# Complex Analysis - Liouville's Theorem

• November 25th 2011, 03:56 PM
kafe
Complex Analysis - Liouville's Theorem
Hello,
I am asked, whether for every analytic non-constant function f there exist z such as $\Re f(z) > |f(z)|^2$.
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, $\Re f(z) \leq |f(z)|^2$ 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!
• November 26th 2011, 05:09 AM
xxp9
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 $\infty$ is an essential singular point of f. According to big picard, we're done.
• November 26th 2011, 07:04 AM
kafe
Re: Complex Analysis - Liouville's Theorem
However, if I can proof that indeed if f has a root then in its neighbourhood I can find z which satisfies $\Re f(z) > |f(z)|^2$, 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 $e^{\frac{1}{f(z)}$: suppose $\Re f(z) \leq |f(z)|^2$, or in other words: $\Re ( \frac{1}{f(z)} ) \leq 1$. So $| e^{\frac{1}{f(z)}} | = e^{\Re ( \frac{1}{f(z)} )} \leq e$. So it's bounded, ergo f is constant.