nth Roots of Complex Numbers

# 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)
\begin{align} \quad z^n = r^n (\cos (n \theta) + i \sin (n \theta)) \end{align}

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 $\displaystyle{r^{1/n} \left ( \cos \left ( \frac{\theta + 2k \pi}{n} \right ) + i \sin \left ( \frac{\theta + 2k \pi}{n} \right ) \right )}$.

(2)
\begin{align} \quad z^n = r^n (\cos n\theta + i \sin n \theta) \end{align}
• We want $z^n = w$, i.e.:
(3)
\begin{align} \quad z^n = r^n (\cos n \theta + i \sin n \theta) = \rho (\cos \phi + i \sin \phi ) = w \end{align}
• 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:
(4)
\begin{align} \quad r &= \rho^{1/n} \\ \quad \theta &= \frac{\phi + 2k\pi}{n} \end{align}
• Therefore:
(5)
\begin{align} \quad z = \rho^{1/n} \left ( (\cos \left ( \frac{\phi + 2k\pi}{n} \right ) + i \sin \left ( \frac{\phi + 2k\pi}{n} \right ) \right ) \end{align}
• 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:

(6)
\begin{align} \quad z_0 = \cos \left ( \frac{\pi}{6} \right ) + i \sin \left ( \frac{\pi}{6} \right ) \end{align}
(7)
\begin{align} \quad z_1 = \cos \left ( \frac{\pi}{6} + \left ( \frac{2\pi}{3} \right ) \right ) + i \sin \left ( \frac{\pi}{6} + \frac{2\pi}{3} \right ) \end{align}
(8)
\begin{align} \quad z_2 = \cos \left ( \frac{\pi}{6} + \left ( \frac{4\pi}{3} \right ) \right ) + i \sin \left ( \frac{\pi}{6} + \frac{4\pi}{3} \right ) \end{align}