Lemma. Suppose $f$ is holomorphic on some open region and $|f|$ is constant. Then $f$ is a constant function.
Proof. Let $f(z) = u(x, y) + iv(x, y)$. $|f| = c$ means $u^2 + v^2 = c^2$. Setting the gradient of the left-hand side equal to $0$, we have
$$
uu_x + vv_x = 0 \\
uu_y + vv_y = 0
$$
which can be transformed via the Cauchy-Riemann equations to
$$
uu_x + vu_y = 0 \\
-vu_x + uu_y = 0
$$
or in matrix notation
$$
\begin{bmatrix}
u & v\\
-v & u
\end{bmatrix}
\begin{bmatrix}
u_x \\
u_y
\end{bmatrix}
= \bar{0}.
$$
The determinant of the 2×2 matrix on the left is $u^2 + v^2$. If this is zero, then $f = 0$ everywhere since its modulus is constant. On the other hand, if it’s nonzero, then we can multiply both sides by its inverse and get $u_x = u_y = 0$. Similarly (via CR equations), $v_x = v_y = 0$. This implies that $u$ and $v$ are constant, so $f$ is constant.
Theorem (Maximum Modulus Principle). Suppose $f$ is holomorphic on some open region $R$. Then if $f$ is non-constant, $|f|$ takes a maximum on the boundary of R.
Proof. Let $z_0$ attain the maximum modulus. Pick a circle $C \subset R$ with radius $r$ centered on $z_0$. For any $z$ in the interior of $C$, we can use the Cauchy integral formula to get
$$
\begin{eqnarray}
|f(z_0)| &=& \left| \frac{1}{2 \pi i} \int_C \frac{f(w)}{w – z_0} dw \right| \\
&\leq& \frac{1}{2 \pi} \int_C \left| \frac{f(w)}{w – z_0} \right| ds \\
&=& \frac{1}{2\pi} \int_0^{2\pi} \frac{|f(z_0 + re^{it})|}{r} \cdot r dt \\
&\leq& \frac{1}{2\pi} \int_0^{2\pi} |f(z_0)| dt \\
&=& |f(z_0)|.
\end{eqnarray}
$$
So all of the inequalities here are equalities, so $\int_0^{2\pi} |f(z_0)| – |f(z_0 + re^{it})| dt = 0$. But since the integrand is continuous and non-negative, this implies it’s equal to $0$ identically. Since $r$ was arbitrary, it follows that $|f|$ is constant on some open neighborhood of $z_0$. By the above lemma, this implies that $f$ is itself constant.