I would like to ask a question about an exterior conformal mapping. Could you please help me?
Let $L$ be a Jordan curve in $\mathbb{C}.$ Let $G$ be the exterior of $L.$ Then, there exists a conformal mapping $\Phi:G\rightarrow \overline{\mathbb{C}}\setminus \{z:|z|\leq 1\}$ such that $\Phi(\infty)=\infty$ and $\Phi'(\infty)>0.$ My question is
"Is the $n$-th derivative $\left(\frac{1}{\Phi}\right)^{(n)}(z)\not=0$ for all $z\in G$?"
Thank you so much.
Mike
The answer is no. Begin with $$f(z)=\frac{z}{1+z^{-1}}$$ which is univalent in some neighborhood of infinity, denoted $N$. For sufficiently large $R$ we have $\{z:|z|\ge R\}\subset f(N)$. Let $L$ be the preimage of the circle $\{z:|z|=R\}$. Let $G$ be the exterior of $L$. Then the map $\Phi(z)=R^{-1}f(z)$ satisfies all conditions of your problem.
The constant factor $R^{-1}$ does not affect the vanishing of derivatives. Dropping it, we work with $1/f(z)=z^{-1}+z^{-2}$ and find that $$(1/f)^{(n)}(z) = (-1)^n n! z^{-n-2}\left(z+n+1\right)$$ The derivative vanishes at $z=-n-1$, which lies in $G$ when $n$ is large enough.