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 ?
 A: Just power the left equation like the following:
$$(\sqrt{-3-2i} + \sqrt{-3 + 2i})^2 = -3 - 2i - 3 + 2i + 2\sqrt{(-3 + 2i)\times(-3 - 2i)} = -6 + 2 \sqrt{9 - (2i)^2} = -6 + 2\sqrt{13} = 2(\sqrt{13}-3)$$
A: hint
$$z_1=-3+2i=\sqrt {(-3)^2+2^2}e^{it} $$
$$=\sqrt {13}e^{it} $$
$$z_2=-3-2i=\sqrt {13}e^{-it} $$
$$\sqrt {z_1}+\sqrt {z_2}=2\sqrt{\sqrt {13}}\cos (\frac {t}{2}) $$
with $$\cos (t)=\frac {-3}{\sqrt {13}} $$
and
$$\cos (\frac {t}{2})=\sqrt {\frac {1+\cos (t)}{2}   }$$
A: 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*}
A: 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*}
A: Consider the following.
\begin{align}
\sqrt{-3-2i}+\sqrt{-3+2i} &= \sqrt{a} \\
(\sqrt{-3 - 2 i} + \sqrt{-3 + 2 i} )^2 &= a \\
-6 + \sqrt{(-3-2i)(-3+2i)} &= a \\
(-3-2i)(-3+2i) &= (a+6)^2 \\
(a+6)^2 - 13 &=0 \\
a^2 + 12 a + 23 &= 0 \\
\end{align}
from here it is determined that $a = -6 \pm \sqrt{13}$, which yields the desired result.
A: 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.
A: 
$$\sqrt{-3-2i}+\sqrt{-3+2i} \ne \sqrt{2(\sqrt{13}-3)}$$

Two complex values are equal iff they are at the same location on the complex plane.  
$-3 - 2i$ is in quadrant 3.  So $\sqrt{-3 -2i}$ is in quadrant 2, because square roots half the angle.
$-3 + 2i$ is quadrant 2.  So $\sqrt{-3 + 2i}$ is in quadrant 1, not on the positive real axis.
So  $\sqrt{-3 - 2i} +  \sqrt{-3 + 2i}$ is in quadrant 1 or quadrant 2, not on the real axis.  So it cannot be equal to any real number.
