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

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

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



    PART (C)
    -------------------
    Find the maximal radius R>0 for which the function $\displaystyle
    f(z) = (\sin z)^{ - 1}$ 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 $\displaystyle v_x=-v_x$ then $\displaystyle 2v_x=0\Rightarrow v(x,y)=f(y)$

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

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

    Then do these two steps for $\displaystyle u(x,y)$ as well.
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  5. #5
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    For part (c), the Laurent series for $\displaystyle \csc(z)$ 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:

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

    Now, if:

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

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

    Probably an easier way to get the terms though. Also, the radius of convergence extends to the nearest singularity which is $\displaystyle \pi$
    Last edited by shawsend; Aug 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 $\displaystyle
    f(z) = (\sin z)^{ - 1}$ 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 $\displaystyle g(z) = \frac1{\sin z} - \frac1z = \frac{z-\sin z}{z\sin z}$. 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 $\displaystyle \frac1{\sin z} = \frac1z + g(z)$ 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:
    $\displaystyle \csc(z)=\frac{1}{z-\frac{z^3}{3!}+\cdots}$ and the first term when you do that is $\displaystyle 1/z$ then all the rest are positive powers of $\displaystyle z$. 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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