An obscure explanation in Conway's Complex analysis

Question

I don't understand a paragraph in Conway's complex analysis at the beginning of Chapter VI page 128 (Maximum Modulus Theorem).

He says: "Note that in Theorem 1.2 we did not assume that $G$ is connected as in Theorem 1.1. Do you understand how Theorem 1.1 puts the finishing touches on the proof of Theorem 1.2? Or could the assumption of connectedness in Theorem 1.1 be dropped?"

For reference I included both theorems here: 1.1 Maximum Modulus Theorem, First Version. If $f$ is analytic in a region $G$ and $a$ is a point in $G$ with $|f(a)| \geq |f(z)|$ for all $z$ in $G$, then $f$ must be constant. (it is proved with the open mapping theorem).

1.2 Maximum Modulus Theorem, Second Version. Let $G$ be a bounded open set in $\mathbb{C}$ and suppose $f$ is continuous on the closure of $G$ and analytic in $G$. Then the maximum is attained on $\partial G$.

I was thinking that I understood the maximum modulus but since I can understand Conway's little paragraph I must be missing something important. If someone has the book, thanks for any help.

Answer 1

I think this answer may help to this question (even though the question is too old):

In class I was given the proof of 1.2 in the following way (Answering to Conway's question "do you understand(...)":

Let $M=\max\{|f(z)|: z\in \overline{G}\}$ and $N=\max\{|f(z)|: z\in \partial G\}$. Suppose that $M>N$ so there exists $a\in G$ such that $M=|f(a)|$. Let $G_a$ be the connected component of $G$ which contains $a$. By Maximum Modulus (the first version) $f(z)=f(a)$ $\forall z\in G_a$ and by continuity $f(z)=f(a)$ $\forall z\in \partial G_a$. As $\mathbb{C}$ is locally connected, it is true that $\partial G_a\subset \partial G$ so there exists $z\in\partial G$ such that $f(z)=f(a)$, which implies $M=N$, absurd.