# Open Mapping Theorem

Idea: maximum modulus plus Rouche.

# Open Mapping Theorem

Any holomorphic non-constant map is an open map.

## Exercises

Show that there is no bijective conformal map from $$A\coloneqq\left\{{0< {\left\lvert {z} \right\rvert} < 1}\right\}$$ to $$B\coloneqq\left\{{1<{\left\lvert {z} \right\rvert} < 2}\right\}$$.

Suppose toward a contradiction that such an injective map $$f: A\to B$$. Since $$f(A) \subseteq B$$ which is bounded, $$f$$ is bounded and thus any singularities in $$A$$ are removable. So $$f$$ extends holomorphically to $$F: {\mathbb{D}}\to \overline{B}$$, and since $$F$$ is an open map, $$F({\mathbb{D}}) \subseteq B$$. Write $$w_0 \coloneqq F(0) \in B$$, then by surjectivity there is a $$z_0\in A$$ such that $$f(z_0) = w_0$$. Since $$F = f$$ on $$A$$, we also have $$F(z_0) = w_0$$. Since $$A$$ is open, $$z_0$$ is interior, and since $$z_0\neq 0$$ we can separate $$z_0$$ and zero by disjoint open sets $$U$$ and $$V$$. Since $$F$$ is open, $$F(U), F(V)$$ are open in $$B$$ and contain $$w_0$$. Then $$F(U) \cap F(V)$$ is an open set containing $$w_0$$, so there is a disc $$D_{\varepsilon}(w_0)$$ in $$F(U)$$ and $$F(V)$$. But then $$F({\mathbb{D}}_{\varepsilon}(w_0))$$ is an open set intersecting $$U$$ and $$V$$, contradicting that they were disjoint.

Prove that if $$f$$ is holomorphic on a connected open set $$\Omega$$ and $$f^2(z) = \overline{f(z)}$$ then $$f$$ is constant.

Write $$F(z) \coloneqq f^3(z)$$ so that \begin{align*} F(z) = f^2(z) f(z) = \overline{f(z)}f(z) = {\left\lvert {f(z)} \right\rvert}^2 .\end{align*} Since $$F$$ is also holomorphic on $$\Omega$$ and this calculation shows $$F(\Omega) \subseteq \Omega$$, $$F$$ must be constant by the open mapping theorem. So $$f^3(z) = c$$ for some $$c$$, forcing $$f$$ to be constant as well.

#complex/exercise/completed