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Math Help - Residue Theorm - Q1

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    Residue Theorm - Q1

    f(z) = P(z)/Q(z) where P(z) and Q(z) are polynomials in z
    Degree[Q(z)] >= Degree[P(z)]+2

    Prove \oint f(z)dz=0, where integration is over a closed curve which encloses all poles of f(z)

    I have been trying this for long - but really stuck up. Tried Cauchy, Laurent Expansion but not going anywhere. Request help please.
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    At infinity, the quotient \frac{P_n(z)}{Q_m(z)}\sim \frac{c}{z^k},\quad k\geq 2 for some constant c right?
    That is, it becomes dominated by the higher-degree denominator. Then the ML-inequality says what about the limit:

    \lim_{R\to\infty}\left|\oint \frac{P_n(z)}{Q_m(z)}dz\right|\
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    Quote Originally Posted by shawsend View Post
    At infinity, the quotient \frac{P_n(z)}{Q_m(z)}\sim \frac{c}{z^k},\quad k\geq 2
    I'm not sure that statement is rigorous or even correct Ok. There are other ways to show the integral is zero. One nice way is to use the residue at infinity. Another is to place more explicit bounds on numerator and denominator. Both these methods are used to show the case of 1/p(z) in "Basic Complex Analysis" by Marsden and Hoffman and I assume the methods would apply for the more general case of p(z)/q(z).
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    Quote Originally Posted by shawsend View Post
    I'm not sure that statement is rigorous or even correct Ok. There are other ways to show the integral is zero. One nice way is to use the residue at infinity. Another is to place more explicit bounds on numerator and denominator. Both these methods are used to show the case of 1/p(z) in "Basic Complex Analysis" by Marsden and Hoffman and I assume the methods would apply for the more general case of p(z)/q(z).
    Thanks I could work out based on your idea of ML inequality for time being I guess it will convince me.
    Thanks for your follow-up post !
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    Quote Originally Posted by shawsend View Post
    One nice way is to use the residue at infinity.
    How about for fun then? To be honest I didn't even know this until you brought it up. Here goes: If a function is analytic outside some domain, then we can consider the residue at infinity. So outside the hull, H, of zeros for Q_m(z), the quotient is analytic so that we can say:

    \mathop\oint\limits_{|z|>H} \frac{P_n(z)}{Q_m(z)}dz=2\pi i \mathop\text{Res}_{z=0} \left\{\frac{1}{z^2} \frac{P(1/z)}{Q(1/z)}\right\}

    And with some work we get:

    =2\pi i\mathop\text{Res}_{z=0}\left\{\frac{1}{z^2} \frac{z^k(a_0 z^n+a_1 z^{n-1}+\cdots+a_n}{b_0 z^m+b_1 z^{m-1}+\cdots+b_m}\right\}

    but k\geq 2 so that:

    =2\pi i\mathop\text{Res}_{z=0}\left\{\frac{z^c(a_0 z^n+\cdots+a_n}{b_0 z^m+\cdots+b_m}\right\}

    with c\geq 1 and therefore that's zero.
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  6. #6
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    Quote Originally Posted by shawsend View Post
    How about for fun then? To be honest I didn't even know this until you brought it up. Here goes: If a function is analytic outside some domain, then we can consider the residue at infinity. So outside the hull, H, of zeros for Q_m(z), the quotient is analytic so that we can say:

    \mathop\oint\limits_{|z|>H} \frac{P_n(z)}{Q_m(z)}dz=2\pi i \mathop\text{Res}_{z=0} \left\{\frac{1}{z^2} \frac{P(1/z)}{Q(1/z)}\right\}

    And with some work we get:

    =2\pi i\mathop\text{Res}_{z=0}\left\{\frac{1}{z^2} \frac{z^k(a_0 z^n+a_1 z^{n-1}+\cdots+a_n}{b_0 z^m+b_1 z^{m-1}+\cdots+b_m}\right\}

    but k\geq 2 so that:

    =2\pi i\mathop\text{Res}_{z=0}\left\{\frac{z^c(a_0 z^n+\cdots+a_n}{b_0 z^m+\cdots+b_m}\right\}

    with c\geq 1 and therefore that's zero.
    I have started this topic very recently. So my fundamentals are kind of shaky moreover I have no exposure to real analysis so it's more of a problem. I may not be able to even appreciate some of the things you say, for e.g. a question which I have is why are you taking residue only at 0 and why not all points with isolated singularity. I also do not know the definition of Hull.
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  7. #7
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    Hi Aman,

    Maybe I should have said "convex hull" which is just a border containing all the poles. Outside this border, the quotient is analytic and therefore we can evaluate a contour integral outside the hull by taking the (single) residue at infinity which by definition is:

    \int_C f(z)dz=2\pi i \mathop\text{Res}_{z=0}\left\{\frac{1}{z^2}f(1/z)\right\}

    You'll go over that soon or just look it up.
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