$\sqrt{i}$

$\Large i={e}^{(i\frac{\pi}{2})}={e}^{(i\frac{5\pi}{2})}$ $\Large \sqrt i = {i}^{\frac{1}{2}}={e}^{(i\frac{\pi}{4})}={e}^{(i\frac{5\pi}{4})}$

$\Large \rightarrow \color{#3D99F6}{\boxed{\sqrt i={e}^{(i\frac{\pi}{4})}=(\cos(\frac{\pi}{4})+isin(\frac{\pi}{4}))={e}^{(i\frac{5\pi}{4})}=(\cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))}}$

To answer this problem, we will use de Moivre's formula n-th roots. Given any nonzero complex numbers written in polar form $z= |z| \cos \theta + i \sin \theta$ then applicable $\displaystyle \sqrt[n]{z}= \sqrt[n]{\left|z\right|} \cos \dfrac {\theta +2k\pi}{n} + i \sin \dfrac {\theta +2k\pi}{n}$ . for $k=0,1,2,\cdots,n-1$ . Let in polar form: $i= \cos \dfrac\pi2 + i \sin \dfrac\pi2$ enter into the formula above, is obtained $\sqrt{i}= \cos \dfrac {\dfrac 12 +2k\pi}{2} + i \sin \dfrac {\dfrac 12 +2k\pi}{2}$ $\boxed{Note: n=2 ~ and ~ \theta= \frac 12}$ for k=0, is obtained: $\sqrt{i}= \cos \dfrac {\pi}{4} + i \sin \dfrac {\pi}{4}\approx0,7071+0,7071i ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ meanwhile for k=1: $\sqrt{i}= \cos \dfrac {5\pi}{4} + i \sin \dfrac {5\pi}{4}\approx-0,7071-0,7071i$

$\therefore \boxed{\sqrt{i}= \cos \dfrac\pi4 + i\sin \dfrac\pi4 ~ and ~\sqrt{i}= \cos \dfrac{5\pi}{4} + i\sin \dfrac{5\pi}{4}}$