Let $S$ denote the closed sector $0 \leq \arg (z) \leq 2\pi/3$, in the complex plane, including the vertex at $z = 0$. Show that the function, $g: S \rightarrow \mathbb{C}$ , given by $z\mapsto z^3$, is surjective but not injective.
Another exam prep question.
Since we are given the argument sector, is the best way to evaluate this using Polar form? $z = r \mathrm{cis}(-)$ (data)
so $r^3 \mathrm{cis} (0) \leq z^3 \leq r^3 \mathrm{cis} (2\pi)$
So I guess this shows that the function is surjective. But how do I go about showing it is not injective?
Many thanks!