Finding real numbers such that $(a-ib)^2 = 4i$ Prove that $(a^2 - b^2) = 0$ I sometimes find myself overcomplicating my life... overthinking simple concepts. 

Here I don't use what's given, i.e., $$(a − ib)^2 = 4i$$

So I might say let $a = 1$ and $b = 1$ 
then $a = b$ and $a^2 = b^2$ thus $a^2 - b^2 = 0$
Now that seems fine but I'm given the complex number $(a - ib)^2 = 4i$
Now I know $i^2 = -1$
So here's my attempt
$(a - ib)(a - ib) = 4i$
$a^2 - abi - abi + (bi)^2 - 4i = 0$
$a^2 - 2abi - b^2 - 4i = 0$
$ a^2 - b^2 -(ab + 2)2i = 0$
I've obviously not grasped this correctly as this certainly is not what I have been asked to prove. 
Please could I have come guidance on how to solve this simple equation. 
I noticed I left out what I am intending to prove which has been included in the title now namely:

Prove that $a^2 - b^2 = 0$

Thanks!

You are on the right track, but you are overthinking it. The basic idea behind these sorts of questions is the fact that if $z = x + iy$ and $w = q + ip$ where $z,w \in \mathbb{C}$ and $x,y,q,p \in \mathbb{R}$ then $z = w$ iff $x = q$ and $y = p$.
Equipped with this we may deal with your problem. Indeed first we compute the left side 
$$ (a - ib)^2 = (a^2 - b^2) + i (-2ab) = x + iy$$
and 
$$ 4i = 0 + 4i = p + i q$$
Now we must have
$$x = p \iff a^2 - b^2 = 0, \ y = q \iff 4 = -2ab$$
Using these two equations can you now solve for $a$ and $b$?
