Minimum of cosine on circles Consider the following function
$f(\delta) = \inf_{t \in \mathbb R} \sup_{\varepsilon \in (\delta/2,2\delta)}\sup_{z \in B_{1+\varepsilon}(0)}  \left\vert \left(1+\cos(2\pi z)\right)^2 -t \right\vert$
where $B_{r}(0)$ is the complex circle of radius $r$ centred at $0$.
It is clear that $f(\delta)>0$ for any $\delta>0$, but I wonder if one can have a quantitative lower bound.
 A: Here's an answer with informal reasoning (I used graphing software).
Firstly I plot the function $h(\theta) = (1 + \cos(s e^{i\theta}))^2$ where $s := 2\pi(1+\epsilon)$ on the Cartesian plane (as a parametric function) for $\theta \in [0, \pi]$ (this function has period $\pi$) on https://www.desmos.com/calculator/ksjcpazwa9 for $s = 7 \Leftrightarrow \epsilon = \frac{7}{2\pi} - 1$. The function has similar shape for $s = 8,9,10 ... $ (plot this to check).
Here is the function to copy-paste on the site:
\left(\left(1\ +\ \cos\left(7\cos\left(t\right)\right)\ \frac{\left(e^{\left(7\sin\left(t\right)\right)}\ +\ e^{\left(-7\sin\left(t\right)\right)}\right)}{2}\right)^{2\ }-\ \left(\sin\left(7\cos\left(t\right)\right)\ \frac{\left(e^{\left(7\sin\left(t\right)\right)}\ -\ e^{\left(-7\sin\left(t\right)\right)}\right)}{2}\right)^{2},2\left(1\ +\ \cos\left(7\cos\left(t\right)\right)\ \frac{\left(e^{\left(7\sin\left(t\right)\right)}\ +\ e^{\left(-7\sin\left(t\right)\right)}\right)}{2}\right)\left(\sin\left(7\cos\left(t\right)\right)\ \frac{\left(e^{\left(7\sin\left(t\right)\right)}\ -\ e^{\left(-7\sin\left(t\right)\right)}\right)}{2}\right)\right)

and specify the parameter bounds as $0 \le t \le \pi$ (the site accepts the parameter variable as $t$, which is $\theta$ for us).
From the plot, we see that $h(\theta)$ has atleast two distinct real values, let the greatest one be $x_1(\epsilon)$ and the least one be $x_2(\epsilon)$. Then we can try and compute: $g(\epsilon) = \inf_{t \in \mathbb R} \sup_{z \in B_{1+\varepsilon}(0)}  \left\vert \left(1+\cos(2\pi z)\right)^2 -t \right\vert = \inf_{t \in \mathbb R} \sup_{\theta \in [0,\pi]}  \left\vert h(\theta) -t \right\vert $ where $h(\theta)$ contains the parameter $\epsilon$. Then from the shape of the curve we can 'see' that:
$g(\epsilon) = \frac{x_1(\epsilon) - x_2(\epsilon)}{2}$ with minimum $t$ occurring at $\frac{x_1(\epsilon) + x_2(\epsilon)}{2}$. (Take $3$ cases, one with  $t \ge x_1$, second with $t \le x_2$ and third with $x_2 < t < x_1$). Also we note that $x_1(\epsilon) - x_2(\epsilon)$ increases with $\epsilon$, hence we have that $f(\delta) = \frac{x_1(2\delta) - x_2(2\delta)}{2}$.
Now computing $x_1(2\delta)$ and $x_2(2\delta)$ involves finding the greatest and least real values of $(1 + \cos(s' e^{i\theta}))^2$ where $s' = 2\pi(1 + 2\delta)$ - this shouldn't be hard to find.
Also given the 'informal' intuition above, I think it should be possible to formalize the argument.
