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Thread: [SOLVED] Residue of f'/f with zero of order m

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    Red face [SOLVED] Residue of f'/f with zero of order m



    Happy spring break.

    I feel like I get it. But I'm not quite there... Thanks so much!
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    Behold, the power of SARDINES!
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    Quote Originally Posted by davismj View Post


    Happy spring break.

    I feel like I get it. But I'm not quite there... Thanks so much!
    Well lets use your definition of $\displaystyle f(z)=g(z)(z-z_0)^m$ so then
    $\displaystyle f'(z)=g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}$

    Now $\displaystyle \frac{f'(z)}{f(z)}=\frac{g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}}{g(z)(z-z_0)^m}=\frac{g'(z)}{g(z)}+\frac{m}{z-z_0}$
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    Quote Originally Posted by TheEmptySet View Post
    Well lets use your definition of $\displaystyle f(z)=g(z)(z-z_0)^m$ so then
    $\displaystyle f'(z)=g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}$

    Now $\displaystyle \frac{f'(z)}{f(z)}=\frac{g'(z)(z-z_0)^m+mg(z)(z-z_0)^{m-1}}{g(z)(z-z_0)^m}=\frac{g'(z)}{g(z)}+\frac{m}{z-z_0}$
    I tried that but erased it because I don't see any conclusion of the g/g' issue.

    $\displaystyle g(z) = \frac{f(z)}{(z-z_0)^m}$, taking that derivative and dividing and what not ends up a mess...
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    Behold, the power of SARDINES!
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    Quote Originally Posted by davismj View Post
    I tried that but erased it because I don't see any conclusion of the g/g' issue.

    $\displaystyle g(z) = \frac{f(z)}{(z-z_0)^m}$, taking that derivative and dividing and what not ends up a mess...
    $\displaystyle g(z) \text{ and } g'(z)$ are both analytic and $\displaystyle g(z_0) \ne0$ therefore their residue is zero.

    i.e $\displaystyle \frac{g'(z)}{g(z)}=\sum_{n=0}^{\infty}a_n(z-z_0)$
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    Quote Originally Posted by TheEmptySet View Post
    $\displaystyle g(z) \text{ and } g'(z)$ are both analytic and $\displaystyle g(z_0) \ne0$ therefore their residue is zero.

    i.e $\displaystyle \frac{g'(z)}{g(z)}=\sum_{n=0}^{\infty}a_n(z-z_0)$
    I really appreciate your help, but please forgive me; my book gave very little motivation or background on residues. I have a vague idea of what they are, I'm not really sure where we're supposed to find them, etc.

    Also, I'm not sure why $\displaystyle \frac{g'(z)}{g(z)}=\sum_{n=0}^{\infty}a_n(z-z_0)$. If anything, wouldn't it be equal to $\displaystyle \frac{1}{(z-z_0)}\sum_{n=0}^{\infty}a_n$. How does this relate to residues?
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  6. #6
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    Here's an exercise that might help to motivate "residue".

    Integrate $\displaystyle \oint (z- z_0)^n dz$ around a circle with center at $\displaystyle z_0$ and radius R where n can be any integer. Do that by writing $\displaystyle z= Re^{i\theta}+ z_0$ and integrating with respect to $\displaystyle \theta$.

    You should see that the integral is 0 for all n except n= -1, and, in that case, is $\displaystyle 2\pi i$.

    Now, suppose f(z) has a "pole of order n" at $\displaystyle z_0$. That means that f(z) is not analytic, so cannot be written as a Taylor's series about $\displaystyle z_0$ but can be written as a "Laurent" series (with negative powers up to -n).

    Basically, saying "f(z) has a pole of order n at $\displaystyle z_0$" means that $\displaystyle (z- z_0)^n f(z)$ is analytic at $\displaystyle z_0$ but not for any lower power of $\displaystyle z- z_0$. That, in turn, means that we can write $\displaystyle (z- z_0)^nf(z)$ as a Taylor's series about $\displaystyle z_0$: $\displaystyle (z- z_0)^n f(z)= a_0+ a_1(z-z_0)+ a_2(z- z_0)^2+ a_3(z- z_0)^3+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n(z- z_0)^n+ \cdot\cdot\cdot$.

    Dividing both sides by $\displaystyle (z-z_0)^n$, $\displaystyle f(z)= a_0(z- z_0)^{-n}+ a_1(z- z_0)^{-n+1}+ a_2(z- z_0)^{-n+2}+ a_3(z- z_0)^{-n+3}+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n+ \cdot\cdot\cdot$.

    Now integrate the right side term by term around a path with $\displaystyle z_0$ in its interior. As before, the integral of all powers of x, except one, will be 0. The exception is for the "-1" power- we will have:
    $\displaystyle \oint f(z)dz= a_{n-1}(2\pi i)$. That number is the "residue of f(z) at $\displaystyle z= z_0$".
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Here's an exercise that might help to motivate "residue".
    Basically, saying "f(z) has a pole of order n at $\displaystyle z_0$" means that $\displaystyle (z- z_0)^n f(z)$ is analytic at $\displaystyle z_0$ but not for any lower power of $\displaystyle z- z_0$. That, in turn, means that we can write $\displaystyle (z- z_0)^nf(z)$ as a Taylor's series about $\displaystyle z_0$: $\displaystyle (z- z_0)^n f(z)= a_0+ a_1(z-z_0)+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{n-1}+ a_n(z- z_0)^n+ \cdot\cdot\cdot$.

    Dividing both sides by $\displaystyle (z-z_0)^n$, $\displaystyle f(z)= a_0(z- z_0)^{-n}+ a_1(z- z_0)^{-n+1}+ \cdot\cdot\cdot+ a_{n-1}(z- z_0)^{-1}+ a_n+ \cdot\cdot\cdot$.
    Trying to fix the latex. ^^
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