Suppose $M(z)={az+b\over cz+d}$ is a Möbius map. Then $M'(z)={ad-bc\over (cz+d)^2}$, which is $\neq 0$ for $z\neq -{d\over c}$ or $\infty$. So we can say that $M$ is conformal at these points, so far so good. But what about $z= -{d\over c}$ or $\infty$? I am guessing that it is conformal at $z= -{d\over c}$ because $M'$ is not zero and is not conformal at $\infty$? However in Proposition 2.3 in these notes it is said that Möbius maps are conformal on the entire $\mathbb C \cup \{\infty\}$. I don't understand the rationale behind that. Thank you!
Could someone please explain these notes on Möbius maps to me?
2 Answers
The point $z=\infty$ of the extended $z$-plane has a local z'-coordinate system associated with it where the point $z=\infty$ corresponds to z'=0 and the coordinate variables $z$ and z' are otherwise related via z={1\over z'} resp. z'={1\over z}.
Let's look at the behavior of $M$ near $z=\infty$. To this end we express $M$ in terms of the other coordinate z', resulting in the expression \tilde M(z'):=M\bigl({1\over z'}\bigr)={bz' + a\over d z' +c}\ , which obviously behaves in the expected way near z'=0 (assuming $c\ne 0$).
To account for the other exceptional point $z_0:=-{d\over c}$ (assuming $c\ne 0$) we have to use the proper coordinate variable w'={1\over w} near $M(z_0)=\infty$. This means we should express the image point $M(z)$ for $z$ near $z_0$ by means of its w'-coordinate: w'={1\over w}={1\over M(z)}={cz + d \over a z+b}=:\hat M(z)\ . For $z$ near $z_0$ the resulting expression $\hat M(z)$ behaves in the expected way, as $ad-bc\ne0$ by assumption.
$M$ is conformal at $\infty$ because M'(1/z)=\frac{bc-ad}{(dz+c)^2}, which is not $0$ at $z=0$. This is the way of thinking M'(\infty).