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!
 A: 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 $$\{a^2-b^2=0,~~ -(ab+2)2=0\}$$
A: You're on the right track: If $u + iv = 0$ for real numbers $u, v$, we must have $u = v = 0$. So, your equation $$a^2 - b^2 - (a b + 2) 2 i$$ is equivalent to the system
$$
\left\{
\begin{array}{rcl}
0 &=& a^2 - b^2 \\
0 &=& 2(a b + 2)
\end{array}
\right.
$$
To solve this system, note first the we can rewrite the first equation as $$0 = (a - b) (a + b),$$ so $b = \pm a$.
A: 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 $b$?
A: 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 
$a^2-b^2=0\ \ ........(1)$
$ab=-2\ \ ........(2)$
A: 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 $a-bi$.
A: 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
$b=\mp\sqrt{2}$. 
