I would like to show $\sin(z)$ is an injective map on the set $S=\{-\pi<x<\pi\space\text{and}\space y>0\}.$ Normally, one would show that if $\sin(z_1)=\sin(z_2)$, then $z_1=z_2$.

If $\sin(z_1)=\sin(z_2)$ then $e^{iz_1}-e^{-iz_1}=e^{iz_2}-e^{-iz_2}$. So making the substitution $u=e^{iz_1}$ and $w=e^{iz_2}$, we have the relation: $$u-\frac{1}{u}=w-\frac{1}{w}$$

Here is where I am stuck. I am having trouble moving from the relation to the fact that $z_1=z_2$ to show $\sin(z)$ is injective.

Hints are best! Thanks!

  • $\begingroup$ You are having hard time "digesting the notion of sign in $\mathbb{C}$" because this notion has no sense ! You cannot speak about monotony for a complex function. $\endgroup$ – TheSilverDoe Sep 26 '18 at 12:50
  • $\begingroup$ I didn’t think so, thanks for confirming that! @TheSilverDoe $\endgroup$ – coreyman317 Sep 26 '18 at 12:51

Your first approach does not work because complex numbers do not have a sign, and monotonicity makes no sense.

For your second approach, consider $u$ and $w$ with $u-\frac1u=w-\frac1w$. We want to show $u=w$ or $u=-\frac1w$.

We have $u-w=-\frac1w+\frac1u=-\frac{u-w}{uw}$ so $(u-w)(1+\frac1{uw})=0$. Hence $u=w$ or $u=-\frac1w$. Your choice of $S$ should imply that you must have $u=w$.

  • $\begingroup$ In the first line of the third paragraph, wouldn’t it be $u-\frac{1}{u}=w-\frac{1}{w} \implies u-w=\frac{1}{u}-\frac{1}{w}$? $\endgroup$ – coreyman317 Sep 26 '18 at 13:05
  • $\begingroup$ Yes, thank you, fixed. $\endgroup$ – Kusma Sep 26 '18 at 13:27

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