What is $\sqrt[4]{-1}$ So, we all know that $\sqrt{-1}=i$, that $\sqrt{-2}=\sqrt2i$, and so on. And that, for example, $\sqrt[3]{-27}=-3$. But I was wondering; what would $\sqrt[4]{-1}$ be? Since all $n^{th}$ roots of $-1$ that are odd result in $-1$ (Because $-1^{1,3,5,7,9,\dots}=-1$), I wondered what variation of $i$ would even roots of $-1$ give? (Like $\sqrt{-1},\sqrt[4]{-1},\sqrt[6]{-1},\dots$)

Why would $\sqrt{-1}=i$ and not $\sqrt{-1}=-i$? There is no real reason to choose. 
What we do is to enumerate the roots; there are two square roots, three cubics roots, four fourth roots, etc. 
From $-1=\cos\pi+i\sin\pi$, we can deduce via De Moivre's formula (or using $-1=e^{i\pi}$), that the four fourth roots of $-1$ are 
$$
\cos\frac{(2k+1)\pi}4+i\sin\frac{(2k+1)\pi}4,\ \ k=0,1,2,3.
$$
Explicitly, you have 
$$
\cos\frac\pi4+i\sin\frac\pi4=\frac1{\sqrt2}+i\,\frac1{\sqrt2},
$$
$$
\cos\frac{3\pi}4+i\sin\frac{3\pi}4=-\frac1{\sqrt2}+i\,\frac1{\sqrt2},
$$
$$
\cos\frac{5\pi}4+i\sin\frac{5\pi}4=-\frac1{\sqrt2}-i\,\frac1{\sqrt2},
$$
$$
\cos\frac{7\pi}4+i\sin\frac{7\pi}4=\frac1{\sqrt2}-i\,\frac1{\sqrt2}.
$$
Note that similarly there is no single cubic root of $-1$, but three of them:
$$
\cos\frac{(2k+1)\pi}3+i\sin\frac{(2k+1)\pi}3,\ \ k=0,1,2.
$$
