How could one argue using the Maximum Modulus Principle that if a function $f$ is holomorphic on a connected open set $U \subset \mathbb{C}$ such that Im$f$ has a local maximum in $U$, then $f$ must be constant?

Clearly, if Im$f$ has a local minimum, Im$(1/f)$ has a local maximum. Moreover, $f$ is locally constant at a point $z_0$ if there exists an open set $U$ such that $z_0 \in U$ and $f$ is constant on $U$.

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    $\begingroup$ Note that $\operatorname{Im} (1/f) \ne 1/ \operatorname{Im} (f)$, and $1/f$ is not even holomorphic in $U$ if $f$ has zeros. $\endgroup$
    – Martin R
    Nov 23, 2017 at 15:51
  • $\begingroup$ Thanks for your remark. This means that $1/$Im$f$ has a local max, right? $\endgroup$
    – FunnyBuzer
    Nov 23, 2017 at 15:57
  • $\begingroup$ No, because it can be positive or negative. And that would not help anyway. $\endgroup$
    – Martin R
    Nov 23, 2017 at 16:03

1 Answer 1


Define $g$ on $U$ as $g(z) = e^{-if(z)}$. Then $|g(z)| = e^{\operatorname{Im} f(z)}$, so that the Maximum Modulus Principle can be applied to $g$. It follows that $g$ is constant and since $$ 0 = g'(z) = -if'(z) e^{-if(z)} \,, $$ $f$ is constant as well.

Alternatively one can apply the "maximum modulus principle for harmonic functions" to $\operatorname{Im} f(z)$.

Or, if $\operatorname{Im} f(z)$ has a maximum at $z_0$ in $V \subset U$ then $f(V)$ does not contain an open neighbourhood of $f(z_0)$, i.e. $f$ is not not an open mapping.


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