In Edwards "Riemann's zeta function", regarding the proof of the zeta functional equation p.14, it is stated without proof that $1/(e^z-1)$ is bounded (I have to assume independently of $n$) on the circle $C(0,(2n+1)\pi$). If the result is quite straightforward when considering the square $\pm n\pm i(2n+1)\pi$ instead, I struggle to prove this fact for the circle. Since the function is not holomorphic, I cannot use the maximum principle and view the circle as part of the square.
The result is clear at the following four points of the circle:
At $z=\pm i(2n+1)\pi, 1/(e^z-1)=-1/2$.
At $z=(2n+1)\pi, 1/(e^z-1)=1/(e^{(2n+1)\pi}-1)\le 1$.
At $z=-(2n+1)\pi, 1/(e^z-1)=1/(e^{-(2n+1)\pi}-1)$ of modulus $\le 2$ for $n$ large.
Thanks for any help.
A brute force answer:
Observe that $$|e^z - 1| = (e^{\operatorname{Re} z} - 1)^2 + 2e^{\operatorname{Re} z}(1 - \cos(\operatorname{Im}z)).$$
Divide the circle into two parts: $|\operatorname{Re} z| \geq \frac{\sqrt{3}}{2}\pi$ and $|\operatorname{Re} z| \leq \frac{\sqrt{3}}{2}\pi$.
If $|\operatorname{Re} z| \geq \frac{\sqrt{3}}{2}\pi$, then $$|e^z - 1| \geq (\operatorname{Re} z)^2 \geq \frac{3}{4} \pi^2.$$
If $|\operatorname{Re} z| \leq \frac{\sqrt{3}}{2}\pi$, then $$\sqrt{((2n+1)\pi)^2 - 3\pi^2/4} \leq |\operatorname{Im} z| \leq (2n+1)\pi,$$ which we can also write as $$\frac{\pi}{2} \leq \sqrt{((2n+1)\pi)^2 - 3\pi^2/4} - 2n\pi \leq |\operatorname{Im} z| - 2n\pi \leq \pi.$$ Therefore $$\cos(\operatorname{Im} z) \leq 0$$ and so $$|e^z - 1| \geq 2e^{-\sqrt{3}\pi / 2}.$$