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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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What do the inequalities in "r^3 cis (0) <= z^3 <= r^3 cis (2Pi)" mean? You did not show that the function is surjective. Do you know how to use polar form to find roots? –  Jonas Meyer Jun 1 '12 at 5:54
    
I guess what I was getting at is that z^3 can take any argument between 0 degrees and 360 degrees, so the range is surjective. All complex numbers can be covered? Yes, I know the formula for finding roots of complex numbers in polar form. How would I use that here? Thanks for any help you can give me. –  JackReacher Jun 2 '12 at 7:42
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To show a map $f:A\to B$ is surjective means to show that for all $b\in B$ there exists $a\in A$ such that $f(a)=b$. So suppose that $w$ is a complex number. Can you find $z\in S$ such that $z^3=w$? –  Jonas Meyer Jun 2 '12 at 7:48
    
ok, so z = w^(1/3). Let w = r cis (theta) (sorry I don't know how to do the symbols here). the z = w^(1/3) = r^(1/3) cis ((Theta + 2pi*k)/3) where k = 0,1,2. How would I proceed from here? –  JackReacher Jun 2 '12 at 8:22
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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 for suggestions on how to get used to LaTeX.) –  Jonas Meyer Jun 2 '12 at 8:34

1 Answer 1

up vote 1 down vote accepted

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

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Yes, they give the same point in the codomain. Thanks! –  JackReacher Jun 3 '12 at 2:33
    
Hence not injective... –  copper.hat Jun 3 '12 at 2:39

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