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

Question

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?

Answer 1

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.

Answer 2

$\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.

Answer 3

For real $z$ we have $|z|^2=z^2$ and $|e^{iz}|$=1, hence

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

Answer 4

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