What is $\sqrt[4]{-1}$
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: $$ \cos\frac{(2k+1)\pi}3+i\sin\frac{(2k+1)\pi}3,\ \ k=0,1,2. $$
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 $\theta\in(-\pi,\pi]$
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, $\sqrt{-1}, \sqrt [4] {-1}, \sqrt [6] {-1},\ldots$