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 should use $(x+y)^2=x^2+y^2+2xy$
$(a-ib)^2=4i$
$a^2+i^2b^2-2iab=4i$
$(a^2-b^2)-2iab=0+4i$
compare real and imaginary parts of both sides 
$a^2-b^2=0\ \ ........(1)$
$ab=-2\ \ ........(2)$
