holomorphic function maxima in Re(f) Let $f \colon U \to \mathbb{C}$ be holomorphic . Prove: if  $D(a,r ) \subset U$, then
$$\left\lvert \operatorname{Re} \bigl(f(a)\bigr) \right\rvert \leqslant \max_{0\leqslant \theta \leqslant 2\pi} \left\lvert \operatorname{Re} \bigl(f(a+re^{i\theta})\bigr)\right\rvert.$$
 A: The function $g \colon z \mapsto e^{f(z)}$ is holomorphic on $U$. Applying the maximum principle to $g$ yields
$$e^{\operatorname{Re} f(a)} = \lvert g(a)\rvert \leqslant \max_{0 \leqslant \theta \leqslant 2\pi} \: \bigl\lvert g\bigl(a + r e^{i\theta}\bigr)\bigr\rvert
= e^{\max \:\{ \operatorname{Re} f(a + re^{i\theta}) : 0 \leqslant \theta \leqslant 2\pi\}}$$
and by taking logarithms
$$\operatorname{Re} f(a) \leqslant \max_{0 \leqslant \theta \leqslant 2\pi} \operatorname{Re} f\bigl(a + re^{i\theta}\bigr)  \leqslant \max_{0 \leqslant \theta \leqslant 2\pi} \bigl\lvert\operatorname{Re} f\bigl(a + re^{i\theta}\bigr)\bigr\rvert\,. \tag{1}$$
Doing the same for $h \colon z \mapsto e^{-f(z)}$ yields
$$\operatorname{Re} f(a) \geqslant \min_{0 \leqslant \theta \leqslant 2\pi} \operatorname{Re} f\bigl(a + re^{i\theta}\bigr)  \geqslant -\max_{0 \leqslant \theta \leqslant 2\pi} \bigl\lvert\operatorname{Re} f\bigl(a + re^{i\theta}\bigr)\bigr\rvert\,. \tag{2}$$
Together, $(1)$ and $(2)$ yield the assertion
$$\lvert \operatorname{Re} f(a)\rvert \leqslant \max_{0 \leqslant \theta \leqslant 2\pi} \bigl\lvert\operatorname{Re} f\bigl(a + re^{i\theta}\bigr)\bigr\rvert\,.$$
The argument becomes much shorter if the maximum and minimum principle for real-valued harmonic functions can be used. Then the observation that the real part of a holomorphic function is harmonic immediately yields
$$\min_{0 \leqslant \theta \leqslant 2\pi} \operatorname{Re} f\bigl(a + re^{i\theta}\bigr) \leqslant \operatorname{Re} f(a) \leqslant \max_{0 \leqslant \theta \leqslant 2\pi} \operatorname{Re} f\bigl(a + re^{i\theta}\bigr)\,,$$
and the observation that
$$\max u(\theta) \leqslant \max \lvert u(\theta)\rvert \qquad\text{and}\qquad \min u(\theta) \geqslant -\max \lvert u(\theta)\rvert$$
finishes it.
