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Thread: Residue Theorem

  1. #1
    lpd
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    Residue Theorem

    Hi. I have this problem.

    a) Use the Residue Theorem to calculate
    $\displaystyle \int_C{\frac{dz}{z^2-2az+1}$
    where $\displaystyle C$ is the unit circle {$\displaystyle z:|z|=1$} and $\displaystyle a$ is real with $\displaystyle a>1$.

    Okay, this is my attempt. I found the roots of $\displaystyle z^2-2az+1$ and came up with $\displaystyle a \pm \sqrt{a^2-1}$, and noticed that we can only use $\displaystyle a-\sqrt{a^2-1}$, and i denote this as $\displaystyle \beta=a-\sqrt{a^2-1}$ (and $\displaystyle \alpha =a+\sqrt{a^2-1}$)

    Thus by the residue theorem
    $\displaystyle \int_C{\frac{dz}{z^2-2az+1}=2\pi i Res[\frac{1}{z^2-2az+1}, a-\sqrt{a^2-1}]$
    $\displaystyle =lim_{z \to \beta}(z-\beta) \frac{1}{(z-\alpha)(z-\beta)} = lim_{z\to\beta}\frac{1}{z-\alpha}$
    $\displaystyle =\frac{1}{\beta-\alpha}=\frac{1}{-2\sqrt{a^2-1}}$
    $\displaystyle \int_C{\frac{dz}{z^2-2az+1}=2\pi i \frac{1}{-2\sqrt{a^2-1}} = -\frac{\pi i}{2\sqrt{a^2-1}}$

    b) Use this to calculate
    $\displaystyle \int_0^{2\pi}\frac{dx}{a-cos(x)}$
    So I write the integral by letting $\displaystyle z=e^{ix}$ and $\displaystyle cos(x)=\frac{1}{2}(z+z^{-1})...$
    and yields $\displaystyle \frac{-2}{i}\int_C \frac{dz}{z^2-2az+1}$
    and i use the previous answer, and I get $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$

    The main problem is...

    c) What happens to your argument (and the values of the integrals) when $\displaystyle a<-1$?
    I'm not entirely sure about this one. My guess is $\displaystyle \alpha$ will now be the simple pole, but when i rework it all out again, it comes to the same conclusion of $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$.. so i'm not sure please help.
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  2. #2
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    Quote Originally Posted by lpd View Post
    Hi. I have this problem.

    a) Use the Residue Theorem to calculate
    $\displaystyle \int_C{\frac{dz}{z^2-2az+1}$
    where $\displaystyle C$ is the unit circle {$\displaystyle z:|z|=1$} and $\displaystyle a$ is real with $\displaystyle a>1$.

    Okay, this is my attempt. I found the roots of $\displaystyle z^2-2az+1$ and came up with $\displaystyle a \pm \sqrt{a^2-1}$, and noticed that we can only use $\displaystyle a-\sqrt{a^2-1}$, and i denote this as $\displaystyle \beta=a-\sqrt{a^2-1}$ (and $\displaystyle \alpha =a+\sqrt{a^2-1}$)

    Thus by the residue theorem
    $\displaystyle \int_C{\frac{dz}{z^2-2az+1}=2\pi i Res[\frac{1}{z^2-2az+1}, a-\sqrt{a^2-1}]$
    $\displaystyle =lim_{z \to \beta}(z-\beta) \frac{1}{(z-\alpha)(z-\beta)} = lim_{z\to\beta}\frac{1}{z-\alpha}$
    $\displaystyle =\frac{1}{\beta-\alpha}=\frac{1}{-2\sqrt{a^2-1}}$
    $\displaystyle \int_C{\frac{dz}{z^2-2az+1}=2\pi i \frac{1}{-2\sqrt{a^2-1}} = -\frac{\pi i}{2\sqrt{a^2-1}}$

    b) Use this to calculate
    $\displaystyle \int_0^{2\pi}\frac{dx}{a-cos(x)}$
    So I write the integral by letting $\displaystyle z=e^{ix}$ and $\displaystyle cos(x)=\frac{1}{2}(z+z^{-1})...$
    and yields $\displaystyle \frac{-2}{i}\int_C \frac{dz}{z^2-2az+1}$
    and i use the previous answer, and I get $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$

    The main problem is...

    c) What happens to your argument (and the values of the integrals) when $\displaystyle a<-1$?


    What did you just write here above?? Is it $\displaystyle a<-1$ ? If so then you'll have to take $\displaystyle a+\sqrt{a^2-1}$ , since the other

    root is out of the zone bounded by the path of integration...

    Tonio



    I'm not entirely sure about this one. My guess is $\displaystyle \alpha$ will now be the simple pole, but when i rework it all out again, it comes to the same conclusion of $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$.. so i'm not sure please help.
    .
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  3. #3
    lpd
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    Yeh. I noticed that. But when I take $\displaystyle a+\sqrt{a^2-1}$ and rework my question. I arrive and the same conclusion in the end. So do I just say, I use the alternative root, but the value of the integral is the same?]

    Edit: or does it take the negative value of $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$
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  4. #4
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    Quote Originally Posted by lpd View Post
    Yeh. I noticed that. But when I take $\displaystyle a+\sqrt{a^2-1}$ and rework my question. I arrive and the same conclusion in the end. So do I just say, I use the alternative root, but the value of the integral is the same?]

    Edit: or does it take the negative value of $\displaystyle \frac{2\pi}{\sqrt{a^2-1}}$


    This time I think you get the opposite sign in $\displaystyle \lim\limits_{z \to \beta}(z-\beta) \frac{1}{(z-\alpha)(z-\beta)} = lim\limits_{z\to\beta}\frac{1}{z-\alpha}$.

    Take, for example, $\displaystyle a=-2\Longrightarrow z^2+4z+1=(z-(-2+\sqrt{3}))(z-(-2-\sqrt{3}))$ ...

    Tonio
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  5. #5
    lpd
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    Quote Originally Posted by tonio View Post
    This time I think you get the opposite sign in $\displaystyle \lim\limits_{z \to \beta}(z-\beta) \frac{1}{(z-\alpha)(z-\beta)} = lim\limits_{z\to\beta}\frac{1}{z-\alpha}$.

    Take, for example, $\displaystyle a=-2\Longrightarrow z^2+4z+1=(z-(-2+\sqrt{3}))(z-(-2-\sqrt{3}))$ ...

    Tonio
    Yes. I do in fact get a negative sign in $\displaystyle lim\limits_{z\to\beta}\frac{1}{z-\alpha}$

    So I think I will end up with $\displaystyle \frac{-2 \pi}{\sqrt{a^2-1}}$, so does that mean I'll get the negative answer when $\displaystyle
    a<-1$. whereas when a>1, i will get a postive answer?

    Howcome, when I compute the integral on the calculator for both cases, I get the same answer?
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