# How do you find the fourth roots of i?

Jan 24, 2017

Fourth roots of $i$ are $0.9239 + 0.3827 i$,
$- 0.3827 + 0.9239 i$,
$- 0.9239 - 0.3827 i$ and
$+ 0.3827 - 0.9239 i$

#### Explanation:

We can use the De Moivre's Theorem, according to which

If $z = r {e}^{i \theta} = r \left(\cos \theta + i \sin \theta\right)$

then ${z}^{n} = {r}^{n} {e}^{i \times n \theta} = r \left(\cos n \theta + i \sin n \theta\right)$

As $i = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)$ can be written as $i = \cos \left(2 n \pi + \frac{\pi}{2}\right) + i \sin \left(2 n \pi + \frac{\pi}{2}\right)$

$\sqrt[4]{i} = {i}^{\frac{1}{4}} = \cos \left(\frac{2 n \pi}{4} + \frac{\pi}{8}\right) + i \sin \left(\frac{2 n \pi}{4} + \frac{\pi}{8}\right)$ or

= $\cos \left(\frac{n \pi}{2} + \frac{\pi}{8}\right) + i \sin \left(\frac{n \pi}{2} + \frac{\pi}{8}\right)$

Note that $n = 0 , 1 , 2 , 3$ and after $n = 3$, it will repeat.

This gives us four roots of $i$, which are

$\cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)$,

$\cos \left(\frac{\pi}{2} + \frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{2} + \frac{\pi}{8}\right) = - \sin \left(\frac{\pi}{8}\right) + i \cos \left(\frac{\pi}{8}\right)$

$\cos \left(\pi + \frac{\pi}{8}\right) + i \sin \left(\pi + \frac{\pi}{8}\right) = - \cos \left(\frac{\pi}{8}\right) - i \sin \left(\frac{\pi}{8}\right)$

and $\cos \left(\frac{3 \pi}{2} + \frac{\pi}{8}\right) + i \sin \left(\frac{3 \pi}{2} + \frac{\pi}{8}\right) = \sin \left(\frac{\pi}{8}\right) - i \cos \left(\frac{\pi}{8}\right)$

and to get exact values we can use $\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2} = 0.3827$ and $\cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2} = 0.9239$

Hence, fourth roots of $i$ are $0.9239 + 0.3827 i$, $- 0.3827 + 0.9239 i$, $- 0.9239 - 0.3827 i$ and $+ 0.3827 - 0.9239 i$