# Mobius Transformation viewed as a mapping on $\bar {\mathbb C}$

The question I have on hand is as follows:

Suppose that a Mobius Transformation z $$\to \frac{az + b}{cz + d}$$ (viewed as a mapping on $$\bar {\mathbb C}$$) maps $$\infty \to \infty$$. What information does this yield about any of the coefficients a,b,c and d?

I have come up with the following answer:

With the understanding that T($$\infty$$) = $$\frac{a}{c}$$ and T($$\frac{-d}{c}$$) = $$\infty$$, I had the equation:

$$\frac{a}{c}$$ = $$\frac{-d}{c}$$

ac = -dc

To comply with the condition that ad - bc $$\neq$$ 0,

c = 0, a $$\neq$$ 0, d $$\neq$$ 0, b = K, for K $$\in \mathbb R$$

Any opinions on whether I came up with somewhat a proper deduction? Thanks

• Can you quote the exact question? – Martin R Apr 13 at 6:47
• @MartinR hi I've added the question in – Wei Xiong Yeo Apr 13 at 6:51
• $T(\infty ) = a/c$ which is well-defined because $(a,c) \ne (0,0)$. Thus $T(\infty ) =\infty \implies c=0$ and $T(z) = \alpha z+\beta$ – reuns Apr 13 at 6:58

The formula $$T(z)={az+b\over cz+d}$$is not valid for $$z=\infty$$ or $$T(z)=\infty$$. For image and preimage of the point $$\infty$$ exception handling applies. Your arguments therefore do not catch the fish.

If $$c\ne0$$ then the denominator $$cz+d$$ vanishes at $$z_0=-{d\over c}\in{\mathbb C}$$, hence $$T(z_0)=\infty$$ by the "exception handling rules". Since we don't want that we have $$c=0$$ as a necessary condition for $$T(\infty)=\infty$$. If $$c=0$$ then the condition $$ad-bc\ne0$$ enforces $$ad\ne0$$. We therefore can write $$T(z)={a\over d}z+{b\over d}=a'z+b',\quad a'\ne0\ .$$ Conversely, any map $$T(z)=a'z+b'$$ with $$a'\ne0$$ is a Moebius transformation satisfying $$T(\infty)=\infty$$.

A Mobius transformation preserves the set of generalised circles, particularly the set containing circles and sets of the form $$L \cup \{ \infty \}$$ where $$L$$ is a line in the complex plane (note that none of the circles contain $$\infty$$). As such, the set of lines must be preserved by a Mobius transformation that maps $$\infty$$ to $$\infty$$, which means that the Mobius transformation must be an affine map. In particular, it must take the form $$z \mapsto az + b$$, with $$a \neq 0$$.