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.
 A: As I explained here. there is  a very simple formula for denesting such radicals, namely
Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
$\ 3+2\sqrt{-10}\ $ has norm $= 49.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 7\,\ $ yields $\ {-}4+2\sqrt{-10}\:$
with $\, {\rm\ \sqrt{trace}}\, =\, 2\sqrt{-2}.\ \ \ \rm \color{brown}{Dividing\ this\ out}\ $ of the above we obtain $\,\ \sqrt{-2} + \sqrt 5$ 
Checking it we find  $\,\ (\sqrt{-2} + \sqrt 5)^2 =\, -2+5 + 2\sqrt{-2}\sqrt 5\, =\, 3+ 2\sqrt{-10}$
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
A: 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 $v$.
A: $z=3+2i\sqrt{10}$ is a complex number with modulus $7=\sqrt{9+40}$ in the first quadrant ($\text{Re}(x),\text{Re}(y)>0$), hence we have
$$ z=7 e^{i\theta} $$
with 
$$\theta = \arctan\frac{2\sqrt{10}}{3}=2\arctan t.$$
Since $\tan(2u)=\frac{2\tan(u)}{1-\tan^2(u)}$, by solving $\frac{2t}{1-t^2}=\frac{2\sqrt{10}}{3}$ with the constraint $t>0$  we get $t=\sqrt{\frac{2}{5}}$, hence the square root of $z$ in the first quadrant is given by:
$$ \sqrt{z} = \sqrt{7} e^{it} = \sqrt{7}\left(\cos\arctan\sqrt{\frac{2}{5}}+i\sin\arctan\sqrt{\frac{2}{5}}\right) $$
or:
$$ \sqrt{z} = \sqrt{7}\left(\sqrt{\frac{5}{7}}+i\sqrt{\frac{2}{7}}\right) = \color{red}{\sqrt{5}+i\sqrt{2}}. $$
Let we check the solution so found:
$$ (\sqrt{5}+i\sqrt{2})^2 = 5-2+2i\sqrt{2\cdot 5} = 3+2i\sqrt{10}.$$
A: 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 $y$.
A: $$ \sqrt{3 + 2\sqrt{10}i} = x + iy
$$
$$ 3 + 2\sqrt{10}i=x^2 -y^2 +2ixy $$
$$ x^2 -y^2 =3$$
$$ 2xy=2 \sqrt{10}$$
$$( x^2 + y^2 )^2= ( x^2 - y^2 )^2 +2xy $$
$$ x^2 + y^2 =7$$
$$ x^2 -y^2 =3$$
A: You want $z=a+bi$ where $a,b\in\mathbb{R}$ such that
$$z^2=a^2-b^2+2abi=3+2\sqrt{10}i.$$
Comparing coefficients, you need $a^2-b^2=3$ and $ab=\sqrt{10}$. So $a=\sqrt{5}$ and $b=\sqrt{2}$, or $a=-\sqrt{5}$ and $b=-\sqrt{2}$. Thus the square root of $3+2\sqrt{10}$:
$$\pm(\sqrt{5}+\sqrt{2}i).$$
