Is a map mapping infinity to infinity conformal at infinity?

Question

The map is the following: $$f(z)= \begin{cases} z*e^z & \text{$z \neq \infty$}\\ \infty & \text{$z= \infty$} \end{cases}$$

This function maps $\infty$ to $\infty$, so how to check if it's conformal at infinity? Any help are appreciated.

Answer 1

There are maps that are conformal at $\infty$. Yours is not. In order to check for conformality introduce a new complex coordinate $\zeta$ via $z={1\over \zeta}$ in the neighborhood of $z=\infty$ as well as in the neighborhood of the assumed value $w=\infty$. In your example this would mean looking at the function $$g(\zeta):={1\over f(1/\zeta)}=\zeta\cdot e^{-1/\zeta}\qquad(\zeta\ne0)$$ and $g(0):=0$ in the neighborhood of $\zeta=0$.

On the other hand, the function $h(z):=z$ of course is conformal at $z=\infty$, as are Moebius transformations $$T(z):={az+b\over cz +d},\qquad ad-bc\ne0\ .$$