I am taking an exam tomorrow on Complex Analysis, and this is a problem I have come across multiple times on old exams and I've been stumped.
"Show that there is no function f analytic on the punctured plane $\mathbb{C}-\{0\}$ that satisfies $$|f(z)| \ge \frac{1}{\sqrt{|z|}} \,\,\, \forall z \neq 0."$$
I probably tried a number of things back in June when I first saw it, but can't find my work on that right now. When I was going through again today, my initial approach was to assume by contradiction that there is such an f. I set $g(z)=f(1/z)$. Then g would have a pole at infinity. Bringing that information back to f, there's a pole at 0. Which I probably should have seen before doing that.
Then there's an entire function h and some integer n such that $f(z)=\frac{h(z)}{z^n}$. Right? So I was playing with h. From the inequality, $|h(z)| \ge |z|^{n-1/2}$. That doesn't look like a contradiction to me, but maybe I'm missing something.
I asked somebody who has passed this exam before, and they suggested setting $g(z)=\frac{1}{f(z)}$. That gives the inequality $|g(z)| \le \sqrt{|z|}$. Again, I look at singularities. Here, the only singularity is at infinity, the same as my other g. I feel like it's just giving the same information as my first attempt, and I am still seeing no contradiction.
I attempted to search for this on here, but couldn't find a way to formulate it that got me any relevant results through the search bar. If this is a repeat, just point me to the old one and trash mine.
