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If a holomorphic on the whole complex plane funtion $f$ has a bounded real part then it is constant.

Could I use here Cauchy-Riemann equations here? They seem to lead me nowhere.

maq
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    "Holomorphic" by itself is meaningless. Holomorphic on what set? The function $f(z) = z$ is holomorphic on ${z \in \Bbb{C} : -1 < \Re(z) < 1}$, has bounded real part in that set, but is clearly not constant. – Eric Towers Nov 20 '16 at 13:42
  • Thanks. I meant holomorphic on the whole complex plane. – maq Nov 20 '16 at 13:45
  • The real part of a holomorphic function is harmonic function. Therefore, if a harmonic function is bounded then... – A.Γ. Nov 20 '16 at 13:46

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\begin{align*} f \in H(\Bbb{C}) &\wedge \Re(f) < M \in \Bbb{R} \\ &\implies \mathrm{e}^f \text{entire} \wedge|\mathrm{e}^f| < \mathrm{e}^M \\ &\underset{\text{Louisville }}\implies \exists c \in \Bbb{C} \smallsetminus \{0\}, \mathrm{e}^f = c \\ &\implies f = \ln c \text{.} \end{align*}

A very good way to get at bounded real or imaginary parts is to consider $|\mathrm{e}^f|$ and $|\mathrm{e}^{\mathrm{i}f}|$.

Eric Towers
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