What is $-1$ to the power of a fraction? If I have any negative number (does not have to be $-1$), how would I determine that value to the power of $1/2$?
I learned that $x^\frac{1}{2}$ is equal to $√x,$ so wouldn't $-1^\frac{1}{2}$ be equal to $i$? Why does the calculator return $-1$?
 A: Try not to worry too much about all the notation, and look at this page, DLMF 6.6, from an online reference of special functions.
Notice that in several items we have "$(-1)^n$" and also one "$(-1)^{n-1}$".  These have parentheses in order to accurately express what is meant, a power of $-1$.  Without the parentheses, the order of operations specifies that the power would happen before the negation, so one would not get a power of $-1$.
If one is typing this on a calculator, there are a number of common errors.
-1^1/2     = -((1/1)^2) = -1
-1^(1/2)   = -(1^(1/2)) = -1
(-1)^1/2   = ((-1)^1)/2 = -1/2

If you want the quantity $-1$ raised to the power $1/2$, you need to enter
(-1)^(1/2)

Depending on your calculator (and possibly on a mode setting on your calculator) evaluation of this expression will produce an error (some form of domain error, since $-1$ is not in the domain of the real square root function) or a result that is equivalent to the complex number $0 + 1\mathrm{i}$ produced by the complex square root function.
A: There are two square roots of $-1$ and no reason to prefer one of them to the other. You can call one of them $i$ and the other $-i,$ or switch those two roles around so that the one that you called $-i$ before is now $+i$ and vice-versa, and nothing changes. And there are three cube roots of $-1,$ one of those being $-1$ itself and the others $(1\pm i\sqrt3)/2.$ If you want $(-1)^x$ to change in a smooth way as $x$ changes, then you'll need to pick one of those latter two to be the value of $(-1)^{1/3}.$ And what then is $(-1)^{2/3}\text{?}$ Should it be the square of $(-1)^{1/3} = \sqrt[3]{-1}= -1\text{?}$ That would make it $+1.$ Or should it be the square of $(1+i\sqrt3)/2,$ which is $(-1+i\sqrt3)/2\text{?}$
