# How do you find the nth root of -1?

##### 1 Answer
Oct 21, 2016

${n}^{t h}$ root of $- 1$ is $\cos \left(\frac{\pi}{n}\right) + i \sin \left(\frac{\pi}{n}\right)$

#### Explanation:

According to De Moivre's theorem, if a complex number $z$ in polar form is given by

$z = r \left(\cos \theta + i \sin \theta\right)$, then

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

Let ${z}^{n} = - 1 = 1 \left(\cos \pi + i \sin \pi\right)$ as $\cos \pi = - 1$ and $\sin \pi = 0$

Hence $z$ the ${n}^{t h}$ root of $- 1$ will be

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