Doubt in proof of maximum modulus theorem

Question

In the proof of second version of maximum modulus theorem(Conway 2nd ed) it says that if $f$ is not constant then the result follows from Theorem $1.1$. Below that it asks that could the assumption of connectedness in Theorem $1.1$ be dropped? I think it cannot be dropped for example let $$G=B(0,1)\bigcup B(3,1)$$ and define $f$ as $2$ on $B(0,1)$ and $1$ on $B(3,1)$, then $f$ is analytic on $G$ and $f(0)\geq f(z)$ for all $z\in G$. Now i don't understand how Theorem $1.1$ is applied in the proof. Thanks!!!

Answer 1

Right, as you say connectedness is crucial in Theorem 1.1.

But 1.2 is fine if $G$ is not connected. By compactness and continuity, $|f|$ does have a maximum on the closure of $G$, at $a$ say. Suppose that $a\in G$. Then applying 1.1 to the component $U$ of $G$ containing $a$ gives $f$ constant on $U$. By continuity $f$ is constant on the closure of $U$. Let $b$ be a point in the boundary of $U$. Then $b$ is on the boundary of $G$ and $|f(z)| \le |f(a)|=|f(b)|$ for all $z\in G$.