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Application of Liouville's Theorem Complex Analysis

  1. Mar 7, 2012 #1

    jsi

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    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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!
     
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  3. Mar 7, 2012 #2

    micromass

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    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)).
     
  4. Mar 7, 2012 #3

    jsi

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    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?
     
  5. Mar 7, 2012 #4

    jsi

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    I haven't made any progress on this and I'm still pretty confused...
     
  6. Mar 7, 2012 #5

    Office_Shredder

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    Don't try to re-write g like that. Just use euler's formula on it
     
  7. Mar 7, 2012 #6

    jsi

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    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?
     
  8. Mar 7, 2012 #7

    Dick

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    Personally, I would think about g(z)=1/(f(z)-(n+1)).
     
  9. Mar 7, 2012 #8

    Office_Shredder

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    You might want to re check that. Euler's formula only applies to the imaginary part of the number.
     
  10. Mar 7, 2012 #9

    Dick

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    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?
     
  11. Mar 7, 2012 #10

    jsi

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    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...
     
  12. Mar 7, 2012 #11

    Dick

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    Ok, I see didn't ask enough questions. Isn't g(z)=1/((n+1)-f(z)) also entire as well as bounded?
     
  13. Mar 7, 2012 #12

    jsi

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    Yes, so then |g(z)| must be less than or equal to some constant?
     
  14. Mar 7, 2012 #13

    Dick

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    If g(z) is entire and bounded then it's constant. What does that tell you about f(z)?
     
  15. Mar 7, 2012 #14

    jsi

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    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?
     
  16. Mar 7, 2012 #15

    Dick

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    Sure, if g(z)=C (a constant), then you can solve for f(z). Giving another constant.
     
  17. Mar 7, 2012 #16

    jsi

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    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?
     
  18. Mar 7, 2012 #17

    Dick

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    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)).
     
  19. Mar 8, 2012 #18

    jsi

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    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?
     
  20. Mar 8, 2012 #19

    jsi

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    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?
     
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