Define the function $f(r,\theta) =e^{2 r \cos(\theta)} - 2 e^{ r \cos(\theta) } \cos\left( r \sin( \theta ) \right) + 1$.
Suppose that I have the following constraints on $r$ and $\theta$;
$0 \leq \theta \leq \tfrac{\pi}{2}$
$2\pi N + \Gamma \leq r \leq 2 \pi (N+1) - \Gamma$ for some $N \in \mathbb{Z}^{+}$ and some tiny $0 < \Gamma \ll 2\pi$
By playing around with mathematica, I have come up with what I believe to be a valid lower bound: $$ f(r,\theta) \geq 4 \sin^{2}\left( \tfrac{\Gamma}{2} \right) $$
I've done this mostly as a guess; I noticed I should get $f(r,\theta)=0$ only when $(r,\theta) = (2\pi n, \tfrac{\pi}{2})$. I then essentially minimized the function $f(r,\tfrac{\pi}{2}) = 4 \sin^{2}(\tfrac{r}{2})$ for $r$ as bounded above.
I have two questions:
Is this inequality actually true?
Is there a more analytic/mathematically precise way of showing this inequality is true?
Thanks in advance.
EDIT: If it helps; $f(r,\theta) = e^{z} - 1 = f(z)$ for $z = r e^{i \theta}$ as given above. If anyone is interested where this is coming from, I have a quarter-circlular contour of radius $r$ over a function $\propto \frac{1}{f(z)}$. I need the above bound as part of my proof that this integral falls to $0$ when I make $r$ large.