Finding the square root of a complex number - why two solutions instead of four?

I want to find the square roots of a complex number, $w = a+ib \in \mathbb{C}$, i.e. I'm looking for solutions, $z = x + iy$, for the equation $z^2 = w$.

This question has been asked here a couple of times, but I still don't get why there are only the two solutions,
$$z = \pm \left( \sqrt{\frac{|w|+a}{2}} + i \text{sgn}(b)\sqrt{\frac{|w|-a}{2}}\right)$$ .

How do I see that my intermediate results $$x=\pm \sqrt{\frac{|w|+a}{2}}\\ y=\pm \sqrt{\frac{|w|-a}{2}}$$ cannot be combined to yield four solutions? I feel like I am missing something quite elementary here.

Answer

The identification $(x+\mathrm i\mkern1mu y)^2=a+\mathrm i\mkern1mu b$ leads to the equality $2xy=b$, hence the signs of $x$ and $y$ are the same if $b> 0$, opposite if $b<0$. This shows you cannot combine the signs of the real and imaginary parts of $z$ in an arbitrary way.