Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.