<s>[INST] What is the square root of $3 + 2\sqrt{10}i$? I need to compute the square root of $3 + 2\sqrt{10}i$.
I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this: 
$$
  \sqrt{3 + 2\sqrt{10}i} = x + iy
    \quad \longrightarrow \quad
  3 + 2\sqrt{10}i = x^2 - y^2 +2xyi
$$
and so forth, but my answer isn't correct.
 [/INST] Your "and so on" could go like this:
$$\begin{cases}x^2-y^2=3,\\2xy=2\sqrt{10}.\end{cases}$$
Then squaring and adding both,
$$x^4-2x^2y^2+y^4+4x^2y^2=(x^2+y^2)^2=49,$$ so that
$$x^2+y^2=\pm7.$$
Solving with the help of the first,
$$x^2=5,y^2=2\text{ or }x^2=-2,y=-5.$$
This leaves the possibilities
$$x=\pm\sqrt5,y=\pm\sqrt2.$$
By the second equation, we know the signs are synchronized, hence
$$\sqrt5+i\sqrt2\text{ or }-\sqrt5-i\sqrt2.$$

More generally,
$$\begin{cases}x^2-y^2=u,\\2xy=v.\end{cases}$$
yields
$$u^2=\frac12(\sqrt{v^2+u^2}+u),v^2=\frac12(\sqrt{v^2+u^2}-u),$$
and 
$$x=\pm\sqrt{\frac12(\sqrt{v^2+u^2}+u)},
y=\pm\sqrt{\frac12(\sqrt{v^2+u^2}-u)},$$
where the sign of $xy$ must match the sign of $v$.
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