Is 4th root of $-1$ the same as $i^2$? I am using mathway to check my algebra problems and tried entering in the 4th root of $-1$ (or $(-1)^{1/4}$.   I get the same term back, but I thought that since the square root of $-1$ is $i$, that the answer should be $i^2$ given that 
$$
(-1)^{1/4} 
= (-1)^{1/2} * (-1)^{1/2}
= i * i
= i^2 $$
Would someone please confirm whether my reasoning is correct?  Unfortunately my textbook doesn't go into complex numbers at length and the web resources don't sufficiently cover nth roots of negative 1.
Thanks in advance!
 A: Let's try applying that logic in a more familiar setting:
$$
16^{1/4} \overset{?}= 16^{1/2}\times 16^{1/2} = 4 \times 4 = 4^2
$$
Perhaps now you can see where you went wrong.
A: First notice that $(-1)^{\frac{1}{2}} = \pm i$ and $i^2 = -1$. It is a common mistake to think that $(-1)^{\frac{1}{2}} = i$.
And your second mistake only has to do with powers: $x^{a+b} = x^{a}x^{b}$ but $x^{a\cdot b} = (x^{a})^b \neq x^{a}x^b$
Your third mistake: $x^{a+b} = x^{a}x^{b}$ is not always true in the complex plane. See below.
A: Either you use the expoential form: $\mathrm i=\mathrm e^{\mathrm i\pi}$,hence the fourth roots of $-1$ are the complex numbers $\mathrm e^{\mathrm i\theta}$, with
$$4\theta\equiv \pi\mod 2\pi,\enspace\text{whence}\enspace\theta\equiv \frac\pi4\mod\frac\pi2.$$
Alternatively, anyone having made some computations with complex numbers knows $(1+\mathrm i)^2=2i$, hence $(1+\mathrm i)^4=-4$, so that a fourth root of $-1$ is:
$$\frac{1+\mathrm i}{\sqrt2}.$$
The other roots are obtained, as usual, multiplying this one with the fourth roots of $1$.
A: Other answers pointed out your mistake in splitting the exponential. The more interesting part is, that a $4$th root in the complex plane formally has $4$ solutions. Actually, you are finding the solutions of the polynomial $z^4=-1$ which can easily be solved in the polar form, as $z=e^{(i\pi+2n\pi)/4}$ for $n=0,1,2,3$ and cyclically repeating. If you are not familiar with this notation, just write $z=a+bi$ and find all $4$ solutions by hand.
