plotting $\frac{-\pi}{2}<x<\frac{\pi}{2} $ and $ 0<y<1$ under mapping $w=\sin(z)$

Question

i need to plot this $\frac{-\pi}{2}<x<\frac{\pi}{2} $ and $ 0<y<1$ under $w=\sin(z)$ mapping so what i did is $ y=0 , \frac{-\pi}{2}<x<\frac{\pi}{2} => -1<u<1 , v=0 $
$ y=1 , \frac{-\pi}{2}<x<\frac{\pi}{2} => (\frac{u^2}{\cosh(1)^2})+(\frac{v^2}{\sinh(1)^2})=1 $
$ x=\frac{\pi}{2} , 0<y<1 => 1<u<\cosh(1) , v=0$ $x=\frac{-\pi}{2} , 0<y<1 => -1<u<-\cosh(1) , v=0 $
now how should i plot

Answer 1

I think you've got it, all you need to do is to combine these into a plot.

I guess you want something like this?

mapping

edit

*error in the picture, it should be the upper half of the ellips.

Because when $$\gamma = \frac{\pi}{2}t + i \quad\text{then}\quad \sin(\Gamma) = -\cosh 1 \sin\frac{\pi}{2}t +i\sinh 1 \cos\frac{\pi}{2}t$$

Meaning if $t = 0$ then this evaluates as $0+i\sinh1$