 Solution

The Cauchy integral formula says that if $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D(z_0;R)}$, then for $z\in D(z_0;R)$,In particular, at the centre,for every $0<r<R$.

If $f(z_0)=0$, the assumed bound gives $|f(z)|\le 0$, so $f=0$. Otherwise define $g=f/f(z_0)$. Then $g(z_0)=1$ and $|g|\le 1$. Applying the centre formula and taking real parts givesThe continuous nonnegative function $1-\mathrm{Re}g$ has integral zero, so it vanishes around the circle. Equality $\mathrm{Re}g=1$ together with $|g|\le 1$ 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.