definition of linear fractional transformation - why demand $ad-bc \ne 0$?

Question

I am reading Ash & Novinger's Complex Variables. In Definition 4.4.1. (Page 18), the authors define Linear Fractional Transformations on $\hat{\mathbb C}$.

Definition (Linear Fractional Transformation). If $a,b,c,d \in \mathbb C$ such that $ad-bc \ne 0$, the linear fractional transformation $T: \hat{\mathbb C} \to \hat{\mathbb C}$ associated with $a,b,c,d$ is defined by $$T(z)=\begin{cases} \frac{az+b}{cz+d} & z\ne \infty \\ a/c & z=\infty \\ \infty & z=-d/c \\ \end{cases}$$

The authors furthermore go on to say that

Note that the condition $ad-bc \ne 0$ guarantees that $T$ is not constant.

My question is that: The authors defined $T$ for $ad-bc\ne 0$, so, I cannot possibly assume $ad-bc =0$ and proceed to show that $T$ is constant (but where?). If I quit being pedantic for a moment and suppose that $ad-bc=0$ then I need to show that that $T$ is constant (where?). First, assume that $c\ne 0$. Then we have that $ad-bc=0$ implies that $$T(z) = \dfrac{(a/c) (cz + d)}{cz + d} = \dfrac{a}{c}$$ for $z\in \mathbb C \setminus \{ -d/c \}$.

But my attempt of proof may fail if I assume $d=0$. So what are the authors exactly trying to say here?

Answer 1

The reason that we want $ad-bc\neq 0$ is to ensure that the transformation is invertible.

A good framework to view this is as an action of $M_2(\mathbb{C})$ ($2\times 2$ matrices with complex coefficients) on $\hat{\mathbb{C}}$. Indeed, we define $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}.z = \frac{az+b}{cz+d}. $$ The condition $ad-bc\neq 0$ is actually saying that the determinant is non-zero, i.e. the matrix lies in $\text{GL}(2,\mathbb{C})$, and so the action indeed becomes a group action. In particular, the action is invertible, i.e. $M^{-1}.(M.z) = (M^{-1}M).z = I.z = z$.