Show $\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{R^2-1}$ for $R>1$ and $z\in\text{Range}(\gamma_R)$

Define the semicircular arc $$\gamma_R$$ by $$\gamma_R(t)=Re^{it}$$, where $$0\leq t\leq\pi$$ and $$R>1$$ is a real constant. Let $$\gamma$$ be the join of $$\gamma_R$$ and the line segment from $$-R$$ to $$R$$.

Show that, if $$z\in\text{Range}(\gamma_R)$$, then $$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{R^2-1}.$$

My attempt:

Consider the inequality, \begin{align} |z^2+1|&\geq||z^2|-1|\\ &=|R^2-1| \ \ \ \text{(suppose z=R>1, such that z\in\text{Range}(\gamma_R))} \\ &=R^2-1 \\ \frac{1}{|z^2+1|}&\leq\frac{1}{R^2-1} \\ \frac{|e^{iz}|}{|z^2+1|}&\leq\frac{|e^{iz}|}{R^2-1} \\ \left|\frac{e^{iz}}{z^2+1}\right|&\leq\frac{e^{-\Im(z)}}{R^2-1} \\ \left|\frac{e^{iz}}{z^2+1}\right|&\leq\frac{1}{R^2-1} \ \ \ \text{(as z=R is a real constant by assumption)} \end{align} Is this correct? I am unsure if $$z=R$$ is a valid step. While this step does agree with the condition $$z\in\text{Range}(\gamma_R)$$, I'm unsure if the equality must work $$\forall z\in\text{Range}(\gamma_R)$$

Your deductions that $$|z^2+1|\ge R^2-1$$ and $$|e^{iz}|\le e^{-\Im(z)}$$ both seem fine. However, your last step is quite dodgy. To finish off your proof, write $$e^{-\Im(z)}=e^{-R\sin(t)}$$ and use the conditions on $$R$$ and $$t$$ to conclude that $$e^{-R\sin(t)}\le 1$$.
• Oh I see. I was thinking that because $R$ is a real constant, then $\Im(z)=0$. Is this not correct? – user557493 Sep 24 '18 at 1:33
• Since $R>1$ and $0\le t\le\pi$, we have $-R\sin(t)\le 0$, and thus $e^{-R\sin(t)}\le 1$. – eloiprime Sep 24 '18 at 1:35
• If $R=2$ and $t=\pi/2$, then $\Im(z)=-R\sin(t)=-2\ne 0$. – eloiprime Sep 24 '18 at 1:40
• Well, the imaginary axis from $0$ to $R$ doesn't look like much of a semicircular arc$\ldots$ – eloiprime Sep 24 '18 at 3:52
• Yep, very silly mistake. Too much math and not enough sleep. I've also posted a second part of this question. I am having trouble find the integral of the LHS of the above results. That is, $$\int_{\gamma_R} \frac{e^{iz}}{z^2+1} \ dz.$$ – user557493 Sep 24 '18 at 3:57