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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!

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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.

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$M$ 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 $M'(\infty)$.

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