This is exercise 2 from Chapter 4 of Stein and Shakarchi's Complex Analysis.
Suppose that for fixed $a>0$, $f$ is holomorphic in the horizontal strip $$S_a = \{z \in \mathbb{C}: |\Im(z)|<a\}.$$
and there exists a constant $A>0$ such that $$|f(x+iy)| \le \frac{A}{1+x^2}\; \text{for all} \; x\in \mathbb{R} \; \text{and} \; |y|<a.$$ Then I would like to show that $f^{(n)}$ satisfies the above whenever $0 \le b < a$.
The idea is to use Cauchy inequalities.
Choose $b<a$ and fix some $\epsilon>0$ small enough that $b+2 \epsilon < a$. Consider the disc $D$ centered at $z_0 = x_0 + iy_0$ with radius $\epsilon$ and let $C$ be the boundary circle.
Then from Cauchy inequalities we have $|f^{(n)}(x_0 + iy_0)| \le \frac{n!}{\epsilon^n} \sup_{x+iy \in C} |f(x+iy)| $
For each $x+iy \in C$, we have $|f(x+iy)| \le \frac{A}{1+x^2}$ from the given condition on $f$ and the fact that $|y|<a$ by the chosen radius $\epsilon$.
However, I cannot figure out how to bound the supremum of these by $\frac{A'}{1+x_0^2}$, which is what I need for the proof. How could I show this?