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Math Help - Problem #5 Contour Integration

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    Problem #5 Contour Integration

    Let's up the difficulty for this one, shall we?

     \int _{|z|=1}  \frac{6z^{98}}{23z^{99} -2z^{81} + z^4 - 7}~dz
    (where z is a complex variable)

    (Source: Mathematics Stack Exchange )

    Hint: What does Rouche's theorem say about the zeros of the denominator?

    -Dan
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    Forum Admin topsquark's Avatar
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    Re: Problem #5 Contour Integration

    Sorry about the delay. I had a hiccup with the proof.

    Anyway, let's calculate
    \int _{|z| = 1} \frac{6 z^{98}}{23z^{99} - 2 z^{81} + z^4 - 7}~dz

    This problem, like just about all problems with integration over complex numbers, is going to depend on just where the zeros of the polynomial in the denominator are. I don't know about you but I would have some slight difficulty in finding the 99 zeros of the denominator. But we may make an appeal to Rouche's theorem. Since we want to integrate over the circle |z| = 1 let's see what we can do on this boundary.

    ala Rouche: |-2 z^{81} + z^4 - 7| < |23 z^{99}| on (most of) the disk |z| \leq 1, so all the zeros of 23z^{99} - 2 z^{81} + z^4 - 7 lie within the unit circle. (Here's where my glitch occurred: The zeros of the denominator actually lie in a washer on the unit disk. The point is that none of the zeros occur outside the unit disk.)

    Now for the fun part. We get the same result integrating over |z| = 1 as we do for any area that encloses |z| = 1 since there are no singularities outside of the unit circle. So let's integrate over the circle |z| = R, where R is very large. In fact let's go for broke and take the limit as R goes to infinity:
    \int _{|z| = 1} \frac{6 z^{98}}{23z^{99} - 2 z^{81} + z^4 - 7}~dz = \lim_{R \to \infty} \int _{|z| = R} \frac{6 z^{98}}{23z^{99} - 2 z^{81} + z^4 - 7}~dz

    Now, the integrand only has singularities inside the unit circle and the boundary (at R) is continuous so we may switch the limit and the integral. I'm also going to let z = R e^{i \phi} and integrate over \phi

    \int _{|z| = 1} \frac{6 z^{98}}{23z^{99} - 2 z^{81} + z^4 - 7}~dz = \int _0 ^{2 \pi} \lim_{R \to \infty} \frac{6 R^{98} e^{98 i \phi}}{23R^{99} e^{99 i \phi} - 2 R^{81} e^{81 i \phi} + R^4 e^{4i \phi} - 7}~R i e^{i \phi} d \phi

    or
    \int _{|z| = 1} \frac{6 z^{98}}{23z^{99} - 2 z^{81} + z^4 - 7}~dz = (i) \int _0 ^{2 \pi} \lim_{R \to \infty} \frac{6 R^{99} e^{99 i \phi}}{23R^{99} e^{99 i \phi} - 2 R^{81} e^{81 i \phi} + R^4 e^{4i \phi} - 7}} d \phi

    It looks awful, but upon taking the limit we see that only the first term in the denominator survives, so putting the "intermediate step" in for clarity:
    \int _{|z| = 1} \frac{6 z^{98}}{23z^{99} - 2 z^{81} + z^4 - 7}~dz = (i) \int _0 ^{2 \pi} \lim_{R \to \infty} \frac{6 R^{99} e^{99 i \phi}}{23R^{99} e^{99 i \phi}} d \phi = (2 \pi i) \frac{6}{23}

    -Dan
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