<s>[INST] 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 ?
 [/INST] Note that the right hand side is real, so if we take complex conjugates it remains unchanged.
$$
\begin{align}
\sqrt{-3-2i}+\sqrt{-3+2i} &= \sqrt{2(\sqrt{13}-3)} \\
& = \sqrt{-3+2i}+\sqrt{-3-2i} \\ 
\end{align}
$$
Multiply the 2 conjugate forms to obtain 
$$
\begin{align}
&(\sqrt{-3-2i}+\sqrt{-3+2i})\cdot (\sqrt{-3+2i}+\sqrt{-3-2i}) \\
&= \sqrt{13}+(-3-2i)+(-3+2i)+\sqrt{13} \\ 
&= 2(\sqrt{13}-3) \\
&= \left(\sqrt{2(\sqrt{13}-3)}\right)^2
\end{align}
$$
Alternatively just rearrange the terms of the right-hand side rather than looking at it as a complex conjugate.
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