nth Roots of Complex Numbers Examples 1

Recall from the nth Roots of Complex Numbers page that if $z = r (\cos \theta + i \sin \theta) \in \mathbb{C}$, $z \neq 0$, then the $n$ many $n^{\mathrm{th}}$ roots of $z$ are given by the following formula for each $k \in \{ 0, 1, ..., n - 1 \}$:

(1)

\begin{align} \quad z_k = r^{1/n} \left ( \cos \left ( \frac{\theta}{n} + \frac{k}{n} 2\pi \right ) + i \sin \left ( \frac{\theta}{n} + \frac{k}{n} 2\pi \right ) \right ) \end{align}

We will now look at some examples of computing the $n^{\mathrm{th}}$ roots of some complex numbers.

Example 1

Compute the square roots of $1 + i$.

Since we want to compute the square roots of $1 + i$ we have that $n = 2$. We have that $r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2} = 2^{1/2}$. Lastly, we have that $\displaystyle{\theta = \mathrm{Arg}(z) = \frac{\pi}{4}}$. Therefore the square roots of $1 + i$ are:

(2)

\begin{align} \quad z_0 &= 2^{1/4} \left ( \cos \left ( \frac{\pi/4}{2} \right ) + i \sin \left ( \frac{\pi/4}{2} \right ) \right ) = 2^{1/4} \left ( \cos \left ( \frac{\pi}{8} \right ) + i \sin \left ( \frac{\pi}{8} \right ) \right ) \\ \quad z_1 &= 2^{1/4} \left ( \cos \left ( \frac{\pi/4 + 2\pi}{2} \right ) + i \sin \left ( \frac{\pi/4 + 2\pi}{2} \right ) \right ) = 2^{1/4} \left ( \cos \left ( \frac{9\pi}{8} \right ) + i \sin \left ( \frac{9\pi}{8} \right ) \right ) \end{align}

Example 2

Compute the cube roots of $1 + i$.

This is similar to the previous example. Since we want to compute the cube roots of $1 + i$ we have that $n = 3$. We have again that $r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2} = 2^{1/2}$ and that $\displaystyle{\theta = \mathrm{Arg}(z) = \frac{\pi}{4}}$. Therefore the cube roots of $1 + i$ are:

(3)

\begin{align} \quad z_0 &= 2^{1/6} \left ( \cos \left ( \frac{\pi/4}{3} \right ) + i \sin \left ( \frac{\pi/4}{3} \right ) \right ) = 2^{1/6} \left ( \cos \left ( \frac{\pi}{12} \right ) + i \sin \left ( \frac{\pi}{12} \right ) \right ) \\ \quad z_1 &= 2^{1/6} \left ( \cos \left ( \frac{\pi/4 + 2\pi}{3} \right ) + i \sin \left ( \frac{\pi/4 + 2\pi}{3} \right ) \right ) = 2^{1/6} \left ( \cos \left ( \frac{9\pi}{12} \right ) + i \sin \left ( \frac{9\pi}{12} \right ) \right ) \\ \quad z_2 &= 2^{1/6} \left ( \cos \left ( \frac{\pi/4 + 4\pi}{3} \right ) + i \sin \left ( \frac{\pi/4 + 4\pi}{3} \right ) \right ) = 2^{1/6} \left ( \cos \left ( \frac{17\pi}{12} \right ) + i \sin \left ( \frac{17\pi}{12} \right ) \right ) \end{align}