Is $\left|\frac{\tanh^{2}\left(z\right)}{z^{2}}\right|\le\frac{1}{\left|z\right|^{2}}$?

Question

For context, I am trying to solve a real integral, and part of the process seems like I have to evaluate a contour integral over $\Gamma$, which is a semicircle contour with a large radius $R$ traversed in the positive direction. Keep in mind I made up all of this, so I don't even know if I am going in the right direction. But anyways, my gut says the following statement is true:

"Let $f(z) := \frac{\left(e^{2z}-1\right)^{2}}{z^{2}\left(e^{2z}+1\right)^{2}}$ and let $\Gamma$ be a semicircle contour with radius $R$ above the real axis. Show that

$$\left|\int_{\Gamma}\frac{\left(e^{2z}-1\right)^{2}}{z^{2}\left(e^{2z}+1\right)^{2}}dz\right| \leq \frac{\pi}{R}.$$

To try to prove this, I can use the ML-Inequality and find $M$ so that $|f(z)| \leq M$ on the semicircle contour. Then

$$|f(z)| = \left|\frac{\left(e^{2z}-1\right)^{2}}{z^{2}\left(e^{2z}+1\right)^{2}}\right| = \frac{|\operatorname{tanh^2{(z)}}|}{\left|z^2\right|} = \frac{|\operatorname{tanh{(z)}}|^2}{\left|z\right|^2} \leq \frac{1}{|z|^2} = M.$$

We know that $L$ is the length of the contour and that if $z$ is on $\Gamma$, then $|z| = R$. So,

$ML = \frac{1}{|z|^2}\left(\frac{2\pi R}{2}\right) = \frac{\pi R}{R^2} = \frac{\pi}{R}.$

Question: Is it true that $\left|\tanh^{2}\left(z\right)\right|\ \le\ 1$ If so, how can I prove it? I did a little digging on this site and found this: Integral $\int_0^\infty\frac{\tanh^2(x)}{x^2}dx$. The user robjohn says that "$|\text{integrand}| \leq \frac{1}{|z|^2}$" on the sides of the contour integral he was solving, but it's not clear to me as to why.

Attempt: I tried

$$\left|\tanh^{2}\left(z\right)\right|\ =\ \left|\tanh\left(z\right)\right|^{2}\ =\ (\Re(\tanh\left(z\right)))^2 + (\Im(\tanh\left(z\right)))^2,$$

and then I replaced $z$ with $Re^{i\theta}$, let $\theta \in \left[0,\pi\right]$ since I'm dealing with a semicircle above the real axis, and tried taking the real and imaginary parts, but I was just getting a mess. I figured I must be missing something trivial and that I am overthinking things.

I know Desmos only deals with real numbers, but visually, I can see on their graphs that $\left|\frac{\tanh^{2}\left(x\right)}{x^{2}}\right|\le\frac{1}{\left|x\right|^{2}}$, but I am not sure if the same can apply for moduli of complex numbers.

Any help and hints are appreciated.

Answer 1

It's best to think of it geometrically. Let $e^{2z} = \zeta$. Where is $|\zeta - 1|\leq |\zeta + 1 |$? Precisely the points where moving forward one instead of backward one moves you further from the origin. So the points $A = \{\zeta \mid \Re(\zeta) \geq 0.5\}$. Where is $e^{2z} \in A$? $$e^{2z} = e^{2x + 2iy} = e^{2x}e^{2iy} = re^{i\theta}$$

So think about for which $\theta$ this would be true. ${2y \in (-\pi, \pi)}$.

Since you described $R$ as a large radius, $\Im(z)$ will sometimes be outside of that range.

Answer 2

Here’s a proof that the inequality is true for $z \in \mathbb{R}$. Note that the inequality is not true $\forall z \in \mathbb{C}$ such as the point $z=\frac{\pi}{2} i$ and @robjohn was using an imaginary value along which that inequality did hold. In fact it is only true when $\cos (2\Im (z))>0$.

Write $$\left|\frac{\tanh^2 z}{z^2}\right|=\frac{\left|\left(e^{2z}-1\right)^2\right|}{|z^2| \left|\left(e^{2z}+1\right)^2\right|}=\frac{|e^{2z}-1|^2}{|z|^2 |e^{2z}+1|^2}$$ Now apply the reverse triangle inequality on the denominator, namely for any $a,b \in \mathbb{C}$ $$\frac{1}{|a-b|}\leq \frac{1}{||a|-|b||}$$ and since both sides are positive then $$\frac{1}{|a-b|^2} \leq \frac{1}{||a|-|b||^2}$$ Set $a=e^{2z}$ and $b=-1$ so we have $$\frac{|e^{2z}-1|^2}{|z|^2 |e^{2z}+1|^2}\leq \frac{|e^{2z}-1|^2}{|z|^2 ||e^{2z}|-1|^2}$$

Now if $z>0$, $|e^{2z}|=e^{2\Re (z)}=e^{2z}$ allowing for the denominator to cancel with the numerator and giving the desired inequality $$\left|\frac{\tanh^2 z}{z^2}\right| \leq \frac{1}{|z|^2}$$ when $z>0$ which implies it is true for $z<0$ since both sides of the inequality are even. The case $z \to 0$ can be seen trivially. Since both sides of the inequality are strictly positive for $z \in \mathbb{R}$ we have finally $$\frac{\tanh^2 x}{x^2} \leq \frac{1}{x^2}$$ $\forall x \in \mathbb{R}$. $\square$