Consider the entire function $f(z)=z(z-i)$. Put $$S=\bigg\{\frac{1}{|f(z)|}\ |\ |z|\geq 2\bigg\}.$$ At what value(s) of $z$ is the maximum of the set $S$ attained?
My Idea: As $f(z)$ is an entire function, we need to find the minimum value of $f(z)$ in $|z|\geq 2$. This must be attained on the boundary. But the boundary is whole $2e^{i\theta}$ for $0\leq \theta \leq 2\pi$. How to reach to the exact answer?
Since $f$ is analytic and $\lim_{z\to\infty}f(z)=0$, the maximum of $|f|$ must be attained at the boundary of your region (by the maximum principle), that is, when $|z|=2$. But then$$\max\frac1{|f(z)|}=\max\frac1{2|z-i|}=\frac12\frac1{\min|z-i|},$$and the minimum of $|z-i|$ (when $|z|=2$) is attained at $2i$.