Tough complex numbers problem Let $c$ be a complex number. Suppose there exist distinct complex numbers $r$, $s$, and $t$ such that for every complex number $z$, we have
  $$(z - r)(z - s)(z - t) = (z - cr)(z - cs)(z - ct).$$
Compute the number of distinct possible values of $c$.
OK, so I thought $1, \omega,$ and $\omega^2$ were the only answers, but apparently I'm missing something. Any help?

Let $p(z)$ denote the cubic on the left-hand side; the right-hand side is then $c^3p(z/c)$. Write $p(z)=z^3+Az^2+Bz+C$ so$$z^3+Az^2+Bz+C\equiv z^3+cAz^2+c^2Bz+c^3C\\\implies(c-1)A=(c^2-1)B=(c^3-1)C=0.$$If $r,\,s,\,t$ are all nonzero, $C\ne0$ so $c^3=1$ as per your reasoning. For $c=\exp\frac{\pm 2\pi i}{3}$ to work, we need $A=B=0$, so $p(z)=z^3+C$, with three distinct roots as required, provided $C\ne0$.
Suppose without loss of generality $r=0$ so $s+t=-A,\,B=st\ne0,\,C=0$, and $c^2=1$ so $c=\pm1$. Only one of these values for $c$ are new to us, namely $c=-1$, whence $A=0$ and $t=-s$. Indeed, $z(z-s)(z+s)=z(z+s)(z-s)$ works for any $s\ne0$.
So, there are four possible values of $c$.
