If $a$ is an $n$-th root of $z$ , $b$ is an $n$-th root of $w$, and $c$ is an $n$-th root of $zw$, then $ab=c$.

Question

How do I prove that for all natural numbers $n$ and complex numbers $a, b, c, z, w$ if $a$ is an $n$-th root of $z$ , $b$ is an $n$-th root of $w$, and $c$ is an $n$-th root of $zw$, then $ab=c$.

Thanks.

Answer 1

A trivial counter example would be $n=2, z=w=a=b=1$ and $c=-1$.

Answer 2

You have $a^n=z$ , $b^n=w$ and $c^n=zw$ so $\left(\frac{ab}c\right)^n=1\quad$ (the case $z=0$ or $w=0$ being trivial).

Thus all we can deduce is $\frac{ab}c$ is a n-root of unity, but not necessarily $1$.