Constant $A$: $|e^{(2m+1)\pi e^{i\theta}}-1| \geq A > 0$ for all $m\in \mathbb{N}_{>0}$, $\theta \in [0, 2\pi)$ 
Problem: Prove that there exists some real constant $A$ such that
$$|e^{(2m+1)\pi e^{i\theta}}-1| \geq A > 0$$
for any natural numbers $m \geq 1$ and any real number $0 \leq \theta < 2\pi$.

Context:
I am currently working on my bachelor's thesis and trying to show that the following contour integral tends towards zero for fixed $\Re(z) < 0$ for large natural numbers $m$:
$\int_{|w| = (2m+1)\pi} \frac{(-w)^{z-1}}{e^w - 1}dw$. I could already find the following:
\begin{equation}
 |\int_{|w| = (2m+1)\pi} \frac{(-w)^{z-1}}{e^{w} - 1} \thinspace dw| \leq\thinspace 2\pi (2m+1)\pi \max_{|w| =(2m+1)\pi}|\frac{(-w)^{z-1}}{e^{w} - 1}| 
\end{equation}
but to continue I am trying to show the following inequality for any natural numbers $m \geq 1$ and any real number $0 \leq \theta < 2\pi$:
$$|e^{(2m+1)\pi e^{i\theta}}-1| \geq A > 0$$
for some real constant $A$. I do not need to know the specific constant but just that there is one. With this I could conclude:
\begin{equation}
 |\int_{|w| = (2m+1)\pi} \frac{(-w)^{z-1}}{e^{w} - 1} \thinspace dw| \leq\thinspace 2\pi (2m+1)\pi \max_{|w| =(2m+1)\pi}|\frac{(-w)^{z-1}}{e^{w} - 1}| 
    \leq  \thinspace2\pi (2m+1)\pi \frac{((2m+1)\pi)^{\Re(z)-1}}{A}
    =  \thinspace 2\pi \frac{((2m+1)\pi)^{\Re(z)}}{A} \xrightarrow{m\rightarrow\infty} 0.
\end{equation}
The only problem is that I have no clue where to start, could someone give me an idea of how to maybe find something? I know that graphically $A \approx 1$ but besides that I am stuck.
Edit: I tried approaching this with the inverse triangle inequality but that won't get us there because if we write:
$|e^{(2k+1)\pi e^{i\theta}}-1|\geq |e^{(2k+1)\pi \cos\theta}-1| $
then we can no longer find such an $A$ since $\theta = \frac{\pi}{2}$ will result in |1-1| = 0. So I am asking if someone maybe knows a different approach.
 A: From the triangle inequality, we can write,
$$\vert{e^{(2k+1) \pi e^{i \theta}}-1}\rvert \ge \vert \vert{e^{(2k+1) \pi e^{i \theta}}}\rvert -1 \rvert$$
Consider,
$$e^{(2k+1) \pi e^{i \theta}}$$
You can write this in the form $e^{x+iy}$ where x and y are real numbers using Euler's Formula.
Then using the fact $\vert{e^{iy}}\rvert = 1$,
You can find boundary values for $e^{(2k+1) \pi e^{i \theta}}$. Then use that with the above inequality to find the value of A
Hope this helps...
A: Note that $e^{i\theta}=1$ so $|(2k+1)\pi e^{i\theta}|=(2k+1)\pi$ hence $$|(2k+1)\pi e^{i\theta}-2mi\pi| \ge \pi$$ for all $k,m$ integers as for $|2k+1| > 2|m|$ we apply the triangle inequality to get $$|(2k+1)\pi e^{i\theta}-2mi\pi|\ge (|2k+1|-2|m|)\pi \ge \pi$$ and same for $2|m| >|2k+1|$
But now if $z$ is a complex number for which $|z-2mi\pi| \ge \pi$ for all integers $m$, let $m_0$ st $|\Im (z-2m_0i\pi)|$ minimal when $m$ goes through the integers and clearly $|\Im (z-2m_0i\pi)| \le \pi$
But if $\pi/2 \le |\Im (z-2m_0i\pi)| \le \pi$ we have that $\cos \Im (z-2m_0i\pi) \le 0$ so $\Re e^z= \Re e^{z-2m_0i\pi} \le 0$ hence $|e^z-1| \ge \Re (1-e^z) \ge 1$
If now $|\Im (z-2m_0i\pi)| \le \pi/2$, using that $|z-2m_0i\pi| \ge \pi$ gets us $|\Re z| =|\Re (z-2m_0i\pi)| \ge \pi \sqrt 3/2$ so we have either $|e^z-1| \ge e^{\Re z}-1 \ge e^{\pi \sqrt 3/2}-1 >1$ when $\Re z \ge \pi \sqrt 3/2$ or $|e^z-1| \ge 1-e^{\Re z} \ge 1- e^{-\pi \sqrt 3/2}$ when $\Re z \le -\pi \sqrt 3/2$ so we can take $A=1- e^{-\pi \sqrt 3/2}$ and we are done!
A: Here's an explicit formula
for that expression.
Write
$n$ for $2m+1$
and
$t$ for $\theta$.
$\begin{array}\\
|e^{n\pi e^{it}}-1| 
&=|e^{n\pi (\cos(t)+i\sin(t)}-1|\\
&=|e^{n\pi \cos(t)}e^{n\pi i\sin(t)}-1|\\
&=|e^{n\pi \cos(t)}
(\cos(n\pi \sin(t))+i\sin(n\pi \sin(t))-1|\\
&=|e^{n\pi \cos(t)}
\cos(n\pi \sin(t))
-1
+ie^{n\pi \cos(t)}\sin(n\pi \sin(t))|\\
&=|e^{u}
\cos(v)
-1
+ie^{u}\sin(v)|
\qquad u=n\pi \cos(t), v=n\pi \sin(t)\\
|e^{n\pi e^{it}}-1|^2
&=|e^{u}
\cos(v)
-1
+ie^{u}\sin(v)|^2\\
&=(e^{u}\cos(v)-1)^2
+(e^{u}\sin(v))^2\\
&=e^{2u}\cos^2(v)-2e^{u}\cos(v)+1
+e^{2u}\sin^2(v)\\
&=e^{2u}-2e^{u}\cos(v)+1\\
&=e^{2u}-2e^{u}\cos(v)+\cos^2(v)+1-\cos^2(v)\\
&=(e^{u}-\cos(v))^2+\sin^2(v)\\
\end{array}
$
