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!

We could use the general approach for solving $z^n = Re^{i\theta}$
First, let $z$ = $a-bi$ 
Then we have $$z^2 = 4i = 4e^{({\pi \over 2})i} = 4e^{({\pi \over 2}+2k\pi)i},  k = 0, 1$$
$$z = 2e^{({\pi \over 4}+{k\pi)i}}, k=0,1$$
You can then convert $z$ back to cartesian form and compare with $a-bi$.
