Let $f(z)=\exp\left(\dfrac{z+1}{z-1}\right),\quad (z\in\mathbb{D}=\{z: |z|<1\})$
(a) Show that $f(\mathbb{D})=\mathbb{D}\setminus\{0\}$
(b) Show that for all $z\in \mathbb{D}\setminus\{0\}$, the set $$f^{-1}(w)=\{z\in\mathbb{D}\mid f(z)=w\}$$ is infinite and countable.
My attemp for (a):
Let $z=re^{i\theta}, (0\leqslant r<1)$. SO \begin{eqnarray*} |f(re^{i\theta})|&=&\left|\exp(-\dfrac{1+re^{i\theta}}{1-re^{i\theta}})\right|\\ &=& \exp(-Re \dfrac{1+re^{i\theta}}{1-re^{i\theta}})\\ &=& \exp(-Re \dfrac{(1+re^{i\theta})(1-re^{i\theta})}{(1-re^{i\theta})(1-re^{i\theta})}) \\ &=& \exp(-Re\dfrac{1-r^2-2i\sin(\theta)}{1-r^2-2r\cos(\theta)})\\ &=& \exp(- \dfrac{1-r^2}{1-r^2-2r\cos(\theta)})\\ &=& \exp(-\dfrac{1-r^2}{(1-r)^2+4r\sin^2(\frac{\theta}{2})})<1 \end{eqnarray*} So $$\forall z\in \mathbb{D}:0 <|f(z)|<1$$ Now How can I show $f(\mathbb{D})=\mathbb{D}\underline{\backslash \{0\}}$ ?
And What about (b)?
Let $g: \mathbb{D} \rightarrow \mathbb{C}$ be the function $g(z)=\frac{z+1}{z-1}$. It can be shown that $g(\mathbb{D}) = \{x+iy \in \mathbb{C} : x<0\}$. In your above work you already managed to show that $g(\mathbb{D}) \subseteq \{x+iy \in \mathbb{C} : x<0\}$. To show that other direction, I would use the fact that $g$ is a Mobius transformation.
Thus given $re^{i\theta} \in \mathbb{D} \setminus \{0\}$. We have that $\ln(r)+i\theta \in \{x+iy \in\mathbb{C}:x<0\}$ (since $0<r<1$). Hence there exists $w \in \mathbb{D}$ such that $g(w)=\ln(r)+i\theta$. We now have
$f(w)=e^{(g(w))}=e^{\ln(r)+i\theta}=e^{\ln(r)}e^{i\theta}=re^{i\theta}$
Thus $\mathbb{D} \setminus \{0\} \subseteq f(\mathbb{D})$. This completes the second half of part (a).
For part (b), you can combine what is above with the fact that $e^{w}=e^z$ if, and only if, $w-z=2\pi ki$ for some $k \in \mathbb{Z}$.
I hope this helps!