# How can I show $\sin(z)$ is injective on an infinite strip?

I would like to show $$\sin(z)$$ is an injective map on the set $$S=\{-\pi0\}.$$ 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!

• 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. – TheSilverDoe Sep 26 '18 at 12:50
• I didn’t think so, thanks for confirming that! @TheSilverDoe – coreyman317 Sep 26 '18 at 12:51

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