$\sqrt{i}$

$\displaystyle i=e^{i\pi/2} \Rightarrow \sqrt{i}=e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$

Actually, $i$ has two square roots, just as any real number does. These are the two roots of $z^2=e^{\pi i/2}$. One of the roots is $e^{\pi i/4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ as noted previously. The other is $e^{5\pi i/4}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$. Doubling the argument of both of these complex numbers and squaring the modulus indeed gives $e^{\pi i/2}$.

let $\sqrt{i}$ = $x$

we have $x^{2}$ = $i$

converting i to polar form, we have

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

using Demoivre's Theorem

we have $x_{1} = \cos \frac{\pi}{4} + i\sin \frac{\pi}{4} = \frac{\sqrt2}{2} + \frac{\sqrt2}{2} i$ and $x_{2} = \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} = -\frac{\sqrt2}{2} -\frac{\sqrt2}{2} i$

$\sqrt{i}$ is a complex number, so it can be written in the form $a+bi$ where $a$ and $b$ are real numbers.

Squaring gives $(a^2-b^2)+2abi$. It follows that $2ab=1$ and $a^2=b^2$.

From the second we have that $a=b$ or $a=-b$.

Substituting in the first, we get $a^2=\frac{1}{2}$ or $a^2=-\frac{1}{2}$

So $a=\pm \frac{1}{2} \sqrt{2}$ or $a=\pm \frac{1}{2} \sqrt{2}*i$.

and $b=\pm \frac{1}{2} \sqrt{2}$ or $b=\mp \frac{1}{2} \sqrt{2}*i$.

Using these in our original expression $a+bi$, we get only two distinct solutions:

$\frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2}*i$ and $- \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{2}*i$.

I think usually the first is chosen as THE root of $i$, but only because it's closest to $(1,0)$.