# Liouville's Theorem

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1. Jan 19, 2015

### retrostate

1. The problem statement, all variables and given/known data
Show that if f is an entire function that satisfies

|1000i + f(z)| ≥ 1000, for all z ∈ C, then f is constant.

2. Relevant equations
(Hint: Consider the function g(z) = 1000/1000i+f(z) , and apply Liouville’s Theorem.)

3. The attempt at a solution
Ok, so I assume that as f is entire, then for it to be a constant, it must be bounded (Liouville’s Theorem).

Am I right in thinking that as g(z) is the reciprocal of |1000i + f(z)|, Then

1000/1000i+f(z) ≤ 1000/1000 =1

This is as far as I’ve got, I’ve sat here for hours, so any help would be very much appreciated….Thank you

2. Jan 19, 2015

### Dick

You should write that a little more carefully. Use parentheses where you need them and don't drop the absolute value. But yes,

|1000/(1000i+f(z))|<=1. So the function g(z)=1000/(1000i+f(z)) is bounded, yes? Is it entire? Why? What does that let you say about f(z)?

3. Jan 19, 2015

### retrostate

I would say g(z) is differentiable on the whole C - {-1000i}, and so would be entire within the region {z:Imz > - 1000}. And is bounded above at g(z)=1,Therefore, by Liouvilles Theorem f(z) is a constant? i.e. if f(z)=0 then g(z)=i < 1…?

4. Jan 19, 2015

### Dick

To apply Liouville's theorem to g(z) it has to be analytic EVERYWHERE. Since g(z) is a quotient of analytic functions it's analytic everywhere that the denominator doesn't vanish. Does the denominator ever vanish? Remember you are given |1000i+f(z)|>=1000.

5. Jan 19, 2015

### retrostate

Ahhh, no the denominator can't vanish because we are dividing 1000 by an inequality which is greater or equal to 1000, therefore, the maximum value g(z) can take is 1 if |1000i+f(z)|=1000.
So, as the denominator does not vanish it must be analytic everywhere

6. Jan 19, 2015

### Dick

Ok, so if g(z) is bounded and entire what does Liouville tell you about it? What does that imply for f(z)?

7. Jan 19, 2015

### retrostate

That f(z) is constant if and only if g(z) is constant, and because g(z) is bounded and entire, by Liouvilles theorem it is constant, which implies f(z) is constant

8. Jan 19, 2015

### Dick

Sure, that's it!

9. Jan 19, 2015

### retrostate

Thank you Dick, I can't tell you how much I appreciate the time you have spent helping me.