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Complex analysis

  1. Feb 16, 2009 #1
    Hi,
    I'm studying complex analysis right now, I would like to use this thread to ask questions when I read books. Many questions will be very stupid, so please bear with me.
    Also, English is my second language.

    text: Complex Analysis (2nd edition)
    author: Stephen D. Fisher

    [question deleted] this first question is very stupid. i figured it out. thank you. i will come back for other questions.

    Thanks.
     
    Last edited: Feb 16, 2009
  2. jcsd
  3. Feb 17, 2009 #2
    Variation of Maximum principle: "If u(x,y) is harmonic and nonconstant on a domain D, then |u(x,y)| has no local maximum in D". the proof of this is left as an exercise. i want to prove it.

    case 1: u(x,y) is complex-valued function, then since u is harmonic, it is analytic on D. By the Maximum Modulus Principle, |u| has no local max in D
    case 2: u(x,y) is real-valued function, then u has no local max and no local min in D. How do you go from this to |u|?

    Thanks.
     
  4. Feb 17, 2009 #3

    HallsofIvy

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    If u is harmonic on D, then there exist another harmonic function, v(x,y), such that f(z)= u(z)+ iv(z) where z= x+ iy, is analytic on D.
     
  5. Feb 17, 2009 #4
    Not true 100%. For example [itex]D=\mathbb{C}\backslash\{0\}[/itex] and [itex]u(z)=\log |z|[/itex].
     
  6. Feb 17, 2009 #5

    HallsofIvy

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    True! I was assuming a simply connected domain.
     
  7. Feb 17, 2009 #6
    then? we're using this fact to prove when the case u is complex-valued function right? I got that part down, put i'm stuck when u is real-valued function.

    actually when I read further down, the book suggest to use mean-value property to prove above statement.
     
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