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Define the region $S:= \{z\in \mathbb{C}:0<\Re(z)<\frac{\pi}{2},\Im(z)>0\}$.
Find the image of $w = \cos z$.

I tried expanding $\cos z = \cos (x+iy)$ and equating real and imaginary parts, but that seemed tedious.

So I tried:
$$\cos z = \frac{e^{iz} + e^{-iz}}{2} = w \\ e^{2iz} - 2e^{iz}w + 1 = 0 \\ e^{iz} = \frac{2w \pm \sqrt{4w^2 - 4}}{2} = w \pm \sqrt{w^2 - 1} \\ iz = \mathrm{Ln}(w\pm\sqrt{w^2 - 1}) = \ln(|w\pm \sqrt{w^2 - 1}|) + i\mathrm{Arg}(w\pm\sqrt{w^2 - 1}) + 2k\pi i \\ z = -i\ln(|w\pm\sqrt{w^2 - 1}|) + \mathrm{Arg}(w\pm\sqrt{w^2 - 1}) + 2k\pi.$$
Using the conditions in the region,
$$\Im(z) = -\ln(|w\pm \sqrt{w^2 - 1}) > 0\\ |w\pm \sqrt{w^2 - 1}| < 1$$

Also $$0<\Re(z) = \mathrm{Arg}(w\pm\sqrt{w^2 - 1}) + 2k\pi < \frac{\pi}{2}$$

I'm not sure how to interpret these...... Is there a better way to do it?
(I think what I'm supposed to get are a bunch of ellipses and hyperbolae?)

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$\cos z=\cos (x+iy)=\cos x \cosh y -i \sin x \sinh y$

When $0<x<\dfrac{\pi}{2}$ and $y>0$ we have

$0< \cos x< 1 \to 0< \cos x \cosh y < \cosh y$

and as $y\to +\infty$ this means that the real part of $\cos z$ is in $(0;\;+\infty)$

Imaginary part is $(- \sin x \sinh y)$ and as $0<\sin x <1$ in $\left(0;\;\dfrac{\pi}{2}\right)$ and $\sinh y \to +\infty$ as $y\to +\infty$ the imaginary part is always strictly less than zero.

The image of the given region is the quadrant of the Argand plane where $x>0;\;y<0$

hope this helps

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