Complex Analysis - Liouville's Theorem

Hello,

I am asked, whether for every analytic non-constant function f there exist z such as $\displaystyle \Re f(z) > |f(z)|^2$.

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, $\displaystyle \Re f(z) \leq |f(z)|^2$ 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!

Re: Complex Analysis - Liouville's Theorem

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 $\displaystyle \infty$ is an essential singular point of f. According to big picard, we're done.

Re: Complex Analysis - Liouville's Theorem

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 $\displaystyle \Re f(z) > |f(z)|^2$, 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 $\displaystyle e^{\frac{1}{f(z)}$: suppose $\displaystyle \Re f(z) \leq |f(z)|^2$, or in other words: $\displaystyle \Re ( \frac{1}{f(z)} ) \leq 1$. So $\displaystyle | e^{\frac{1}{f(z)}} | = e^{\Re ( \frac{1}{f(z)} )} \leq e$. So it's bounded, ergo f is constant.