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Math Help - Residue Theorem

  1. #1
    lpd
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    Residue Theorem

    Hi. I have this problem.

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

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

    Thus by the residue theorem
    \int_C{\frac{dz}{z^2-2az+1}=2\pi i Res[\frac{1}{z^2-2az+1}, a-\sqrt{a^2-1}]
    =lim_{z \to \beta}(z-\beta) \frac{1}{(z-\alpha)(z-\beta)} = lim_{z\to\beta}\frac{1}{z-\alpha}
    =\frac{1}{\beta-\alpha}=\frac{1}{-2\sqrt{a^2-1}}
    \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
    \int_0^{2\pi}\frac{dx}{a-cos(x)}
    So I write the integral by letting z=e^{ix} and cos(x)=\frac{1}{2}(z+z^{-1})...
    and yields \frac{-2}{i}\int_C \frac{dz}{z^2-2az+1}
    and i use the previous answer, and I get \frac{2\pi}{\sqrt{a^2-1}}

    The main problem is...

    c) What happens to your argument (and the values of the integrals) when a<-1?
    I'm not entirely sure about this one. My guess is \alpha will now be the simple pole, but when i rework it all out again, it comes to the same conclusion of \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
    \int_C{\frac{dz}{z^2-2az+1}
    where C is the unit circle { z:|z|=1} and a is real with a>1.

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

    Thus by the residue theorem
    \int_C{\frac{dz}{z^2-2az+1}=2\pi i Res[\frac{1}{z^2-2az+1}, a-\sqrt{a^2-1}]
    =lim_{z \to \beta}(z-\beta) \frac{1}{(z-\alpha)(z-\beta)} = lim_{z\to\beta}\frac{1}{z-\alpha}
    =\frac{1}{\beta-\alpha}=\frac{1}{-2\sqrt{a^2-1}}
    \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
    \int_0^{2\pi}\frac{dx}{a-cos(x)}
    So I write the integral by letting z=e^{ix} and cos(x)=\frac{1}{2}(z+z^{-1})...
    and yields \frac{-2}{i}\int_C \frac{dz}{z^2-2az+1}
    and i use the previous answer, and I get \frac{2\pi}{\sqrt{a^2-1}}

    The main problem is...

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


    What did you just write here above?? Is it a<-1 ? If so then you'll have to take 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 \alpha will now be the simple pole, but when i rework it all out again, it comes to the same conclusion of \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 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 \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 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 \frac{2\pi}{\sqrt{a^2-1}}


    This time I think you get the opposite sign in \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, 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 \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, 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 lim\limits_{z\to\beta}\frac{1}{z-\alpha}

    So I think I will end up with \frac{-2 \pi}{\sqrt{a^2-1}}, so does that mean I'll get the negative answer when <br />
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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