I've been stuck on a specific part of a problem for the past few days. For $k \geq 1$ we consider the meromorphic function $f_k : z \rightarrow \frac{1}{z^{2k}(1-e^z)}$. Its poles are the $2 \pi m$ with $m \in \mathbb{Z}$ and part of the previous questions were about findings the residues at these poles. The residue at $0$ being a number $b$ and the ones at $2 i \pi m$ ($m \neq 0$) being $\frac{(-1)^k}{(2 \pi m)^{2k}}$.
The part that got me stuck is the following : we let $r_N = (2N+\frac{1}{2})\pi $ and consider the line integral $$\int_{\Gamma _N} f_k(z) dz$$
where $\Gamma_N$ denotes the circle of center $0$ and radius $r_N$. The residue theorem gives the value of that integral, and the end goal of the exercise is to show $$b = 2 \sum_{m=1}^{\infty} \frac{(-1)^{k+1}}{(2 \pi m)^{2k}}$$
For that effect we want to show that the aforementioned integral goes to $0$ when $N \rightarrow \infty$, and I have no idea how to do it. The first thing I've tried was to find an upper bound for $\lvert f_k(z)\rvert$ that goes to $0$ but I have to find a suitable lower bound for $\lvert 1-e^z \rvert$ on $\Gamma_N$. Plotting that in geogebra gives me good hope that such a lower bound exists (in fact $\frac{1}{2}$ seems to be a lower bound for every $N$) but... I can't manage to prove it ! Direct computation of $\lvert 1-e^z \rvert ^2$ gives me $$e^{2 r_N \cos (t)} - 2 e^{r_N \cos (t)}\cos (r_N \sin (t)) + 1$$ with $0 \geq t \geq 2 \pi$ and the derivative of that gives me $$2r_Ne^{r_N \cos (t)}(\sin ((r_N + 1)\sin (t))-e^{r_N \cos (t)}\sin (t))$$ and it just seems to complicated to be the way to go (I can't even figure out the zeroes of the derivative...)
I've also tried the more simple inequalites $\lvert z \rvert \geq \lvert \Re (z) \rvert$ and $\lvert z \rvert \geq \lvert \Im (z) \rvert$ but to no avail. In any case I still end up with sinuses inside cosinuses and complicated formulas. There's also the fact that $r_N$ is $(2N+\frac{1}{2})\pi$ and not simply $(2N + 1)\pi$ that makes me think that I'm missing something, but at that point I think I'm stuck too deep in a way that will end up not working...
Thanks in advance to those who will read all of that!
Let $D(z)=|e^z-1|$. Then, we have
$$\begin{align} D(z)&=|e^{z}-1|\\\\ &=|e^x\cos(y)-1+ie^x\sin(y)|\\\\ &=\sqrt{e^{2x}+1-2e^x\cos(y)}\\\\ &=\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)}\\\\ \end{align}$$
Now on $|z|=(2N+1)\pi$, split the analysis into cases where $(2N+1/2)<|y|<(2N+1)\pi$ and $|y|<(2N+1/2)\pi$.
CASE $1$: $\displaystyle (2N+1/2)<|y|<(2N+1)\pi$
Note first that $\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)}\ge 2|\sin(y/2)|$ for all $x$ and $y$. So, when $(2N+1/2)<|y|<(2N+1)\pi$, $$\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)}\ge \sqrt2$$
CASE $2$: $\displaystyle (2N+1/2)>|y|$
Note that we also have $\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)}\ge |e^x-1|$ for all $x$ and $y$. So, for $|y|<(2N+1/2)\pi$, $|x|>\pi\sqrt{2N+3/4}$ and $|e^x-1|\ge 1-e^{-\pi\sqrt{2N+3/4}}$. For $N\ge 1$, $|e^x-1|>0.994$.
Hence, we conclude that $D(z)>0.994$ for all $|z|=(2N+1)\pi$, $N\ge 1$.