Solve $z^6=(z-1)^6$. In the answer of Surb here : How to solve for the complex number $z$?
I don't understand the subtlety. To me it's natural to do $$z^6=(z-1)^6\iff \left(\frac{z}{z-1}\right)^6\iff \frac{z}{z-1}=e^{ki\pi/3}$$
for $k=0,...,5$. I can see that the case $k=0$, but I don't understand why... We do as usual: $$v^6=1\iff v=e^{ki\pi /3}$$
and now replace $v$ by $\frac{z}{z-1}$, we should get the result. So why it doesn't work here ? Since, by the way, $\frac{z}{z-1}$ is well defined for $z\neq 1$ and that $z=1$ is not solution of $z^6=(z-1)^6$.

There are $6$ possible values for $v$ that you get from the equation $v^6=1$.  But to go from a value of $v$ to a value of $z$ which solves the original equation, you have to solve $v=\frac{z}{z-1}$ for $z$, and when $v=1$ there is no such solution.  For if $v=1$, then we would have to have $1=\frac{z}{z-1}$, or $z=z-1$, and this is impossible.
