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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 $\forall z\in\text{Range}(\gamma_R)$

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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 $e^{-R\sin(t)}\le 1$.

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  • $\begingroup$ Oh I see. I was thinking that because $R$ is a real constant, then $\Im(z)=0$. Is this not correct? $\endgroup$ – user557493 Sep 24 '18 at 1:33
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    $\begingroup$ Since $R>1$ and $0\le t\le\pi$, we have $-R\sin(t)\le 0$, and thus $e^{-R\sin(t)}\le 1$. $\endgroup$ – eloiprime Sep 24 '18 at 1:35
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    $\begingroup$ If $R=2$ and $t=\pi/2$, then $\Im(z)=-R\sin(t)=-2\ne 0$. $\endgroup$ – eloiprime Sep 24 '18 at 1:40
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    $\begingroup$ Well, the imaginary axis from $0$ to $R$ doesn't look like much of a semicircular arc$\ldots$ $\endgroup$ – eloiprime Sep 24 '18 at 3:52
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    $\begingroup$ 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, $$\int_{\gamma_R} \frac{e^{iz}}{z^2+1} \ dz.$$ $\endgroup$ – user557493 Sep 24 '18 at 3:57

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