TITLE: Let $f(z) = \frac{1+e^z}{1-e^z} $. Determine the image of $f(A)$
QUESTION [5 upvotes]: On all the complex analysis-exams written by my professor, a question of this nature always pops up:

Let $$f(z)=\frac{1+e^z}{1-e^z}$$
and $A=\{z:\Re(z)<0, \ -\pi<\Im(z)<\pi\}.$ Determine the image $f(A)$.
  Hint: use the fact that $f(z)=M(e^z)$ where $M(z)=\frac{1+z}{1-z}.$

I'm wondering if anyone can somehow break down a solution for problems like these in steps. Like, step 1: check this, step 2: compute this and so on. I already have the profesors solutions but they are too cryptic for me. I need this problem simplified somehow so I can understand what needs to be done.
My initial thought is simply that we have an area $A$, which is an infinite rectangular area that is to the left of the complex halfplane but bounded by $-\pi$ and $\pi$ on the imaginary axis. So if every point inside of $A$ undergoes the transformation $f(z),$ they will form a different figure, i.e the image $f(A).$
However, I don't understand how I'm supposed to do the arithmetic here and use Möbius transforms.
Any help is greatly appreciated, but please no usage of fancy math that is outside the scope of complex analysis.

REPLY [2 votes]: First, determine the image of $A$ under $z \mapsto e^z$. It is fairly easy to see that this will be the interior of the unit disk (minus the part of the negative real axis within the unit disk) by writing $e^{x+iy} = e^x \cdot e^{iy}$ . Then use the fact that Möbius transformations map circles and lines to circles and lines on the boundary of this region.