To estimate a holomorphic function. Let $f:\mathbb{C}\to \mathbb{C}$ be a holomorphic map such that
$$|f(z)|\leq 1, \forall z\in\mathbb{R},$$
and $$|f(z)|\leq C e^{|z|}, \forall z\in\mathbb{C},$$ for some $C>0.$
Can we conclude that $$|f(z)|\leq C e^{|Im(z)|}, \forall z\in\mathbb{C},$$ for some $C>0?$
 A: Assume $C>1$ wlog. Consider $g(z)=f(z)e^{iz}$. On the real axis $|g(z)| \le 1$ and on the positive imaginary axis $|g(z)|=|f(z)|e^{-|z|} \le C$, while in the upper plane $|g(z)| \le Ce^{|z|}$ since $|e^{iz}| \le 1$ there.
Applying Phragmen Lindelof on the two quadrants in the upper plane with angles $\pi/2$ we get $|g(z)| \le C$ (see proof below) in the upper half-plane hence $|f(z)| \le Ce^{\Im z}, \Im z \ge 0$
(for example if $z$ is in the first qudrant - let $\epsilon >0, h(z)=e^{-\epsilon (ze^{-i\pi /4})^{3/2}}g(z), \rho \le |z| \le R, 0 \le \arg z \le \pi /2$ where the $3/2$ power is defined as usual using the principal value of the logarithms; then if $z=re^{it}, 0 \le t \le \pi/2, \Re (ze^{-i\pi /4})^{3/2}=r^{3/2}\cos (\frac{3(t-\pi/4)}{2}) \ge r^{3/2} \cos \frac{3\pi}{8}$ hence $|h(z)| \le |g(z)|e^{-\epsilon r^{3/2} \cos \frac{3\pi}{8}}$ so on the segment boundaries of the domain $|h(z)| \le C$ while on the arc boundaries $|h(z)| \le e^{-\epsilon R^{3/2} \cos \frac{3\pi}{8}+cR}, |h(z)| \le e^{-\epsilon \rho^{3/2} \cos \frac{3\pi}{8}+c\rho}$ and obviosuly both exponential tend to $1$ as $R \to \infty, \rho \to 0$ so $|h(z)| \le C$ on the domain for large $R$ and small $\rho$ hence on the full quadrant. Letting now $\epsilon \to 0$ we see that the same result holds for $g$ again first on some bounded domain as before and then letting $R \to \infty, \rho \to 0$; same proof with small changes clearly applies if $z$ is in the second quadrant, we just rotate by $-3\pi /4$ in the definition of $h$)
Considering now $g_1(z)=f(z)e^{-iz}$ we get same as above except that we use the negative imaginary axis/lower half-plane, so overall we get $|f(z)| \le Ce^{\Im z}, \Im z \ge 0$ and $|f(z)| \le Ce^{-\Im z}, \Im z \le 0$ which is precisley the required result so done!
