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Math Help - Contour Integral proof

  1. #1
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    Contour Integral proof

    The problem:

    Let P(z) be a polynomial of degree greater than 1, with all of it's zeroes inside the circle C_{R}=\{z:|z|=R\}. show that \int_{C_{R}} \frac{dz}{P(z)} = 0.

    My idea for a proof:

    Since every polynomial over C can be writen as P(z)=(z-a_1)(z-a_2)...(z-a_k), where a_1,...a_k are all the zeroes of P(z) (not necessarily unique). Than we can write \int_{C_{R}} \frac{dz}{P(z)} = \int_{C_{R}} \frac{b_1}{z-a_1} dz + ... + \int_{C_R}} \frac{b_k}{z-a_k} dz.

    Now we can say that for every a_i we can find a circle C_{a_{i}} such that C_{a_{i}}\subset C_{R}, a_i\notin C_{a_{i}}
    and therfore \frac {1}{z-a_i} is analytic in C_{a_i}} so we get \int_{C_{a_{i}}} \frac{dz}{z-a_i} =0, but since C_{a_{i}}\subset C_{R} we have \int_{C_{R}} f = \int_{C_{a_i}} f = 0 and that shows what we needed to show.

    I would appreciate some thoughts on this proof as I am not 100% sure it is valid.
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  2. #2
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    consider the coordinate change z \mapsto \frac{R^2}{z}, we have
    \oint_{C_R} \frac{dz}{P(z)}= R^2 \oint_{C_R} \frac{dz}{z^2P(R^2/z)}
    If P(z)=a_n\Pi_{k=1}^n(z-z_k)
    P(R^2/z)=a_n\Pi_{k=1}^n(R^2/z-z_k)=\frac{a_n}{z^n}\Pi_{k=1}^n(R^2-z_k z)

    Obviously R^2/z_k, k = 1,2, ..., n are all the roots of P(R^2/z), and |R^2/z_k|>R for any k. That is all the roots of P(R^2/z) are outside of the circle C_R,

    So R^2 \oint_{C_R} \frac{dz}{z^2P(R^2/z)}
    = R^2 \oint_{C_R} \frac{z^{n-2}dz}{a_n\Pi_{k=1}^n(R^2-z_k z)}

    Since n>=2 the integrand is analytic inside the circle. DONE.
    Last edited by xxp9; February 3rd 2011 at 07:45 PM.
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  3. #3
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    thanks for the reply, I found a hole in my proof as well so your direction is appreciated.
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  4. #4
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    Thinking about this some more, what happens if the root is zero?
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  5. #5
    Senior Member roninpro's Avatar
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    Here is an alternative:

    Let P(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_1z+z_0. Since all of the roots are inside the circle C_R, if R'>R, then

    \displaystyle \oint_{C_R} \frac{1}{P(z)}\text{ d}z=\oint_{C_{R'}} \frac{1}{P(z)}\text{ d}z

    by Cauchy's integral theorem. (In other words, expanding the circle does not change the value of the integral.) Note that when |z| is sufficiently large, then |a_n/2||z|^n\leq |P(z)|. Then, when R is large,

    \displaystyle \left|\oint_{C_R} \frac{1}{P(z)}\text{ d}z\right|\leq \oint_{C_R} \left|\frac{1}{P(z)}\right| |\text{d}z|\leq \oint_{C_R} \frac{1}{|a_n/2||z|^n}|\text{d}z|=\frac{2\pi R}{|a_n/2|R^n}=\frac{2\pi}{|a_n/2|R^{n-1}}

    (Keep in mind that since n\geq 2, we have n-1\geq 1.) Taking R\to \infty, the right hand side goes to zero, so the integral must be zero.
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  6. #6
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    Quote Originally Posted by roninpro View Post
    Here is an alternative:

    if R'>R, then

    \displaystyle \oint_{C_R} \frac{1}{P(z)}\text{ d}z=\oint_{C_{R'}} \frac{1}{P(z)}\text{ d}z

    by Cauchy's integral theorem. (In other words, expanding the circle does not change the value of the integral.) Note that when |z| is sufficiently large, then |a_n/2||z|^n\leq |P(z)|
    Thanks for the reply.

    One question. Dosent \frac{1}{P(z)} needs to be analytic inside the circle to use Cauchys integral theorem. since the roots of P(z) are inside the circle it clearly isn't. Am I missing something here?
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  7. #7
    Senior Member roninpro's Avatar
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    Quote Originally Posted by skyking View Post
    Thanks for the reply.

    One question. Dosent \frac{1}{P(z)} needs to be analytic inside the circle to use Cauchys integral theorem. since the roots of P(z) are inside the circle it clearly isn't. Am I missing something here?
    Another way to state my claim is that the value of the contour integral is zero in the region of the expansion (so that expanding the circle does not affect the original integral value). Since P(z) has no roots in that region, 1/P(z) is analytic there, so it is okay to use Cauchy's integral theorem.
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