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

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?

• The authors say "$ad-bc\ne 0$ implies $T$ non-constant" but you are asking about what happens when $ad-bc=0$: why? Commented Mar 23, 2023 at 16:24
• Note $ad-bc=0 \iff a/b = c/d$ (you should check what happens if $b$ or $d$ is zero). Now take a common factors of $b$ from the numerator and $d$ from the denominator of $T(z)$ out. You will see that the terms with $z$ cancel out and you are left with the constant function $b/d$ (so it's not invertible). Commented Mar 23, 2023 at 16:31

## 1 Answer

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. $$M^{-1}.(M.z) = (M^{-1}M).z = I.z = z$$.