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Homework Help: [Complex analysis] Coefficients of Laurent series

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    I have some past exam questions that I am confused with


    2. Relevant equations

    [itex]a_{n} = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\, dz[/itex]

    3. The attempt at a solution

    I'm not sure how to approach this, I'm completely lost and just attempted to solve a few:

    a) it says [itex]f(z)[/itex] has a pole of order 5, so [itex] f(z) = \frac{g(z)}{z^5}, g(z)\neq0 [/itex]

    so then I guess the condition is [itex]a_{4} = \frac{g^{(4)}(0)}{4!} [/itex]? But that's just applying the formula for the coefficients...

    c) [itex] f(\frac{1}{z}) = \frac{g(\frac{1}{z})}{z^5} => f(z) = z^5g(z) [/itex]

    so the coefficients are [itex] a_{n} = \frac{1}{2\pi i} \oint_\gamma z^5g(z) dz [/itex]?

    d) [itex] \frac{1}{f(z)} = \frac{g(z)}{z^5} => f(z) = \frac{z^5}{g(z)} [/itex]

    so, [itex] a_{n} = \frac{1}{2\pi i} \oint_\gamma \frac{z^5}{g(z)} dz [/itex]

    g) [itex]a_{-1} = \frac{1}{2\pi i} \oint_ \gamma f(z) dz = \frac{1}{2\pi i} = Res(f; c)*I(\gamma; c) = -Res(f; c)[/itex]

    h) [itex] \frac{a_{n}}{16} = 4^{n}a_{n} => 0 = a_{n}(4^{n} - 4^{-2}) => a_{n} = 0 [/itex] or [itex] n = -2 [/itex]

    for e) and f), I'm not sure what the relevance of the essential singularity is

    Well, I think you can see I'm clearly lost, would appreciate if you could help me out.
  2. jcsd
  3. Apr 10, 2012 #2
    You need to look at these by sayin' for example, (a), if
    [tex]f(z)=\sum_{n=0}^{\infty} a_n z^n+\sum_{n=1}^{\infty}\frac{b_n}{z^n}[/tex]
    and it has a pole of order 5, then that must mean it at least has a non-zero term for the [itex]\frac{1}{z^5}[/itex] term and all the other terms [itex]\frac{1}{z^n}[/itex] for n>5 must be zero else the pole would have a higher order. Ok, I'll do some of (b) same dif:

    [tex]f(z)=\sum_{n=0}^{\infty} a_n z^n+\sum_{n=1}^{\infty}\frac{b_n}{z^n}[/tex]
    and [itex]f(z)-7e^{1/z} [/itex] has a pole of order 5 and I know that [itex]e^{1/z}=\sum_{n=0}^{\infty} \frac{1}{z^n n!}[/itex], then that must mean the terms of f(z) would have to be what to cancel all that out except for at least the [itex]\frac{1}{z^5}[/itex] term?

    Ok then, do that same thing with all the rest.
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