How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$ 
is true ?

Start by squaring the LHS and RHS
\begin{align*}
 (\sqrt{-3-2i} + \sqrt{-3+2i} )^2 &= \sqrt{2(\sqrt{13}-3)}^2 \\
 (-3-2i) + 2\sqrt{(-3-2i)(-3+2i)} + (-3 + 2i) &= 2(\sqrt{13}-3) \\
 -6 + \sqrt{4 (9 + 4)} &= 2(\sqrt{13}-3) \\
 -6 + 2\sqrt{13} &= 2(\sqrt{13}-3) \\ 
 2(\sqrt{13}-3) &= 2(\sqrt{13}-3)
\end{align*}
