The question:

Simplify $\displaystyle (\sqrt{3 + 4i}+\sqrt{3-4i})^2$ where we assume $\displaystyle \sqrt{z}$ has a non negative real part.

My attempt:

$\displaystyle (3+4i) + 2(9-16i^2) + (3-4i)$ (perfect square)

$\displaystyle 3 + 4i + 18 + 32 + 3 - 4i$

= 56

That's not the answer though, it's supposed to be 16. I'm not sure what I'm going wrong. Any assistance would be great!