Question

Find all the values of the given root :
$\sqrt{i}$

Answer 1

$i=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}$

$=\cos (2n\pi +\frac{\pi }{2})+i\sin (2n\pi +\frac{\pi }{2})$

$=\cos (4n+1)\frac{\pi }{2}+i\sin (4n+1)\frac{\pi }{2}$

$\therefore \sqrt{-1}=\cos (4n+1)\frac{\pi }{4}+i\sin (4n+1)\frac{\pi }{4}$

where n = 0,1

$\therefore \sqrt{-1}=\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}or\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}$

$=\frac{1}{\sqrt{2}}(1+i)or\frac{1}{\sqrt{2}}(-1-i)$

or $\sqrt{i}=\pm \frac{1}{\sqrt{2}}(1+i)$