$f: U \to \Complex $ is holomorphic and therefore continuous, so $|f|$ will also be continuous on $U$ . $K\subset U$ is compact and since $|f|$ is continuous on $K$ it must attain a maximum and a minimum value there.

Suppose the maximum of $|f|$ is attained at $z_0$ in the interior of $K$ .

By definition there will exist $r>0$ such that the set $S_r = \left\{z\in \Complex: |z-z_0|^2\le r^2 \right\} \subset K $ .

Consider $C_r$ the boundary of the previous set parameterized counter-clockwise. Since $f$ is holomorphic by hypothesis, Cauchy integral formula says that

\begin{equation} \label{cauchy} f(z_0)= \frac{1}{2\pi i} \oint_C \frac{f(z)}{z-z_0} dz \end{equation} A canonical parameterization of $C_r$ is $z=z_0 + r e^{i\frac{\theta}{r}}$ , for $\theta \in [0,2 \pi r]$ .

\begin{equation} \label{int1} f(z_0)= \frac{1}{2\pi r} \int_0^{2\pi r} f(z_0 + r e^{i \frac{\theta}{r}}) d \theta \end{equation} Taking modulus on both sides and using the estimating theorem of contour integral

$$|f(z_0)| \le \operatorname{max}_{z \in C_r}|f(z)|$$

Since $|f(z_0)|$ is a maximum, the last inequality must be verified by having the equality in the $\le$ verified.

The proof of the estimating theorem of contour integral implies that equality is only verified when

$$ \frac{f(z_o + r e^{i \frac{\theta}{r}})}{r e^{i \frac{\theta}{r}}} = \lambda \overline{i e^{i \frac{\theta}{r}}} $$

where $\lambda\in \mathbb{C}$ is a constant. Therefore, $f(z_o + r e^{i \frac{\theta}{r}})$ is constant and to verify equation its value must be $f(z_0)$ .

So $f$ is holomorphic and constant on a circumference. It's a well known result that if 2 holomorphic functions are equal on a curve, then they are equal on their entire domain, so $f$ is constant.

One way to see this in this particular circumstance is using equation to calculate the value of $f$ on a point $\xi \in $ interior $S_r$ different than $z_0$ . Bearing in mind that $f(z)=f(z_0)$ is constant in $C_r$ the formula reads $f(\xi)=\frac{f(z_0)}{2 \pi i}\oint_{C_r} \frac{1}{z-\xi}dz = f(z_0)$ . So $f$ is really constant in the interior of $S_r$ and the only holomorphic function defined in $K$ that is constant in the interior of $S_r$ is the constant function on all $K$ .

Thus if the maximum of $|f|$ is attained in the interior of $K$ , then $f$ is constant. If $f$ isn't constant, the maximum must be attained somewhere in $K$ , but not in its interior. Since $K$ is compact, by definition it must be attained at $\partial K$ .