The identity $ \sqrt[n]{z}\sqrt[n]{w} = \sqrt[n]{zw}$ for complex numbers

Question

In the general case, when $z$ and $w$ are two complex numbers, we have that

$ (1) \sqrt[n]{z}\sqrt[n]{w} \neq \sqrt[n]{zw}$

For example, $\sqrt{-1}\sqrt{-1} \neq \sqrt{-1.-1} = 1$.

However, there is a condition by which (1) is true:

$ (2) \sqrt[n]{z}\sqrt[n]{w} = \sqrt[n]{zw} \iff -\pi < \arg z + \arg w \leq \pi$

In this case, the principal value of the right hand side coincides with the left hand side.

In the example above given, $\arg -1 + \arg -1 = 2\pi $ and hence the identy $(2)$ does not hold.

The question is: how to prove$(2)$? And how to prove the case where the radicals have different index, say $m$ and $n$?

Answer 1

I assume the definition of $\sqrt[n]{z}$ is $\exp\left(\frac{\ln(z)}{n}\right)$, where $\ln(z)$ is the logarithm with a branch cut along the negative reals. [That means that $\ln\left(re^{i\theta}\right)=\ln(r)+i\theta$ for $\theta\in(-\pi,\pi]$, $r$ real.]

So then using the convention that $z=r_ze^{i\theta_z}$ with $\theta_z\in(-\pi,\pi]$, you have: $$\begin{align} \sqrt[n]{z}\sqrt[n]{w} &=\sqrt[n]{r_ze^{i\theta_z}}\sqrt[n]{r_we^{i\theta_w}}\\ &=\exp\left(\frac{\ln(r_ze^{i\theta_z})}{n}\right)\exp\left(\frac{\ln(r_we^{i\theta_w})}{n}\right)\\ &=\exp\left(\frac{\ln(r_z)+{i\theta_z}}{n}\right)\exp\left(\frac{\ln(r_w)+{i\theta_w}}{n}\right)\\ &=\exp\left(\frac{\ln(r_z)+{i\theta_z}+\ln(r_w)+{i\theta_w}}{n}\right)\\ &=\exp\left(\frac{\ln(r_zr_w)+i(\theta_z+\theta_w)}{n}\right)\\ &=\sqrt[n]{r_zr_w}\exp\left(\frac{i(\theta_z+\theta_w)}{n}\right) \end{align}$$

Meanwhile $$\begin{align} \sqrt[n]{zw} &=\sqrt[n]{r_ze^{i\theta_z}r_we^{i\theta_w}}\\ &=\sqrt[n]{r_zr_we^{i(\theta_z+\theta_w)}}\\ &=\exp\left(\frac{\ln(r_zr_we^{i(\theta_z+\theta_w)})}{n}\right)\\ &=\exp\left(\frac{\ln(r_zr_w)+i(\theta_z+\theta_w)^\dagger}{n}\right)\\ &=\sqrt[n]{r_zr_w}\exp\left(\frac{i(\theta_z+\theta_w)^\dagger}{n}\right)\\ \end{align}$$

Where the $\dagger$ indicates the corresponding angle in $(-\pi,\pi]$. The two expressions are equal if and only if $(\theta_z+\theta_w)^\dagger=\theta_z+\theta_w$. That is, they are equal if and only if the sum of the principal arguments of $z$ and $w$ is itself a principal argument.