# How do you find the fifth root of 32(cos((5pi)/6)+isin((5pi)/6))?

Feb 23, 2017

Fifth root of $32 \left(\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right)$ is $2 \left(\frac{\sqrt{3}}{2} + i \frac{1}{2}\right)$. Other four roots can also be obtained. See below for details.

#### Explanation:

According to DeMoivre's theorem if $z = r \left(\cos \theta + i \sin \theta\right)$

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

It has also been proved that it is true for all $n \in Q$

Hence, if $z = 32 \left(\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right)$

root(5)z=z^(1/5)=32^(1/5)(cos((5pi)/6xx1/5)+isin((5pi)/6xx1/5)

= $2 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)$

= $2 \left(\frac{\sqrt{3}}{2} + i \frac{1}{2}\right)$

In fact you should find five fifth roots, using

$z = 32 \left(\cos \left(2 n \pi + \frac{5 \pi}{6}\right) + i \sin \left(2 n \pi + \frac{5 \pi}{6}\right)\right)$, where $n$ is an integer and then

$\sqrt[5]{z} = 2 \left(\cos \left(\frac{2 n \pi}{5} + \frac{\pi}{6}\right) + i \sin \left(\frac{2 n \pi}{5} + \frac{\pi}{6}\right)\right)$

As may be seen the above root is obtained by putting $n = 0$ and other roots can be obtained by putting $n = 1 , 2 , 3 \text{ or } 4$