This is correct for complex numbers in disk $\mathbb{D}(0,1)$ $$\sqrt{(1-z^{2})^{2}} = |1-z^{2}|$$
How to prove them?
I need compute $\frac{-(1+z^{2}) + |1-z^{2}|}{2z}$
For real numbers I use fact that $z \in [-1,1]$ and it's easy.
How to simplify for complex numbers?
The formula is incorrect. Let $z=(-5+4i)/7$. Then $z^2=(9-40i)/49$ so that $$ (1-z^2)^2 = 2\left(\frac{40}{49}\right)^2 i, $$ which cannot have a real number as a square root.
However, if you meant to write $$ \sqrt{|1-z^2|^2}=|1-z^2| $$ then this is correct, since $|1-z^2|$ is a positive real number it is a consequence of the identity $\sqrt{a^2}=a$ which holds for all positive real numbers $a$.