# How do you find the 4th root of 2 (cos (pi/4) + isin(pi/4)) ?

##### 1 Answer
Jul 22, 2016

The principal $4$th root is:

$\sqrt[4]{2 \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)} = \sqrt[4]{2} \left(\cos \left(\frac{\pi}{16}\right) + i \sin \left(\frac{\pi}{16}\right)\right)$

There are $3$ other $4$th roots.

#### Explanation:

In general, if $n$ is a positive integer and $\theta \in \left(- \pi , \pi\right]$ then the principal $n$th root is given by:

$\sqrt[n]{r \left(\cos \theta + i \sin \theta\right)} = \sqrt[n]{r} \left(\cos \left(\frac{\theta}{n}\right) + i \sin \left(\frac{\theta}{n}\right)\right)$

So in our example:

$\sqrt[4]{2 \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)} = \sqrt[4]{2} \left(\cos \left(\frac{\pi}{16}\right) + i \sin \left(\frac{\pi}{16}\right)\right)$

Note that the primitive Complex $4$th root of $1$ is $i$, so the other $4$th roots are:

$i \sqrt[4]{2} \left(\cos \left(\frac{\pi}{16}\right) + i \sin \left(\frac{\pi}{16}\right)\right) = \sqrt[4]{2} \left(\cos \left(\frac{5 \pi}{16}\right) + i \sin \left(\frac{5 \pi}{16}\right)\right)$

${i}^{2} \sqrt[4]{2} \left(\cos \left(\frac{\pi}{16}\right) + i \sin \left(\frac{\pi}{16}\right)\right) = \sqrt[4]{2} \left(\cos \left(- \frac{7 \pi}{16}\right) + i \sin \left(- \frac{7 \pi}{16}\right)\right)$

${i}^{3} \sqrt[4]{2} \left(\cos \left(\frac{\pi}{16}\right) + i \sin \left(\frac{\pi}{16}\right)\right) = \sqrt[4]{2} \left(\cos \left(- \frac{3 \pi}{16}\right) + i \sin \left(- \frac{3 \pi}{16}\right)\right)$