Thread: Complex Analysis - Liouville's Theorem

Hello,
I am asked, whether for every analytic non-constant function f there exist z such as .
I am pretty sure it is true because I read about Picard's theorems, but I cannot use that. I can use Liouville's Theorem and if needed Cauchy.
I've been trying to assume that for every z, and then find a function g(f(z)) such as the condition bounds g, therefore g is constant and so is f, but with no luck so far.


Thank you!

Note that every point in the interval (0,1) of the real axis satisfies your inequation. So you need only to show that the domain of f intersects with this interval.
If f is a polynomial, since every polynomial( except the constants) has a zero point, a neighborhood of 0 is covered by the domain of f, we're done.
Otherwise the power series of f has infinitly many terms, thus is an essential singular point of f. According to big picard, we're done.

Thank you for your prompt reply.

Firstly, it is known that f is analytic everywhere, so we don't even have to use Big Picard.
Secondly, as I said, I cannot use too advanced theorems, I can use Liouville and Cauchy.
However, if I can proof that indeed if f has a root then in its neighbourhood I can find z which satisfies , then I'm done: let f be a non-constant analytic everywhere function. If f has a root, then according to that, I'm done. Otherwise, we can look at the function : suppose , or in other words: . So . So it's bounded, ergo f is constant.