If $\rho=\cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n})$. Prove that if $z=u$ is a solution of the equation $z^n=w$, then the other solutions have the form $u \rho ^j$ for $j=1,2,...,n-1$.
My proof goes as follows: Since I already proved in another exercise that $z=1, z= \rho, z= \rho ^2,...,z= \rho^{n-1}$ are all the diferent solutions of $z^n=1$. Then $$(u \rho^j)^n=u^n(\rho^j)^n=w1=w$$ for every $j=1,2,...,n-1$. Proving what I wanted. Im still not sure if my prove is OK, and if it is wrong can anyone help by sketching the right proof. Thanks
If $w \ne 0$, By fundamental theorem of algebra, we know that there are at most $n$ roots and you have found $n$ of them which are all distinct, hence you are done.
For $w=0$, $z^n=0$ satisfies $|z|^n=0$ hence $|z|=0$ and $z=0$ is the unique solution and is also satisfied the condition mentioned in the question.