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Thread: area defined as variable resedue question..

  1. #1
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    area defined as variable resedue question..

    i need to calculate this integral from plus to minus infinity$\displaystyle f(z)=\frac{z}{e^{2\pi iz^2}-1}\\$

    in this area

    $\displaystyle
    \gamma _r=\left \{ |z|=r \right \},n<r^2<n+1
    $

    i need to find the points which turn to zero in the denominator
    and non zero in the numerator.
    i got two such points
    $\displaystyle z=\pm \sqrt{n}$
    by using this formula
    $\displaystyle res(\sqrt{a})=\frac{p(a)}{q(a)'}$
    $\displaystyle res(\sqrt{n})=\frac{1}{4\pi i}$
    $\displaystyle res(-\sqrt{n})=\frac{1}{4\pi i}$

    the third point is z=0 but for it we have both numerator and denominator 0
    i calculated the residium for it by $\displaystyle res(f(x),a)=\lim_{x->a}(f(x)(x-a))$ formula
    but then
    my prof says some stuff that involves the area
    he says that my points are 0 +1 -1 +2^(0.5) -2^(0.5) etc.. because the denominator goes to zero
    for each point have a residiu and i need to sum the residiums inside.
    but here the area is not defined
    its not like (by radius 3)

    i dont know what point are inside the area
    Last edited by transgalactic; Jan 16th 2010 at 05:30 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    i need to calculate this integral from plus to minus infinity$\displaystyle f(z)=\frac{z}{e^{2\pi iz^2}-1}\\$

    in this area

    $\displaystyle
    \gamma _r=\left \{ |z|=r \right \},n<r^2<n+1
    $
    This simply makes no sense. You say you want to integrate "from plus to minus infinity", but then you say that |z|= r. They can't both be true.

    Perhaps you mean that you want to integrate around the circle z= |r|
    And saying that "$\displaystyle n< r^2< n+1$" simply says that $\displaystyle \sqrt{n}< r< \sqrt{n+1}$

    i need to find the points which turn to zero in the denominator
    and non zero in the numerator.
    i got two such points
    $\displaystyle z=\pm \sqrt{n}$
    by using this formula
    $\displaystyle res(\sqrt{a})=\frac{p(a)}{q(a)'}$
    $\displaystyle res(\sqrt{n})=\frac{1}{4\pi i}$
    $\displaystyle res(-\sqrt{n})=\frac{1}{4\pi i}$
    $\displaystyle e^{2\pi i z^2}- 1$ is 0 only when $\displaystyle e^{2\pi iz^2}= 1$ which happens only when $\displaystyle 2\pi i z^2$ is a multiple of $\displaystyle 2\pi i$: $\displaystyle z^2= m$ for some integer m so the singularities are at $\displaystyle -\sqrt{m}$ and $\displaystyle \sqrt{m}$. If you are integrating around the path |z|= r, with $\displaystyle
    \sqrt{n}< r< \sqrt{n+1}$, then there are two poles for every positive integer $\displaystyle m\le n$

    the third point is z=0 but for it we have both numerator and denominator 0
    i calculated the residium for it by $\displaystyle res(f(x),a)=\lim_{x->a}(f(x)(x-a))$ formula
    but then
    my prof says some stuff that involves the area
    he says that my points are 0 +1 -1 +2^(0.5) -2^(0.5) etc.. because the denominator goes to zero
    for each point have a residiu and i need to sum the residiums inside.
    but here the area is not defined
    its not like (by radius 3)

    i dont know what point are inside the area
    Last edited by HallsofIvy; Jan 17th 2010 at 03:56 AM.
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