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Prove f is constant - liouville's theorem (?)

  1. May 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f=u+iv, where u(z)>v(z) for all z in the complex plane. Show that f is constant on C.


    2. Relevant equations

    none

    3. The attempt at a solution

    Here's my attempt (just a sketch):

    Since f is entire, then its components u(z), v(z) are also entire <- is this necessarily true?

    Since f is entire, the CR equations hold.

    v(z) is bounded for all z, so since v is entire and bounded, by liouville's theorem v is constant. <- kinda shaky on this one, too.

    Since CR equations hold, we know v_y = u_x = 0 and -v_x = u_y = 0.

    So, f' = u_x - iu_y = 0, so f is constant.

    Does this seem okay.. or am I way off base?
     
  2. jcsd
  3. May 20, 2012 #2
    I hate to be this guy.. but does this look okay? I still need help..
     
  4. May 20, 2012 #3
    Don't know if you have Little Picard at your disposal, but that would be the route that I would take.

    It is not true that ##u## and ##v## are entire. They are real analytic, though. Liouville doesn't apply.

    Even if ##v## were entire, it isn't bounded.
     
  5. May 20, 2012 #4
    No, this doesn't look okay at all. In general, neither the real nor imaginary parts of a holomorphic function are themselves holomorphic. In fact, it's not difficult to show that they cannot be holomorphic unless f is constant (basically, note that both u and v are real-valued, and apply the open mapping theorem).

    As for this problem, the way I would go about it is by looking at the function g(z) = 1/(f(z) - i).
     
  6. May 20, 2012 #5
    gopher p: no I can't use Little Picard, unfortunately. I looked it up, though, and it did seem pretty helpful! Thanks for the suggestion!

    Citan Uzuki: yeah, that's what I thought.. my method definitely seemed a tad suspicious to me. I don't quite understand how 1/(f(z)-i) would help.. could you please elaborate a little more?

    Thank you so much, so far!
     
  7. May 20, 2012 #6
    Or, rather, how'd you get 1/(f(z)-i)?
     
  8. May 20, 2012 #7

    Dick

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    If u(z)>v(z) then f(z)=u+iv only has values in half of the complex plane. Sketch the region. i isn't in that region. So |f(z)-i| has a lower bound. Is that enough of a hint?
     
  9. May 20, 2012 #8
    That's perfect. Thank you very much you guys!
     
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