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Math Help - Rouche Theorem - Proof

  1. #1
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    Rouche Theorem - Proof

    This is a step in Rouche Theorem proof.
    f(z) and g(z) are analytic on and inside a closed path C.
    |f(z) - g(z)| < |f(z)| for all z on C

    Let F(z) = g(z)/f(z)

    I need to prove \oint _C \frac{F'(z)}{F(z)} dz = 0

    Please help - not able to follow this step atall.
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  2. #2
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    Also, C cannot pass through any zeros of f or g. Now for any function f(z) analytic in C except for poles:

    \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)

    where I(f(\gamma);0) is the winding number right? That's how many times the contour f(\gamma) goes around the origin.

    Now, by Rouche's theorem:

    |f(z)-g(z)|< |f(z)|,\quad z\in \gamma

    Since the contour C never passes through any zero of either function, we can divide that expression by |f(z)| and end up with:

    |h(\gamma)-1|<1

    That means h(\gamma) is inside the unit circle centered at one right? What then does that say about the expression:

    \int_{\gamma} \frac{h'(z)}{h(z)}dz=2\pi i I(h(\gamma);0)
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  3. #3
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    Quote Originally Posted by shawsend View Post
    Also, C cannot pass through any zeros of f or g. Now for any function f(z) analytic in C except for poles:

    \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)

    where I(f(\gamma);0) is the winding number right? That's how many times the contour f(\gamma) goes around the origin.

    Now, by Rouche's theorem:

    |f(z)-g(z)|< |f(z)|,\quad z\in \gamma

    Since the contour C never passes through any zero of either function, we can divide that expression by |f(z)| and end up with:

    |h(\gamma)-1|<1

    That means h(\gamma) is inside the unit circle centered at one right? What then does that say about the expression:

    \int_{\gamma} \frac{h'(z)}{h(z)}dz=2\pi i I(h(\gamma);0)
    Thanks. The inequality guarantees that the zeros don't lie on C.

    Sorry but I have no clue of winding numbers etc. Isn't there a simpler explanation using maybe just Laurent Series, Cauchy Formula etc?

    I follow your logic - h(\gamma) never circles 0, hence the winding number is zero. My question would then be how do you get -

    \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    The inequality guarantees that the zeros don't lie on C.
    Ok, thanks. I see that now. Don't know how to prove it without winding numbers though.
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  5. #5
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    Thanks shawsend. I will try to read up on your first equation. I follow the steps after that.
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  6. #6
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    Quote Originally Posted by aman_cc View Post
    My question would then be how do you get -

    \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)
    Let me try. I need to understand it better too.

    So the winding number is just the number of times a contour goes around a point in the complex plane. For the contour \gamma around the origin, we can express it as:

    I(\gamma;0)=\frac{1}{2\pi i}\oint_{\gamma} \frac{dz}{z}

    and for the contour traced by f(\gamma), we could write:

    I(f(\gamma);0)=\frac{1}{2\pi i}\int_{f(\gamma)}\frac{1}{w}dw

    but if w=f(\gamma(t)) and \gamma:[a,b]\to \mathbb{C}\; then:

    I(f(\gamma);0)=\frac{1}{2\pi i}\int_{f(\gamma)}\frac{1}{w}dw=\frac{1}{2\pi i}\int_{a}^{b}\frac{f'(\gamma(t))\gamma'(t)dt}{f(\  gamma(t))}

    =\frac{1}{2\pi i}\int_{\gamma} \frac{f'(z)}{f(z)}dz
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