= Solution
The <Cauchy integral formula> says that if $f$ is <holomorphic function>[holomorphic] on a neighbourhood of the closed disc $\overline{D(z_0;R)}$, then for $z\in D(z_0;R)$,
$$
f(z)=\frac1{2\pi i}\int_{|\zeta-z_0|=R}\frac{f(\zeta)}{\zeta-z}\,d\zeta.
$$
In particular, at the centre,
$$
f(z_0)=\frac1{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta
$$
for every $0<r<R$.
If $f(z_0)=0$, the assumed bound gives $|f(z)|\leq0$, so $f=0$. Otherwise define $g=f/f(z_0)$. Then $g(z_0)=1$ and $|g|\leq1$. Applying the centre formula and taking <real part>[real parts] gives
$$
1=\frac1{2\pi}\int_0^{2\pi}\operatorname{Re}g(z_0+re^{i\theta})\,d\theta,
\qquad \operatorname{Re}g\leq|g|\leq1.
$$
The continuous nonnegative function $1-\operatorname{Re}g$ has integral zero, so it vanishes around the circle. Equality $\operatorname{Re}g=1$ together with $|g|\leq1$ forces $g=1$. This holds for every $r<R$, hence $f(z)=f(z_0)$ throughout the disc. This is the equality case underlying the <maximum modulus principle>.
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