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I wish to show that if $z$ is real, then $$\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$$

I have shown this result, although my inequality is the wrong way around.

I considered \begin{align} |z^2+1|&\leq |z^2|+|1| \ \ \ \ \ \ \ \text{(triangle inequality)} \\ &=|z|^2+1 \\ \\ \Rightarrow |z^2+1|&\leq |z|^2+1 \\ \frac{1}{|z^2+1|}&\geq\frac{1}{|z|^2+1} \\ \frac{|e^{iz}|}{|z^2+1|}&\geq\frac{|e^{iz}|}{|z|^2+1} \\ \left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{e^{-\text{Im}(z)}}{|z|^2+1} \\ \left|\frac{e^{iz}}{z^2+1}\right|&\geq\frac{1}{|z|^2+1} \ \ \ \ \ \ \text{(if $z$ is real $\Rightarrow$ Im$(z)=0$)} \\ \end{align}

Where did I go wrong?

Also, I wonder, would this inequality still hold if $z$ was not real?

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    $\begingroup$ It seems to me that there is an equality. Your consideration is not wrong. For complex $z$ the LHS is unbounded while the RHS is bounded, hence the inequality does not hold. $\endgroup$ – Shashi Aug 3 '18 at 9:43
  • $\begingroup$ I am not sure about the last intermediate step, but the circle $|e^{iz}| $ always is $1$. $\endgroup$ – IAmNoOne Aug 3 '18 at 9:54
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We have

$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{\left|e^{iz}\right|}{\left|z^2+1\right|}= \frac{1}{\left|z^2+1\right|}$$

and

$$0\le\left|{z^2+1}\right|= |z|^2+1$$

therefore the result follows.

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    $\begingroup$ Why doesn't $|e^{iz}|=1$? $\endgroup$ – user557493 Aug 3 '18 at 9:52
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    $\begingroup$ @Bell Opsss...yes of course $|e^{iz}|=1$ $\endgroup$ – user Aug 3 '18 at 9:59
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    $\begingroup$ @Bell Yes of course bot inequalities are true and indeed $$(a\ge b) \land (a\le b) \iff a=b$$ $\endgroup$ – user Aug 3 '18 at 13:32
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    $\begingroup$ @Bell Yes exactly! $\endgroup$ – user Aug 3 '18 at 13:40
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    $\begingroup$ @Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye $\endgroup$ – user Aug 3 '18 at 13:47
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For real $z$ we have $|z|^2=z^2$ and $|e^{iz}|$=1, hence

$$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{1}{|z|^2+1}.$$

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If $z$ is real: $|e^{iz}|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that $$\left|\frac{e^{iz}}{z^2+1}\right|=\frac{1}{|z|^2+1}.$$

If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^{iz}| =e^{-y}$ is huge, and so $$\left|\frac{e^{iz}}{z^2+1}\right|\gg\frac{1}{|z|^2+1}.$$

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  • $\begingroup$ So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality? $\endgroup$ – user557493 Aug 3 '18 at 9:53
  • $\begingroup$ The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too. $\endgroup$ – Lord Shark the Unknown Aug 3 '18 at 9:55
  • $\begingroup$ I agree that the $\left|\frac{e^{iz}}{z^2+1}\right|=\frac{1}{|z|^2+1}$, but I don't understand why $\left|\frac{e^{iz}}{z^2+1}\right|\leq\frac{1}{|z|^2+1}$ for real $z$. $\endgroup$ – user557493 Aug 3 '18 at 9:58
  • $\begingroup$ @Bell If $a=b$ then $a\le b$. $\endgroup$ – Lord Shark the Unknown Aug 3 '18 at 10:00
  • $\begingroup$ I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it. $\endgroup$ – user557493 Aug 3 '18 at 10:02
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$\geq $ and $\leq$ don't contradict each other! When $z$ is real $|z^{2}+1|=|z|^{2}+1$. This proves that the stated inequality is actually an equlity for real $z$. To see that the inequality may not hold for complex $z$ take $z=e^{-in}$ where $n$ is a large positive integer.

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