Find the supremum of $h(z)$. 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. 
 A: 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

