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Consider the strip $S := \{z \in \mathbb{C}: -\pi < Re(z) < \pi/2\}$. The function $\sin z$ maps $\{z \in \mathbb{C}: -\pi/2 < Re(z) < \pi/2\}$ to the upper half plane. Does $\sin z$ map $S$ into the complex plane minus the third quadrant?

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Your claim that $\sin z$ maps $\{ \, z \in \mathbb{C} \colon -\pi/2 < \Re (z) < \pi/2 \, \}$ to the upper half plane is already false; $z=-i$ is a counterexample.

By definition,

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


$$ \sin(-i) = \frac{e - e^{-1}}{2i} = \frac{e^{-1} - e}{2} i, $$

which has negative imaginary part.

(Even if it were true, the second statement would be false; the element $-\frac{3\pi}{4} + i$ of $S$ provides a counterexample:

$$ \begin{align*} \sin(-\frac{3\pi}{4}+i) &= \frac{1}{2i} \left( e^{-1} \left( \frac{\sqrt{2}}{2} (-1-i) \right) - e \left( \frac{\sqrt{2}}{2} (-1+i) \right) \right) \\ &=\frac{\sqrt{2}}{4} \left( \left( -e - e^{-1} \right) + i \left( e^{-1} - e \right) \right), \end{align*}$$

which is in the third quadrant.)

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