Inverse of Conformal Map of Semiinfinite strip to the unit circle, why the inverse doesn't work? I am dealing with a composition of two conformal maps $\gamma$ and $\omega$ to transform a vertical strip, S = {(x,y) : −π/2 ≤ x ≤ π/2, y ≥ 0}, to the unit circle. The resulting map is,

my $\gamma$ is the sine function which transforms the strip to upper half plane and $\omega$ transforms the upper half plane to the unit circle. But to find the new boundary conditions I need to use the inverse of this map. I know it is an elementary calculation but taking the inverse and plotting it gives the image below. Does anyone here have an idea about my mistake? In the original map I didn't have y<0 values, but after inverse mapping I get the negative values of y too. I am open for all new ideas.

 A: Your map is fine. Your picture is also fine. So your real mistake lies in the intuition you're using to decide there must be something wrong!
Here's another example to think about: $f(z) = z^2$ is a conformal mapping from the first quadrant onto the upper half plane. However, $f^{-1}(UHP)$ is the first AND third quadrants. We could still use $f$ to get a valid conformal mapping in the other direction, but we'd need to define that map as "$g(z) = $ (the unique $w$ in the first quadrant such that $f(w) = z$."
Using your map, the bottom half strip $T = \{x+iy : -\pi/2<x<\pi/2, y<0\}$ gets mapped conformally onto $\{z \in \mathbb{C} : |z|>1\} \cup \{\infty\}$. The boundary of that set is still the unit circle, so when you plot the inverse image of the unit circle then you end up with the boundary of $S$ and also the boundary of $T$.
Actually, for any $k \in \mathbb{Z}$, your function also conformally maps the strip $T+2k\pi$ onto the unit disc. So if anything, I'm surprised that your plot doesn't show even more boundary segments. (Maybe you just didn't plot a wide enough region to see them though.)
TLDR: Your map is fine, and your mistake is that you're assuming a conformal mapping has to be an injective function on all of $\mathbb{C}$. Actually, a conformal mapping just needs to be injective on the domain you're working with. That means taking the inverse image of a conformal mapping can give you infinitely many different regions! If you want to work with the inverse function for your conformal mapping, then you should use branch cuts etc to make sure that you always choose the unique inverse that lies in your original domain.

Btw quick note: you're often saying "unit circle" when you mean "unit disc". There's an important difference between those (the unit disc is the open region enclosed by the unit circle) and you should try to train yourself to say the one you mean. No big deal this time though since the context made it clear enough what you meant.
