TITLE: Finding the square root of a complex number - why two solutions instead of four?
QUESTION [16 upvotes]: 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.

REPLY [2 votes]: Understanding squaring and taking roots in the complex numbers (or more general anything about multiplication that does not involve addition) is easier to understand in the 'polar coordinate' picture, i.e. using the length/modulus (distance to zero) and argument (angle between the pos real axis and the line from 0 to the number) of the number rather than the real and imaginary part. 
If we multiply two complex numbers their arguments add up and their lengths multiply. Now we look for numbers $w$ such that $w^2 = z$. This means that the length of $w$ multiplied by itself is the length of $z$ and that the argument of $w$ added to itself (i.e. multiplied by 2) equals the argument of $z$. 
It follows that $|w| = \sqrt{|z|}$ (here we can only use the positive root because it is a length) and $\arg(w) = 1/2 \arg(z)$, or - surprise twist - $\arg(w) = 1/2 \arg(z) + \pi$ (because $2\pi = 0$ in terms of angles). This is all there is.
(I am a aware that this is more an answer to the question in the title than to the one in the body of your post)