if $w$ is a complex number, how to show that $w^\left(1/2\right)$ has 2 roots? except the case for $w=0$

Question

I am looking for a convincing argument to show that if $w$ is a complex number, $w^{\frac 12}$ has 2 different roots.

Answer 1

Observe that if $(x+iy)^2=w$, then $((-x)+i(-y))^2=w$ as well.

Answer 2

Consider the polar representation $w=r e^{i(\phi+2k\pi)}$.

Its square root yields 2 distinct roots that repeat ad infinitum.

That is: $$w^{1/2}=\sqrt r e^{i(\phi/2+k\pi)}$$ If $r=0$ we have only 1 root, which is 0. Otherwise we get the 2 distinct roots $\sqrt r e^{i\phi/2}$ and $\sqrt r e^{i(\phi/2+\pi)}$. All other values of $k$ map to these 2 roots due to the $2\pi$ periodicity.