In the complex plane, "i" is directly above the origin at distance 1. In polar coordinates that would be \(\displaystyle r= 1\), \(\displaystyle \theta= \pi/2\). We can write \(\displaystyle i= 1(cos(\pi/2))+ 1(sin(\pi/2)i)\) since \(\displaystyle cos(\pi/2)= 0\) and \(\displaystyle sin(\pi/12)= 1\). Since \(\displaystyle e^{ix}= cos(x)+ isin(x)\), we an also write \(\displaystyle i= e^{i\pi/2}\) so that \(\displaystyle \sqrt{i}= (e^{i\pi/2})^{1/2}= \)\(\displaystyle e^{i\pi/4}= cos(\pi/4)+ i sin(\pi/4)= \)\(\displaystyle \frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}\).
Of course every number has two square roots. To get the other one, use the fact that sine and cosine are "periodic" with period \(\displaystyle 2\pi\), \(\displaystyle cos(\theta+ 2\pi)= cos(\theta)\) and \(\displaystyle sin(\theta+ 2\pi)= sin(\theta)\). So \(\displaystyle i= e^{i(\pi/2+ 2\pi)}= e^{5\pi/2}\) also. But dividing the exponent by 2 breaks that identity. The other value for \(\displaystyle \sqrt{i}\) is \(\displaystyle \sqrt{i}= \cos\left(\frac{5\pi}{2}\right)+ i\sin\left(\frac{5\pi}{2}\right)= -\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}\).
It should be no surprise that the two square roots are negatives of each other: \(\displaystyle (-x)^2= x^2\) even for complex numbers.