for $z,w\in\mathbb{C}$, $\sqrt{zw} = \sqrt{z}\sqrt{w}$?

Question

for $z,w\in\mathbb{C}$, $\sqrt{zw} = \sqrt{z}\sqrt{w}$

I started by writing $z$ and $w$ in polar coordinates, and writing it out, giving another form for the question: $$ \begin{align} y &= r_1r_2cos(\theta_1 + \theta_2) + r_1r_2sin(\theta_1 + \theta_2) i\\ y &= a + bi \end{align} $$ so $$ \sqrt{zw} = \sqrt{r_1(cos(\theta_1)+isin(\theta_2))}\sqrt{r_2(cos(\theta_2)+isin(\theta_2))} $$ but how to prove the original problem without just saying: square both sides?

Answer 1

This is very dangerous...

Even if you choose a consistent definition (e.g. $\sqrt{re^{i\theta}}=\sqrt{r}e^{i\theta/2}$ with $\theta\in[0,2\pi]$), what happens when $z=i$ and $w=-i$?

Answer 2

Another correct formulation.
A complex number may have two square-roots. If you take any square-root of $w$ and multiply it by any square-root of $z$, the product will be a square-root of $zw$.