How to show mapping $w = \sin z$ is one to one?

Question

If I consider the region in $xy$ plane $$\frac{-\pi}{2} \leq x \leq \frac{\pi}{2} \,\text{ and } \,y \geq 0$$
is mapped to upper half $v \geq 0$ of $w$ plane under $w = \sin z$. It can be easily seen that the line $x = c$ where $c$ is between $\frac{-\pi}{2} \leq c \leq \frac{\pi}{2}$ is mapped to a hyperbola and consequently it is easily observed that the above region is mapped to upper half plane in $uv$ plane. I wonder if this map is a one to one. What do you think? If yes how can I prove it ?

Answer 1

$$w = \sin z = \frac{1}{2i} ( e^{iz} - e^{-iz})$$

$$2 i w e^{iz} = e^{2iz} -1$$

$$ e^{2iz} - 2iw e^{iz} -1 = 0$$

$$ e^{iz} = iw \mp \sqrt{1-w^2 }$$

$$z = -i \log \left[ iw \mp \sqrt{1-w^2} \right]$$

We need to see if both or just one of these lie in the half-strip ...

The function $w=\sin z$ is periodic with period $2\pi$. Also, $$\sin (-z) = - \sin z$$ so we only need to show that the function is one-to-one in the half strip $0\le x\le \pi/2$ and $y\ge 0$.