What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?

$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$

WA gives me different result:

$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$

Why the principal root must be equal to $\frac{\sqrt 2}2-i \frac{\sqrt 2}2$?
I used Moivre's formula. But I dont know. How can I must choose value of $k$ in Moivre's formula? I chose $k=0$.
$$z=r\left(\cos x+i\sin x\right)$$
$$z^{\frac 1n}=r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n}\right)$$

There are two square root of $-i$ namely, $$ \pm\frac {\sqrt 2}{2} (1-i)$$
You may verify it by squaring each one $$ (\pm\frac {\sqrt 2}{2} (1-i))^2= \frac {1}{2}(1-2i-1)=-i$$
One  way to find them is of course the polar form of complex number $-i=e^{3i \pi /2}$ as mentioned in Kevin's answer.
The other way is looking at the unit circle and locate the square roots by dividing the angle $3\pi /2 $ by $2$ to find the first one and add $\pi $ to this angle to  find the second one. 
