Liouville's Theorem: Prove f is Constant

[retrostate]

Homework Statement

Show that if f is an entire function that satisfies

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

Homework Equations

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

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

 

[Dick]

Homework Statement

Show that if f is an entire function that satisfies

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

Homework Equations

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

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

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)?

 

[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…?

 

[Dick]

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…?

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.

 

[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

 

[Dick]

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

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

 

[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

 

[Dick]

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

Sure, that's it!

 

[retrostate]

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