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're doing great.  Two complex numbers are equal when the real and imaginary parts agree.  Hence you need to solve the system of equations $$\{a^2-b^2=0,~~ -(ab+2)2=0\}$$
Since
$\sqrt{i}
=\pm\frac{1+i}{\sqrt{2}}
$,
$a-ib
=\pm2\frac{1+i}{\sqrt{2}}
=\pm(1+i)\sqrt{2}
$,
so
$a=\pm\sqrt{2}$
and
$b=\mp\sqrt{2}$.