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Help with a unit disc property for a holomorphic function

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose that f is holomorphic in an open disc U and that Re(f) is
    constant in U. I have to show that f must be constant in U. Also what is the essential
    property of the disc U that it used here? Give an example of an open
    set U for which the conclusion fails.


    2. Relevant equations
    Cauchy–Riemann equations.


    3. The attempt at a solution
    Let f=u+vi where u is a constant.Since f is holomorphic by the Cauchy–Riemann equations->
    u_x=v_y and u_y=-v_x but since u is a constant u_x=u_y=0 => 0=v_y =-v_x...therefore f is constant.
    The disc U has to be open,as in:U(a,r)={z:|z-a|<r}.
    Is this correct?What should I do for the last part?
    Thank you
     
  2. jcsd
  3. Jan 30, 2012 #2
    will U(0,1) be the special case?
     
  4. Jan 30, 2012 #3

    Dick

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    Suppose the open set is the union of two disks that don't intersect?
     
  5. Jan 30, 2012 #4
    I'm sorry but I don't know what to answer to that question.
     
  6. Jan 30, 2012 #5

    Dick

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    Any reason why f can't have one constant value on the first disk and a different constant value on the second disk?
     
  7. Jan 30, 2012 #6
    To be honest I don't think so...that's like having a large supermarket divided in two...one or more products may be on both sides.
     
  8. Jan 30, 2012 #7

    Dick

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    That's a strange analogy. Look. Define f(x)=1 on U(0,1) and f(x)=2 on U(3,1). Let the domain D be the union of the two disks. i) is there any place in the domain where the derivatives of f aren't zero? ii) is f constant on the domain?
     
  9. Jan 30, 2012 #8
    No there isn't...also I think that if there are 2 or more disjoint open sets that make U=> in a contradiction
     
  10. Jan 30, 2012 #9

    Dick

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    I hope that means you get it. So what's the 'essential property' of the domain that would eliminate these cases?
     
  11. Jan 30, 2012 #10
    U is connected
     
  12. Jan 30, 2012 #11

    Dick

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    That's it.
     
  13. Jan 30, 2012 #12
    Thank you:)
     
  14. Jan 30, 2012 #13
    Doesn't such choice of f(x) contradict the given fact in this problem that "Re(f) is constant in U" ?
     
  15. Jan 30, 2012 #14

    Dick

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    Good point. I lost track of the problem. Make that f(x)=i and f(x)=2i.
     
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