Constant $A$: $|e^{(2m+1)\pi e^{i\theta}}-1| \geq A > 0$ for all $m\in \mathbb{N}_{>0}$, $\theta \in [0, 2\pi)$

Question

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.

Answer 1

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...

Answer 2

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!

Answer 3

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} $