C}$ there exists $n$ many $n^{\mathrm{th}}$ roots for each $n \in \mathbb{N}$.

Theorem 1: Let $w \in \mathbb{C}$, $w \neq 0$ with $w = \rho (\cos \phi + i \sin \phi)$. Then there exists $n$ many $n^{\mathrm{th}}$ roots of $w$ given by the formula $\displaystyle{z = \rho^{1/n} \left ( (\cos \left ( \frac{\phi + 2\pi k}{n} \right ) + i \sin \left ( \frac{\phi + 2k\pi}{n} \right ) \right )}$ where each $k \in \{ 0, 1, ..., n - 1 \}$ yields a distinct $n^{\mathrm{th}}$ root. |

*So if $z = r(\cos \theta + i \sin \theta)$ then the $n^{\mathrm{th}}$ roots of $z$ are given by $\displaystyle{r^{1/n} \left ( \cos \left ( \frac{\theta + 2k \pi}{n} \right ) + i \sin \left ( \frac{\theta + 2k \pi}{n} \right ) \right )}$.*

**Proof:**Let $z = r (\cos \theta + i \sin \theta)$ and let $w = \rho (\cos \phi + i \sin \phi)$. Then by De Moivre's Formula for the Polar Representation of Powers of Complex Numbers we have that:

- We want $z^n = w$, i.e.:

- Hence $r^n = \rho$ and $n \theta = \phi + 2k\pi$ for some $k \in \mathbb{Z}$. We solve for $r$ and $\theta$. We have that:

- Therefore:

- This is satisfied for each $k \in \{ 0, 1, ..., n - 1 \}$, i.e., there are $n$ many $n^{\mathrm{th}}$ roots to every nonzero complex number. $\blacksquare$

## Example 1

**Compute the cube roots of $z = i$.**

We have that $r = \mid z \mid = \mid i \mid = 1$ and that $\displaystyle{\theta = \arg(z) = \frac{\pi}{2}}$. Thus, the cube roots, $z_0$, $z_1$, and $z_2$ of $z$ are:

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