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 ?

Note that 
\begin{align*}
(\sqrt{-3-2i}+\sqrt{-3+2i})^2 &= (-3-2i) + 2\sqrt{(-3-2i)(-3+2i)} + (-3+2i) \\
&= -6 + 2\sqrt{13}\\
&= 2(\sqrt{13} - 3).
\end{align*}
