Is the map conformal at infinity?

Question

Consider the map defined as following: $$f(z)=\begin{cases} \frac{z-1}{z} & \text{if $z \neq 0, \infty$ } \\ \infty & \text{if $z=0$ }\\ 1 & \text{if $z=\infty$} \end{cases}$$

To see if this map is conformal at $\infty$, we should check if $f(1/z)$ is conformal at 0. But then we have: $$f(1/z)=\begin{cases} 1-z & \text{if $z \neq 0, \infty$ } \\ \infty & \text{if $z=0$ }\\ 1 & \text{if $z=\infty$} \end{cases}$$ How does this tell me if $f(1/z)$ is conformal at 0? Any help are much appreciated.

Answer 1

You forgot to invert the variable at $z=0$ and at $z=\infty$.

$f(1/z)=\infty$ for $z=\infty$

and

$f(1/z)=1$ for $z=0$.

Then $g(z)=f(1/z)=1-z$ near $z=0$, which is analytic, with derivative $g'(0)=-1\neq0$, and therefore conformal.