Find the image of $f(z) = e^{-\frac{1+z}{1-z}}$ I am trying to solve the following problem: let $f(z) = e^{-\frac{1+z}{1-z}}$, and let $\mathbb{D} = \{z: |z|<1\}$. What is the image of $\mathbb{D}$, and for each $w$ in the image, what are all of its preimages?
So far, I've noted that $-\frac{1+z}{1-z}= \frac{x^2 + y^2 -1}{(1-x)^2+y^2} + \frac{-2y}{(1-x)^2+y^2}i$ for $z=x+iy$. Since the real part of this is negative, I've concluded that $f(\mathbb{D}) \subset \mathbb{D}$. My hypothesis is that $f(\mathbb{D}) = \mathbb{D}-\{0\}$. I tried to prove this by showing that for any $0<a<1$ and $-\pi<b\leq\pi$, there exists a $z$ such that $f(z) = e^a(\cos b+ i \sin b)$ but this lead to an overwhelming amount of algebra, and it seems like there should be a sleeker method.

$f(z)=\exp(g(z))$ where $g(z)=-\frac{1+z}{1-z}$. $g$ maps the unit disk $\mathbb{D}$ onto $\mathbb{H} = \{z : \mathrm{Re}(z) < 0\}$ bijectively; (see Moebius transformation). Then $\exp$ sends $\mathbb{H}$ onto $\mathbb{D}\setminus\{0\}$, as you correctly said.
