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!

  • 1
    $\begingroup$ 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? $\endgroup$ – Jonas Meyer Jun 1 '12 at 5:54
  • $\begingroup$ 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. $\endgroup$ – JackReacher Jun 2 '12 at 7:42
  • 1
    $\begingroup$ 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 $z^3=w$? $\endgroup$ – Jonas Meyer Jun 2 '12 at 7:48
  • $\begingroup$ 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? $\endgroup$ – JackReacher Jun 2 '12 at 8:22
  • 1
    $\begingroup$ 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.) $\endgroup$ – Jonas Meyer Jun 2 '12 at 8:34

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

| cite | improve this answer | |
  • $\begingroup$ Yes, they give the same point in the codomain. Thanks! $\endgroup$ – JackReacher Jun 3 '12 at 2:33
  • $\begingroup$ Hence not injective... $\endgroup$ – copper.hat Jun 3 '12 at 2:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.