= Solution
Choose a disk on which $|f(z)|\leq M=|f(a)|$. If $M=0$ the assertion is immediate. Otherwise multiply $f$ by a fixed unit scalar so that its value at $a$ is the positive real number $M$. The mean-value identity shows the average of $\operatorname{Re}f$ on every sufficiently small circle is $M$. Since $\operatorname{Re}f\leq|f|\leq M$ there, continuity forces equality at every point on each circle. Equality of real part to $M$ together with modulus at most $M$ forces the imaginary part to vanish. Varying the radius covers a neighborhood of $a$. Thus \b[both $|f|$ and $f$ are constant near $a$], proving the local maximum-modulus assertion.
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