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Laurent's Theorem

  1. Aug 24, 2008 #1
    Hi just a bit of help needed here as I don;t know where to start:

    Part (A)
    ----------------------------
    Suppose [itex]f(z) = u(x,y) + iv(x,y)\;and\;g(z) = v(x,y) + iu(x,y)[/itex] are analytic in some domain D. Show that both u and v are constant functions..?

    I guess we have to use the CRE here but not really sure how to approach this..?

    Part (B)
    ----------------------------
    Let f be a holomorphic function on the punctured disk [itex]D'(0,R) = \left\{ {z \in C:0 < |z| < R} \right\}[/itex] where R>0 is fixed. What is the formulae for c_n in the Laurent expansion:
    [itex]
    f(z) = \sum\limits_{n = - \infty }^\infty {c_n z_n }[/itex].

    Using these formulae, prove that if f is bounded on D'(0,R), it has a removable singularity at 0.

    - Well I know that:
    [itex]c_n = \frac{1}
    {{2\pi i}}\int\limits_{\gamma _r }^{} {\frac{{f(s)}}
    {{(s - z_0 )^{n + 1} }}} ds = \frac{{f^{(n)} (z_0 )}}
    {{n!}}[/itex].
    Any suggestions from here?


    PART (C)
    -------------------
    Find the maximal radius R>0 for which the function [itex]
    f(z) = (\sin z)^{ - 1}[/itex] is holomorphic in D'(0,R) and find the principal part of its Laurent expansion about z_0=0

    ??

    Any help would be greatly appreciated.

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 24, 2008 #2

    Dick

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    I'll start you out with the first one. CRE's for the f(z) tell you u_x=v_y and u_y=-v_x. CRE's for g(z) tell you v_x=u_y and v_y=-u_x. What happens when you put both of these together?
     
  4. Aug 24, 2008 #3

    Dick

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    For the second one, you might want to focus your efforts on proving that c_n=0 for n<0.
     
  5. Aug 25, 2008 #4
    hmm so for part (1)
    u_x = v_y = -u_x AND
    u_y = -v_x = v_x

    so u and v are constant because u_x = -u_x and -v_x = v_x

    is that correct?
     
  6. Aug 25, 2008 #5

    Dick

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    Yes. u_x=-u_x means u_x=0. The same for all of the other stuff. All of the partial derivatives are zero. Hence?
     
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