<s>[INST] What is $\sqrt{i}$? If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?   
Is it used or considered often in mathematics? How is it notated?
 [/INST] More generally, if you want to compute all the $n$-th roots of a complex number $z_0$, that is, all the complex numbers $z$ such that
$$
z^n = z_0 \ ,  \qquad \qquad \qquad \qquad [1]
$$
you should write this equation in exponential form: $z = re^{i\theta}, \ , z_0 = r_0 e^{i\theta_0}$. Then [1] becomes
$$
\left( r e^{i \theta}\right)^n = r_0 e^{i\theta}   \qquad \Longleftrightarrow \qquad r^n e^{in\theta} = r_0 e^{i\theta_0}  \ .
$$
Now, if you have two complex numbers in polar coordinates which are equal, their moduluses must be equal clearly:
$$
r^n = r_0 \qquad \Longrightarrow \qquad r = +\sqrt[n]{r_0}
$$
since $r, r_0 \geq 0$.
As for the arguments, we cannot simply conclude that $n\theta = \theta_0$, but just that they differ in an integer multiple of $2\pi$:
$$
n\theta = \theta_0 + 2k\pi   \qquad \Longleftrightarrow \qquad  \theta = \frac{\theta_0 + 2k \pi}{n} \quad \text{for} \quad k = 0, \pm 1 , \pm 2, \dots
$$
It would seem that we have an infinite number of $n$-th roots, but we have enough with $k = 0, 1, \dots , n-1$, since for instance for $k=0$ and $k=n$ we obtain the same complex numbers. Thus, finally
$$
\sqrt[n]{r_0 e^{i\theta_0}} = +\sqrt[n]{r_0} e^{i \frac{\theta_0 + 2k\pi}{n}} \ , \quad k = 0, 1, \dots , n-1 
$$
are all the complex $n$-th roots of $z_0$.
Examples 
(1) For $n=2$, we obtain that every complex number has exactly two square roots:
$$
\begin{align}
\sqrt{z_0} &= +\sqrt{r_0}e^{i\frac{\theta_0 + 2k\pi}{2}} \ , k = 0,1  \\\
           &= +\sqrt{r_0}e^{i\frac{\theta_0}{2}} \quad \text{and} \quad +\sqrt{r_0}e^{i\left(\frac{\theta_0}{2} + \pi \right)}  \ .
\end{align}
$$
For instance, since $i = e^{i\frac{\pi}{2}}$, we obtain
$$
\sqrt{i} = 
\begin{cases}
e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} +i \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}    \\\
e^{i(\frac{\pi}{4} + \pi)} = \cos\frac{5\pi}{4} +i \sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}  \ .
\end{cases}
$$
Also, if $z_0 = -1 = e^{i\pi}$,
$$
\sqrt{-1} = e^{i \frac{\pi}{2}} = i \quad \text{and} \quad e^{i\left( \frac{\pi}{2} + \pi\right)} = e^{i\frac{3\pi}{2}} = -i \ .
$$
(2) For $z_0 = 1 = e^{i \cdot 0}$ and any $n$, we obtain the $n$-th roots of unity:
$$
\sqrt[n]{1} = e^{i\frac{2k\pi}{n}} \ , \quad k= 0, 1, \dots , n-1 \ .
$$
For instance, if $n= 2$, we get
$$
\sqrt{1}= e^{i \cdot 0} = 1 \quad \text{and} \quad  e^{i\pi}= -1
$$
and for $n= 4$, 
$$
\sqrt[4]{1} = e^{i\frac{2k\pi}{4}} \ , \quad k = 0, 1, 2, 3 \ ,
$$
that is,
$$
\sqrt[4]{1} = 1, i, -1 , -i \ .
$$
</s>