How to find $\sqrt{-i}$ correctly? To solve this question, firstly, I let $-i = e^\frac{3i\pi}{2}$. So when I square root it, I will get $\sqrt {-i} = e^\frac{3i\pi}{4}$. When I convert it to Cartesian equation, I get $\sqrt {-i} = \cos(\frac{3i\pi}{4}) + i\sin(\frac{3i\pi}{4})$ which is $-\frac{1}{\sqrt 2} + \frac{i}{\sqrt2}$. But when I checked with google/wolfram'answer, their answer is $\frac{1}{\sqrt 2} - \frac{i}{\sqrt2}$. So I wonder which part am I wrong??
 A: Notice that $-\frac{1}{\sqrt2}+\frac{i}{\sqrt2}=-\left(\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right)$
Since you are finding a square root, this can be both positive and negative, your answer is not wrong, it is just another value.
Also, one part you missed out is that with trigonometric functions, you need to add $2k\pi$ where $k$ is an integer.  Doing so would enable you to get all the solutions to the equation instead of only $1$.
So $-i=e^{i\left(\frac{3\pi}{2}+2k\pi\right)}$
A: There are two square roots. You only found one of them. Negate it to get the other. 

Wolfram Alpha chose to show the other one. 

The correct solution is to show both.

Alternatively, represent $-i$ as 
$$e^{i
\left(
{\Large{-\frac{\pi}{2}}}
\right)
}
$$
and then divide the angle by $2$, to get
$$e^{i
\left(
{\Large{-\frac{\pi}{4}}}
\right)
}
$$
That gives you the other one, but then, if you choose to do that one first, also negate it, so as to get both square roots.
A: When you are asked to find the square root of $-i$ (or the square root of any number), then you are meant to find $x$ such that:
$$x^2 = -i.$$
Now, you know that $-i = e^\frac{3i\pi}{2}$, and hence you must solve this equation:
$$x^2 = e^\frac{3i\pi}{2}.$$
The solutions are $2$:
$$x = \pm \left(e^\frac{3i\pi}{2} \right)^{\frac{1}{2}} = \pm\left(-\frac{1}{\sqrt 2} + \frac{i}{\sqrt2}\right).$$
