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Math Help - Laurent's Theorem

  1. #1
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    Laurent's Theorem

    Hi just a bit of help needed here as I don;t know where to start:

    Part (A)
    ----------------------------
    Suppose f(z) = u(x,y) + iv(x,y)\;and\;g(z) = v(x,y) + iu(x,y) 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 D'(0,R) = \left\{ {z \in C:0 < |z| < R} \right\} where R>0 is fixed. What is the formulae for c_n in the Laurent expansion:
    <br />
f(z) = \sum\limits_{n =  - \infty }^\infty  {c_n z_n }.

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

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



    PART (C)
    -------------------
    Find the maximal radius R>0 for which the function <br />
f(z) = (\sin z)^{ - 1} 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
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  2. #2
    Super Member wingless's Avatar
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    A) Write CRE for both functions.

    B) See here.
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  3. #3
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    PART (A)
    CRE's for the f(z) :
    u_x=v_y and u_y=-v_x.

    CRE's for g(z) :
    v_x=u_y and v_y=-u_x.

    SO:
    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?
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  4. #4
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    If v_x=-v_x then 2v_x=0\Rightarrow v(x,y)=f(y)

    But you can also use the CR equations to show 2v_y=0\Rightarrow v(x,y)=f(x)

    Only way both statements are true is if v(x,y)=k

    Then do these two steps for u(x,y) as well.
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  5. #5
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    For part (c), the Laurent series for \csc(z) is kind of messy. Could certainly use long division to get the first few terms. Here's one way to get the rest of the terms: First look at Mathworld under Bernoulli numbers. There they show how to derive:

    z\cot(z)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n}(2z)^{2n}}{(2n)!}

    Now, if:

    \csc(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}2(2^{2n-1}-1)B_{2n}}{(2n)!}z^{2n-1}

    then split up the series for \csc(z) into two sums, and figure out how the series for z\cot(z) could be modified to represent those two series. Here's a start: what happens when I substitute z\to z/2 in the series for z\cot(z)?

    Probably an easier way to get the terms though. Also, the radius of convergence extends to the nearest singularity which is \pi
    Last edited by shawsend; August 25th 2008 at 10:08 AM. Reason: corrected formula
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  6. #6
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    Quote Originally Posted by mathfied View Post
    PART (C)
    -------------------
    Find the maximal radius R>0 for which the function <br />
f(z) = (\sin z)^{ - 1} is holomorphic in D'(0,R) and find the principal part of its Laurent expansion about z_0=0
    As shawsend has said, R = π. To find the principal part of the Laurent expansion, you might guess that since sin(z) is close to z when z is small, the principal part of 1/sin(z) ought to be 1/z. You can then justify that guess as follows.

    Let g(z) = \frac1{\sin z} - \frac1z = \frac{z-\sin z}{z\sin z}. Since z-sin(z) has degree 3 and z*sin(z) has degree 2, it follows that g(z) has a removable singularity at the origin and is therefore holomorphic in a neighbourhood of the origin. Therefore \frac1{\sin z} = \frac1z + g(z) has the same principal part as 1/z (namely 1/z).
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  7. #7
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    I just realized I over-killed it: You only wanted the singular part. Do what Opalg said or just use long division:
    \csc(z)=\frac{1}{z-\frac{z^3}{3!}+\cdots} and the first term when you do that is 1/z then all the rest are positive powers of z. But hey Mathfied, thanks a bunch. I hadn't known about all that stuff I posted above until I researched it and worked it out myself today. The Bernoulli stuff is interesting I think.
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