# How do you calculate root(n)(i), ninRR?

Jul 20, 2017

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

#### Explanation:

The polar form of $z = i$ is given by

$z = 0 + i 1 = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = {e}^{i \frac{\pi}{2}}$

Now according to De Moivre's theorem

if $z = r \left(\cos \theta + i \sin t \eta\right)$

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

or $\sqrt[n]{z} = {z}^{\frac{1}{n}} = {r}^{\frac{1}{n}} \left(\cos \left(\frac{\theta}{n}\right) + i \sin \left(\frac{\theta}{n}\right)\right)$

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