Exponential type function bounded for the reals I am trying to solve the following problem:

Without using Phragmén-Lindelöf, show that if $f$ if of exponential type and $|f(x)| \leq M$ for all $x$ in the real axis then $|f(z)|\leq Ke^{c|\Im(z)|}$ for some constants $K$ and $c$.

Firstly, I was trying to prove the statement in the upper half plane, so I considered the function $g(z) = f(z) e^{iz}$ in that region. If we show that $g$ is bounded by a constant $K$, then $|f(z)|\leq K e^{|\Im(z)|}$. However, I don't know how to prove that $g$ is bounded, so I guess that I have to consider some variation of this $g$. Can you give me any hint?

A comment that got too long. Just mimic the proof of Phragmen Lindelof because ultimately something like that is needed, so consider a rotation of $g$ by $45$ degrees clockwise, call it $h(z)=g(\omega z), \omega=e^{-i\pi/4}$ hence $h$ is bounded on the sides of the $x=y, x=-y, x >0$ angle (hence for $|\arg z| \le \pi/4$) by some $K$ and $|h(z)| \le e^{|z|}$ inside the angle, then pick some number between $1$ and $2$ eg $3/2$ and take $h_{\epsilon}(z)=h(z)e^{-\epsilon z^{3/2}}$ which is analytic inside the angle with the principal value of the logarithm, and continuous on the edges.
Now $|\arg z| \le \pi/4$ implies $|\arg z^{3/2}| \le 3\pi/8$  so $\cos \arg z^{3/2} \ge \delta>0$ or $\Re z^{3/2} \ge \delta |z|^{3/2}$ inside the angle. This immediately implies that we have $h_{\epsilon}(z) \le Ce^{R-\epsilon R^{3/2}\delta}$ inside the angle, so since for large $R$ we have $e^{R-\epsilon R^{3/2}\delta} \to 0$, we get that on the boundary of the angle cut by the circle of radius $R$ large we have $|h_{\epsilon}(z)| \le K$ since we have that on the straight edges, while the arc bound goes to zero, hence we have that inside by maximum modulus so $|h_{\epsilon}(z)| \le K$; letting $\epsilon \to 0$ gives $|h(z)| \le K$ inside the angle, hence $|g(z)| \le K$ inside the first quadrant; repeat that for the second quadrant with an appropriate rotation to move second quadrant to the above angle and conclude $|g(z)| \le K$ in the upper half plane. Then using $f(z)e^{-iz}$ one deals similarly with lower half plane.
