Thread: Complex Analysis proof

Hey guys, this question was on my final last week and I couldn't get it. If anyone could enlighten me as to what the solution is I'd really appreciate it. Thanks.

Let f(z) be entire (i.e. analytic on all of C) and let |f(z)| > 1. Prove that f is constant.

I know there is a theorem that says the only bounded entire functions are constant functions, so my thought was to show that f is bounded since we assumed it is entire. However, we also assumed that |f(z)| > 1 so it's clearly unbounded. This is where I'm stuck; the only solution I can think of is that my teacher made a typo (and the inequality should read the other way). Any help is appreciated.

consider the function : it is bounded and analytic, thus it is a constant.
further, the range of a nonconstant entire function is the complex field C.

Originally Posted by Shanks

consider the function : it is bounded and analytic, thus it is a constant.
further, the range of a nonconstant entire function is the complex field ...or the whole complex plane minus one single point, according to (little) Picard's Theorem

Tonio

Tonio, THX very much!