Suppose that $$f(z)=\frac{e^{-z}}{z^2+1}.$$ Is it true that $$|f(z)|\leq\frac{2}{R^2+1}$$ for all $z$ such that $\Re(z)\geq 0$ and $|z|>R$?
My attempt:
consider the triangle inequality \begin{align} |z^2+1|&\leq |z|^2+1 \\ \frac{1}{|z^2+1|}&\geq \frac{1}{|z|^2+1} \\ &\leq\frac{1}{R^2+1} \\ \frac{|e^{-z}|}{|z^2+1|}&\leq \frac{|e^{-z}|}{|z|^2+1} \\ &=\frac{1}{e^{\Re(z)}}\frac{1}{R^2+1} \\ &\leq\frac{1}{R^2+1} \end{align}
While my working suggests the statement is incorrect, I wonder if the logic from my second to third line is correct? Any advice would be greatly appreciated.