Question

Find the square root of $i$. Where $i^2=-1$.

Answer 1

We have,

$i$ $=\frac{1}{2}\times 2i$

$=\frac{1}{2}\times \{1+i^2+2i\}$ [ Since $i^2=-1$]

$=\frac{1}{2}\times (1+i{)}^2$.

Then square root of $i$ is $\pm \frac{1}{\sqrt{2}}(1+i)$.