Univalent Möbius transformation 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 $\infty$.