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 $\sqrt{-1},\sqrt[4]{-1},\sqrt[6]{-1},\dots$)
 A: 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:
$$
\cos\frac{(2k+1)\pi}3+i\sin\frac{(2k+1)\pi}3,\ \ k=0,1,2.
$$
A: Method 1 Hint:
Consider
$$z^4=i$$
where 
$$z=a+bi$$
with $a,b\in\mathbb{R}$.
Expand and compare coefficients.
(Tedious)

Method 2:
Consider
$$i=i \cdot1^k= \exp\left({\frac\pi2i}\right)\cdot\left(\exp(2\pi i )\right)^k=\exp\left({\frac\pi2i+2k\pi i}\right)$$
for $k\in\mathbb{Z}$
Let
$$z=\exp({\theta i})$$
such that
$$z^4=i$$
Then use the relation
$$\exp(\theta i)=\cos\theta+i\sin\theta$$
So
\begin{align}
\exp(4\theta i)&= \exp\left({\frac\pi2i+2k\pi i}\right)\\
4\theta i&= \frac\pi2i+2k\pi i\\
\theta&=\frac\pi8+\frac\pi2k\\
\end{align}
Hence, the fourth roots of $i$ are
$$z=\cos\theta+i\sin\theta$$with
$$\theta= -\frac{7\pi} 8,-\frac{3\pi}8,\frac\pi8,\frac{5\pi}8$$
for $\theta\in(-\pi,\pi]$
A: We will use Euler's Identity, $$e^{i\pi} + 1 = 0,\tag1$$ or in particular, $$\begin{align} e^{i\theta} &= \cos\theta + i\sin\theta \\ &= \text{cis}\,\theta.\end{align}\tag2$$

From $(1)$, we get that $e^{i\pi} = -1$, thus $$\begin{align} \left(e^{(i\pi)/2}\right)^{1/2} &= e^{(i\pi)/4} \\ &= \sqrt [4] {-1}.\end{align}$$ Now, using $(2)$, by substituing $\theta = \dfrac{\pi}{4}$, we obtain the following result: $$\begin{align} \sqrt [4] {-1} &= \cos \left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) \\ \\ &= \frac{1}{\sqrt{2}} + i\left(\frac{1}{\sqrt{2}}\right) \\ \\ &= \boxed{ \ \frac{1}{\sqrt{2}}\left(1+i\right). \ }\end{align}$$ See what pattern you can find now with the sequence, $\sqrt{-1}, \sqrt [4] {-1}, \sqrt [6] {-1},\ldots$
