Does the square root of $i$ necessitate quaternions? The square root of i is $\frac{\sqrt{2} + i \sqrt{2}}{2}$. But how is it valid to use a number in expressing the square root of that number?
 A: The set of complex numbers form an algebraically closed field. Hence, there exists solutions to $z^2=i$. You may want to look up the fundamental theorem of algebra. This guarantees that there are no rabbit holes, i.e., you do not need to move away from complex numbers. The problem of solving for square-roots, cube-roots and in general any polynomial equation is solved once and for all by the complex numbers. 
In your case, you want to find a $z$ such that $z^2 = i$. If $z=x+iy$, where $x,y \in \mathbb{R}$, we get that
$$z^2 = (x^2-y^2) + i(2xy) = i \implies x = \pm y \text{ and } xy = \dfrac12 \implies x=y=\dfrac1{\sqrt{2}} \text{ or } x=y= - \dfrac1{\sqrt2}$$
Hence, we get that
$$i^{1/2} = \pm \left(\dfrac{\sqrt2 + i \sqrt2}{2}\right)$$ (Note that one of your solution is incorrect.)
A: Square it and see.  It works fine.  In the complex field, you can take the square root of any number.  If we use the polar representation $z=re^{i\theta}$ ($r, \theta$ real), the square roots will be $\sqrt r e^{\frac {i \theta}{2}}$ and $\sqrt r e^{\frac {i \theta}{2}+i\pi}$ 
This extends to $n^{\text {th}}$ roots.  The $n^{\text {th}}$ root of $z=re^{i\theta}$ is $r^{\frac 1n}e^{i(\frac \theta n+\frac {2k\pi}n)}$ for $k=0,1,2,\ldots n-1$
