We put
$$h(z) = \frac{|\frac{e}{e-1}(e^z-1)|\;-\frac{3}{2}e^{|z|}+1}{\frac{\pi}{\pi^2-1}\sinh(\pi|z|)\;+\frac{1}{\pi^2-1}(\cosh(\pi|z|)-1) -e^{|z|}+1},\; z\in\mathbb{C}\setminus\{0\}. $$
My question is: What is the $\displaystyle\sup_{z\in\mathbb{C}\setminus\{0\}}h(z)$ (in other words, the least $M\leq \infty$ such that $h(z)\leq M$, for all $z\neq0$)?
To Find this I firstly checked that the denominator of $h(z)$ is positive. Then using Wolfram Alpha I get it approximately equals to $0.000759$. Now I want to know that is it correct? Anyone can help me to obtain the supremum (and how to calculate)? thanks a lot.
Hint.
Making $z = r e^{i\phi}$ we have
$$ d(r)={\frac{\pi}{\pi^2-1}\sinh(\pi r)\;+\frac{1}{\pi^2-1}(\cosh(\pi r)-1) -e^{r}+1} = C_0 $$
for $r^2=x^2+y^2$
and
$$ n(r,\phi) = \left|\frac{e}{e-1}(e^{r e^{i\phi}}-1)\right|\;-\frac{3}{2}e^{r}+1 = \frac{e \sqrt{e^{2 r \cos (\phi )}-2 e^{r \cos (\phi )} \cos (r \sin (\phi ))+1}}{e-1}-\frac{3 e^r}{2}+1 $$
has a maximum for $\phi=0$ and also is an increasing function for $\phi = 0$. In fact
$$ n(r,0) = \frac{(e-3) e^r+2}{2(1-e)}\\ d(r) =\frac{\pi \sinh (\pi r)}{\pi ^2-1}+\frac{\cosh (\pi r)-1}{\pi ^2-1} -e^r+1 $$
The conclusion is that the maximum for $\frac{n(r,\phi)}{d(r)}$ should be located along the $x > 0$ axis. We know also that $\lim_{r\to\infty}\frac{n(r,\phi)}{d(r)}\approx e^{-(\pi-1)r}\to 0$. Also we have that
$$ n(r,0) \lt 0\ \ \ \ \text{for}\ \ \ 0\le r \lt \ln \left(\frac{2}{3-e}\right),\ \ \ \text{and}\ \ \ n(r,0) \gt 0\ \ \ \text{for}\ \ \ r > \ln \left(\frac{2}{3-e}\right) $$
Follows a plot showing $\frac{n(r,0)}{d(r)}$
Follows a plot for this surface, with the level curves in black