Let $\displaystyle f:\mathbb{C}\to \mathbb{C}$ be defined as $\displaystyle f(x)=x^\frac{1}{2}$. Then $\displaystyle f$ is not a well defined map, because there exists $\displaystyle x\in\mathbb{C}$ such that $\displaystyle f(x)$ maps to two different elements in $\displaystyle \mathbb{C}$
This doesn't really apply to your proof, but in general be careful when you use the square root with complex numbers, as you'll often get multiple answers. In fact, when you use the square root with real numbers, there are multiple possible answers, but it's convention to ignore the negative parts. I could do the same proof without complex numbers.
We know that $\displaystyle (-6)^2=36$ and the square root and square operations are inverses so they cancel and $\displaystyle \sqrt{36}=\sqrt{(-6)^2}=-6$ .
And we know that $\displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab}$ .
example: $\displaystyle \sqrt{4}\sqrt{9} = \sqrt{4(9)} = \sqrt{36}$
So $\displaystyle 6 = 2(3) = \sqrt{4}\sqrt{9} = \sqrt{(4)(9)} = \sqrt{36} $ .
So $\displaystyle 6=-6$. gg.
Where is the problem with this?