What does the square root of minus $i$ equal?

I'll add an answer simply because I think it would be helpful or interesting to the original poster to eliminate the exponential. Drawing from either answer by Marvis or Andyk, we have $$e^{i3\pi/4} = \frac{\sqrt{2}}{2}(-1+i), ~ e^{i7\pi/4} = \frac{\sqrt{2}}{2}(1-i) \quad\Longrightarrow\quad \sqrt{-i} = \pm\tfrac{\sqrt{2}}{2}(1-i).$$ Note that unlike with real numbers, you cannot select a principal square root; that is, neither choice is "better" than the other. Working backwards to prove this is correct: $$\left(\pm\frac{\sqrt{2}}{2}(1-i)\right)^2 = \frac{1}{2}(1-i)^2= (1-2i+i^2)/2 = -2i/2 = -i.$$ EDITED TO ADD: In a comment above, Peter Tamaroff says "It is not a nice practice to write $\sqrt{a}$ whenever a is either a negative real number, or a complex number." This is because $\sqrt{\cdot}$ is used to refer to the principal square root of a number, which for positive numbers refers to the nonnegative root. That is, if $x^2=9$, then we know that $x$ could be $+3$ or $-3$; but the principal square root is $\sqrt{9}=3$. There is no corresponding convention to define the principal square root of a negative or complex number---though to be frank, for negative numbers I've seen $\sqrt{-a} \triangleq \sqrt{a}\cdot i$ adopted in more casual discussion.

We have $$i = e^{2k \pi i+ i \pi/2}$$ Hence, $$-i = e^{2k \pi i - i \pi/2} \implies (-i)^{1/2} = \left(e^{2k \pi i - i \pi/2}\right)^{1/2} = e^{k \pi i - i \pi/4} = e^{i 3 \pi/4}, e^{i 7 \pi/4}$$ We have two possible values, just like we have two possible values for $x^2 = -1$.