<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] Just using calculus, you are looking for the number $z=x+iy$ such that $z^2=3+2\sqrt{10}i$. Expand $$z^2=x^2-y^2+2ixy=3+2\sqrt{10}i$$ Identify the real and imaginary parts; this gives $$x^2-y^2=3\tag 1$$ $$xy=\sqrt{10}\implies y=\frac {\sqrt{10}}x\tag 2$$ So, equation $(1)$ becomes $$x^2-\frac {10}{x^2}=3\implies x^4-3x^2-10=0\tag 3$$ Which is a quadratic equation in $x^2$; let $t=x^2$ and solve $t^2-3t-10=0$ the roots of which are $??$. One solution must be discarded since the product of the roots is negative; so, only one $t$ must be kept. Now, back to $x^2=t$ will give $x$ and $(2)$ will give the corresponding $y$.
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