Let $f(z)=\frac{(az+b)}{(cz+d)}$ be a Möbius transformation where $c \neq0$. Through the process of actually computing $f^{-1}$ show that $f$ is a univalent function whose domain-set is $\mathbb{C} \setminus \{-d/c\}$ and the range is $\mathbb{C} \setminus \{a/c\}$
I would like to confirm that the following shows that it is univalent.
First computing the inverse:
$$w=\frac{(az+b)}{(cz+d)}\Rightarrow \frac{(dw-b)}{(a-cw)}=z$$
Since we found an inverse, it is bijective and hence one-to-one.
Finally, it seems odd, but I don't see why the range is $\mathbb{C}\setminus \{a/c\}$.
To give a slightly higher level perspective on things....
Let $\bar{\mathbf{C}} = \mathbf{C} \cup \{ \infty \}$ be the Riemann sphere a.k.a. the projective complex numbers.
It turns out that a Möbius transformation is an invertible function on $\bar{\mathbf{C}}$, and is the continuous extension of the corresponding partial function on $\mathbf{C}$.
Your missing point in the range of $f$ is precisely the point that would be the image of $\infty$ if you extended $f$ to all of $\bar{\mathbf{C}}$. Similarly, the missing point in the domain of $f$ is the point whose image is $\infty$.