A cube root of i is

$\text{ Cube root of }i=i^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
$\text{ we know that }i=\text{ cis }\frac{\pi }{2}$
$\begin{array}{l}{i}^{1/3}={\left(\text{ cis }\frac{\pi }{2}\right)}^{1/3}\\ ={\left[\mathrm{cis}\left(2k\pi +\frac{\pi }{2}\right)\right]}^{1/3},k=0,1,2\\ =\mathrm{cis}{\left((4k+1)\frac{\pi }{2}\right]}^{1/3},k=0,1,2\\ =\mathrm{cis}(4k+1)\frac{\pi }{6},k=0,1,2 \left(from Demovier\text{'}s theorem\right)\\ \end{array}$
$\begin{array}{r}k=0\Rightarrow \mathrm{cis}\frac{\pi }{6}=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\\ =\frac{\sqrt{3}+i}{2} , which is one of the cube roots of i\end{array}$