complex numbers - Using De Moivre's formula for finding $sqrt{i}$

In class today, we learned about complex numbers, and the teacher described a simple procedure for finding the square root of $i$ in $z^2=i$. He explained to us to set $z=a+bi$ and square from there, which did work, but it took some work to get through.

Afterwards, he said that there was an easier way, and that was to use De Moivre's formula to find the square roots.

So...

Question: How would you use De Moivre's formula to find the square root of any nonzero complex number?

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I do understand that the formula goes something like this:$$(\cos x+i\sin x)^y=\cos(xy)+i\sin(xy)\tag{1}$$
But I'm not too sure how to apply that to$$z^2=i\tag2$$ for $z$ is another complex number.

Answer

Using the $n$-th roots formula (which is actually an equivalent version of De Moivre's formula )
$$
\sqrt{i}=i^{\frac{1}{2}}=(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})^{\frac{1}{2}}
=\cos(\frac{\pi}{4}+\kappa\pi)+i\sin(\frac{\pi}{4}+\kappa\pi)
$$
for any $\kappa\in\mathbb{Z}$. The above, for $\kappa$ even, gives

$$
\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}
$$
while for $\kappa$ odd, it gives
$$
-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}
$$

For the general case, the formula for the $n$-th roots of the complex number $z=\rho (\cos \phi + i \sin \phi)$, is given by
$$

[\rho (\cos \phi + i \sin \phi)]^{1/n} = \rho^{1/n}\left( \cos \frac{\phi + 2 \pi k}{n} + i \sin \frac{\phi + 2 \pi k}{n} \right), \quad k = 0, 1, \dots,
$$
$\rho$ is the modulus. For the square roots, just set $n=2$.