Complex anaylsis question involving liouville's theorem

[harmonie_Best]

Homework Statement

Show that if f is an entire function which satisfies (a) I am f(z) > - 10 or (b) |f(z)| >= 5, then f is constant.

Homework Equations

liouville's theorem, Cauchy's inequality(?)

The Attempt at a Solution

Want to show that both are bounded as it will satisfy liouville's theorem and prove they are constant,

So there exists an M $$\in$$ R s.t. |f(z)| <= M, for all z $$\in$$ f(z)

Just get really confused now, can't see how to get it so that the sign is the other way round. Do you take the negative of both to change direction of the sign =/

Thank you!

 

[Dick]

Taking the negative won't change the sign of the inequality. It's an absolute value. Set g(z)=1/f(z) and see what Liouville tells you about g(z).

 

[harmonie_Best]

Okay so can I say:

Set g(z) = 1/f(z) which is <= 1/5, therefore g(z) is bounded by 1/5, and by Cauchys integral theorem, g(z) is constant. As g(z) a function including f(z) we can conclude that f(z) is also constant?

 

[Dick]

Okay so can I say:

Set g(z) = 1/f(z) which is <= 1/5, therefore g(z) is bounded by 1/5, and by Cauchys integral theorem, g(z) is constant. As g(z) a function including f(z) we can conclude that f(z) is also constant?

You are leaving the absolute values out. |g(z)|<=1/5, right. So it's constant. And it's by Liouville's theorem, not Cauchy's. If g(x) is constant then since f(z)=1/g(z), f(z) must be constant.

 

[harmonie_Best]

Ohhhh, I get it, thank you =) I need to pay more attention!x