Application of Liouville's Theorem Complex Analysis

In summary: Yes, so then |g(z)| must be less than or equal to some constant?Yes, so then |g(z)| must be less than or equal to some constant?
  • #1
jsi
24
0

Homework Statement



Given: f is an entire function, Re f(z) ≤ n for all z. Show f is constant.

Homework Equations





The Attempt at a Solution



So I thought I'd use Liouville's Theorem which states that, if f(z) is entire and there is a constant m such that |f(z)| ≤ m for all z, then f is constant.

Looking at the problem I thought to rename Re f(z) as g(z) so then g(z) ≤ n for all z. Then take the modulus of both sides and I get |g(z)| ≤ |n| which is exactly what Liouville's Theorem is asking for. Then g(z) is constant and since I said g(z) = Re f(z), Re f(z) must be constant. I'm not sure how to show from there that f(z) instead of Re f(z) is constant. I'm also not sure my logic followed. I'd appreciate some help, thanks!
 
Physics news on Phys.org
  • #2
Your suggestion is to take g(z)=Re(f(z)). But the problem is that g, in general, will not be an analytic function. So Liouville is not applicable.

Something you could do is to apply Liouville on g(z)=exp(f(z)).
 
  • #3
So am I going to want to show g(z) = (exp(f(z)) - exp(f(0))) / z and apply Liouville's Thm which would then show exp(f(z)) = exp(f(0)) which shows f(z) = f(0) then f is constant?
 
  • #4
I haven't made any progress on this and I'm still pretty confused...
 
  • #5
Don't try to re-write g like that. Just use euler's formula on it
 
  • #6
so e^(f(z)) = cos(f(z)) + isin(f(z)) or since it's looking for Re(f(z)) should it be just cos(f(z)) since that's the real part of Euler's formula?
 
  • #7
Personally, I would think about g(z)=1/(f(z)-(n+1)).
 
  • #8
You might want to re check that. Euler's formula only applies to the imaginary part of the number.
 
  • #9
As I've gotten bracketed by posts I'm going to reiterate so it won't get missed. Think about g(z)=1/((n+1)-f(z)). Isn't that bounded?
 
  • #10
Yes, that's bounded, but I'm really not sure how that relates to Liouville's Theorem =/

also, then if I did e^(f(z)), it would give me the imaginary part of the f(z) but since the problem is asking to consider the real part of f(z), e^(Re(f(z))) would just be e^(some real function)? Again, still really confused as to how these relate to Liouville...

My understanding of Liouville is that if you have a function f that is entire and there's a constant such that |f(z)| is less that or equal to that constant, then f(z) is constant. And the proof involves setting a g(z) = (f(z) - f(0))/z which makes g entire then showing g is 0 which shows f(z)-f(0) = 0 implying f(z) = f(0) so f is constant. I can't quite put together how e^f(z) makes showing that easy or how using e^(f(z)) can then show that f(z) is constant...
 
  • #11
jsi said:
Yes, that's bounded, but I'm really not sure how that relates to Liouville's Theorem =/

also, then if I did e^(f(z)), it would give me the imaginary part of the f(z) but since the problem is asking to consider the real part of f(z), e^(Re(f(z))) would just be e^(some real function)? Again, still really confused as to how these relate to Liouville...

My understanding of Liouville is that if you have a function f that is entire and there's a constant such that |f(z)| is less that or equal to that constant, then f(z) is constant. And the proof involves setting a g(z) = (f(z) - f(0))/z which makes g entire then showing g is 0 which shows f(z)-f(0) = 0 implying f(z) = f(0) so f is constant. I can't quite put together how e^f(z) makes showing that easy or how using e^(f(z)) can then show that f(z) is constant...

Ok, I see didn't ask enough questions. Isn't g(z)=1/((n+1)-f(z)) also entire as well as bounded?
 
  • #12
Yes, so then |g(z)| must be less than or equal to some constant?
 
  • #13
jsi said:
Yes, so then |g(z)| must be less than or equal to some constant?

If g(z) is entire and bounded then it's constant. What does that tell you about f(z)?
 
  • #14
If g(z)=1/((n+1)-f(z)) like you stated, then f(z) must be constant as well since it's in a function that is constant?
 
  • #15
jsi said:
If g(z)=1/((n+1)-f(z)) like you stated, then f(z) must be constant as well since it's in a function that is constant?

Sure, if g(z)=C (a constant), then you can solve for f(z). Giving another constant.
 
  • #16
ok that's super easy then! If I were to do it with g(z) = e^(f(z)) would that also be entire so then it would be bounded and therefore constant and would follow the same way for f(z) to be constant? Or is that a little different? And how did the real part of f(z) being less than or equal to n play into how you did it?
 
  • #17
jsi said:
ok that's super easy then! If I were to do it with g(z) = e^(f(z)) would that also be entire so then it would be bounded and therefore constant and would follow the same way for f(z) to be constant? Or is that a little different? And how did the real part of f(z) being less than or equal to n play into how you did it?

Now I'm not very confident that you understand why g(z)=1/((n+1)-f(z)) is bounded and entire. Can you explain why? Look at the real part of ((n+1)-f(z)).
 
  • #18
applying cauchy's theorem to g(f(z)) would show that it is entire because all points are in the radius of |f(z)| less than or equal to n?
 
  • #19
I ended up getting it I think: I set g(z) = e(f(z)) which is an entire function and then |g(z)| = e^(Re(f(z))) and since Re(f(z)) is bounded by n because Re(f(z)) ≤ n as the problem stated, |g(z)| ≤ e^(n) which is a constant so g(z) is constant and since g(z) = e^(f(z)) it follows that ln(g(z)) = f(z) and ln(some constant) = f(z) => f(z) is constant. Is that correct?
 

FAQ: Application of Liouville's Theorem Complex Analysis

1. How is Liouville's Theorem used in complex analysis?

Liouville's Theorem is used to prove that entire functions, which are functions that are analytic everywhere in the complex plane, are either constant or unbounded. This theorem is fundamental in understanding the behavior of complex functions and has many applications in various areas of mathematics and physics.

2. What is the significance of Liouville's Theorem in complex analysis?

Liouville's Theorem is significant because it provides a powerful tool for analyzing and understanding the behavior of entire functions. It also has important applications in areas such as number theory, differential equations, and harmonic analysis.

3. Can Liouville's Theorem be applied to functions with poles?

No, Liouville's Theorem only applies to entire functions, which are analytic everywhere in the complex plane. Functions with poles, which are points where the function is undefined, do not satisfy the conditions of the theorem.

4. How does Liouville's Theorem relate to the fundamental theorem of algebra?

Liouville's Theorem is closely related to the fundamental theorem of algebra. In fact, it can be seen as a generalization of the fundamental theorem of algebra to entire functions. Both theorems state that a non-constant polynomial or entire function must have at least one point where it is unbounded.

5. What are some real-world applications of Liouville's Theorem?

Liouville's Theorem has many real-world applications, particularly in physics and engineering. For example, it can be used to prove the existence of non-periodic solutions to differential equations, which are important in the study of chaotic systems. It also has applications in statistical mechanics, fluid dynamics, and quantum mechanics.

Back
Top