complex analysis and Liouville's theorem problems

**oogie** · December 6th 2009, 12:21 PM

I had these two problems on a problem set and my solutions were incorrect and I haven't been able to figure them out.

Let f(z) be analytic on the complex plane. Prove f is a constant in the following cases:

$|f(z)|\geq7$

and

$Re f(z)\geq 7$ , hint: consider $\exp(f(z))$
both for all z in the complex plane.

Thanks.

**shawsend** · December 6th 2009, 01:30 PM

A non-constant entire function achieves all values in $\mathbb{C}$ at least once with at most one exception: If it's a non-constant polynomial, it attains all values n times, and if it's non-polynomial, it reaches all values infinitely often with at most one exception.

**Jose27** · December 6th 2009, 01:38 PM

Since $\vert f(z)\vert \geq 7$ we have that $g(z)=\frac{1}{f(z)}$ is an entire bounded function. For the second one notice that $\vert e^{f(z)} \vert =\vert e^{Re(f(z))} \vert \geq e^7$ now argue as in the first problem.

**oogie** · December 6th 2009, 01:58 PM

Thanks shawsend. Do you or does anyone have something more mathematical? I was half-expecting a Cauchy integral to be used or something. I have some other problems like this and I'm sure I'd be asked something like this for the final where I'd be expected to be more rigorous.

**oogie** · December 6th 2009, 02:00 PM

Originally Posted by Jose27

Since $\vert f(z)\vert \geq 7$ we have that $g(z)=\frac{1}{f(z)}$ is an entire bounded function. For the second one notice that $\vert e^{f(z)} \vert =\vert e^{Re(f(z))} \vert \geq e^7$ now argue as in the first problem.

Yeah, I inverted the function like you did but apparently it's not complex analytic according to the person who graded my problem set so it was deemed incorrect.

**oogie** · December 6th 2009, 04:01 PM

Can anyone verify that $g(z)=\frac{1}{f(z)}$ is indeed analytic based on the above?
Thanks.

**shawsend** · December 7th 2009, 06:46 AM

I felt Jose did a better job than me showing that under those constraints, the function must be constant. Everything he did is completely analytic and entire. I don't see what the problem was.

**oogie** · December 7th 2009, 09:10 AM

Okay great, thanks to the both of you.

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