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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!

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    You have to show that you can always choose $k$ so that $z$ is in $S$. (To do the symbols here, you put them between dollar signs and use LaTeX. E.g., $r^{1/3}(\cos(\theta)+i\sin(\theta))$ is written as `$r^{1/3}(\cos(\theta)+i\sin(\theta))$`. $\mathrm{cis}(\theta)$ can be written as `$\mathrm{cis}(\theta)$`. See the answers [here](http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto) for suggestions on how to get used to LaTeX.)2012-06-02

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Compute $g(r e^{i \frac{2 \pi}{3}})$ and $g(r)$. What do you notice?

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    Hence not injective...2012-06-03