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Thread: Complex Analysis Residue question

  1. #1
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    Complex Analysis Residue question

    if f has a zero of order 2 at z0 and A=f''(z0) and B=f'''(z0) and h(z)=1/f(z), how does one find a formula for Res(h;z0) in terms of A and B?
    thanks!
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  2. #2
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    Quote Originally Posted by morganfor View Post
    if f has a zero of order 2 at z0 and A=f''(z0) and B=f'''(z0) and h(z)=1/f(z), how does one find a formula for Res(h;z0) in terms of A and B?
    thanks!
    Just expand them in terms of Laurent series:

    $\displaystyle \frac{g(z)}{h(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{(z-z_0)}+\cdots$

    $\displaystyle h(z)=\frac{h''(z_0)}{2}(z-z_0)^2+\frac{h'''(z_0)}{6}(z-z_0)^2$

    Therefore $\displaystyle g(z)=h(z)\left[\frac{b_2}{(z-z_0)^2}+\cdots \right]$

    Multiply it out, take derivatives, equate coefficients and solve for $\displaystyle b_1$

    I'd recommend you find "Basic Complex Analysis" by Marsden and Hoffman. It's a nice read on basic stuff and it goes over this derivation.
    Last edited by mr fantastic; Jun 4th 2009 at 08:26 PM. Reason: Fixed a couple of subscripts
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  3. #3
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    You could try the following:


    $\displaystyle Res[h,z_{0}] = \lim_{z \to z_{0}} \frac {d}{dz} (z-z_{0})^{2} h(z) $

    $\displaystyle = \lim_{z \to z_{0}} \frac {d}{dz} (z-z_{0})^{2} \frac {1}{f(z)} $

    then using the product rule

    $\displaystyle = \lim_{z \to z_{0}} \Big(2(z-z_{0}) \frac {1}{f(z)} + (z-z_{0})^{2} \frac {-1}{[f(z)]^2} f'(z) \Big) $

    And then since the forms of both terms are indeterminate, apply L'Hospital's rule. But multiple applications will be required.
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  4. #4
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    Thank you!
    Shawsend, I'm just confused about where the g(z) came from and what it is - do you mean f(z)?
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  5. #5
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    Quote Originally Posted by shawsend View Post
    Just expand them in terms of Laurent series:

    $\displaystyle \frac{g(z)}{h(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{(z-z_0)}+\cdots$

    $\displaystyle h(z)=\frac{h''(z_0)}{2}(z-z_0)^2+\frac{h'''(z_0)}{6}(z-z_0)^2$

    Therefore $\displaystyle g(z)=h(z)\left[\frac{b_2}{(z-z_0)^2}+\cdots \right]$

    Multiply it out, take derivatives, equate coefficients and solve for $\displaystyle b_1$

    I'd recommend you find "Basic Complex Analysis" by Marsden and Hoffman. It's a nice read on basic stuff and it goes over this derivation.
    I'm sorry my $\displaystyle h(z)$ conflicts with your $\displaystyle h(z)$. Mine was just general terms. You have:
    $\displaystyle \frac{1}{f(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+a_0+\cdots$
    and:
    $\displaystyle f(z)=\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f'''(z_0)}{6}(z-z0)^3+\cdots$
    and therefore
    $\displaystyle 1=\left[\frac{f''(z_0)}{2}+\frac{f'''(z_0)}{6}(z-z_0)+\cdots\right]\cdot\left[b_2+b_1(z-z_0)+a_1(z-z_0)^2+\cdots\right]$
    so that
    $\displaystyle 1=\frac{f''(z_0}{2} b_2+\left(\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}\right)(z-z_0)+\cdots$
    Equating coefficients:
    $\displaystyle 1=\frac{b_2 f''(z_0)}{2}$
    and:
    $\displaystyle 0=\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}$
    Solving for $\displaystyle b_1$
    $\displaystyle b_1=-2/3 \frac{f'''(z_0)}{(f''(z_0))^2}$
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  6. #6
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    Quote Originally Posted by shawsend View Post
    I'm sorry my $\displaystyle h(z)$ conflicts with your $\displaystyle h(z)$. Mine was just general terms. You have:
    $\displaystyle \frac{1}{f(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+a_0+\cdots$
    and:
    $\displaystyle f(z)=\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f'''(z_0)}{6}(z-z0)^3+\cdots$
    and therefore
    $\displaystyle 1=\left[\frac{f''(z_0)}{2}+\frac{f'''(z_0)}{6}(z-z_0)+\cdots\right]\cdot\left[b_2+b_1(z-z_0)+a_1(z-z_0)^2+\cdots\right]$
    so that
    $\displaystyle 1=\frac{f''(z_0}{2} b_2+\left(\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}\right)(z-z_0)+\cdots$
    Equating coefficients:
    $\displaystyle 1=\frac{b_2 f''(z_0)}{2}$
    and:
    $\displaystyle 0=\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}$
    Solving for $\displaystyle b_1$
    $\displaystyle b_1=-2/3 \frac{f'''(z_0)}{(f''(z_0))^2}$
    This might be a stupid question, but how do we know that $\displaystyle f'(z_{0}) = 0 $ ?
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  7. #7
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    If $\displaystyle f(z)$ has a zero of order 2 at $\displaystyle z_0$, then $\displaystyle f(z)$ can be written as:

    $\displaystyle f(z)=a_2(z-z_0)^2+a_3(z-z_0)^3+\cdots$

    but that's the Taylor series which is unique so that:

    $\displaystyle a_2=\frac{f''(z_0)}{2}$

    $\displaystyle a_3=\frac{f'''(z_0)}{6}$

    and so on. That means the first two terms of the Taylor series must be zero or $\displaystyle f(z_0)=0$ and $\displaystyle f'(z_0)=0$.
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  8. #8
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    Quote Originally Posted by shawsend View Post
    If $\displaystyle f(z)$ has a zero of order 2 at $\displaystyle z_0$, then $\displaystyle f(z)$ can be written as:

    $\displaystyle f(z)=a_2(z-z_0)^2+a_3(z-z_0)^3+\cdots$

    but that's the Taylor series which is unique so that:

    $\displaystyle a_2=\frac{f''(z_0)}{2}$

    $\displaystyle a_3=\frac{f'''(z_0)}{6}$

    and so on. That means the first two terms of the Taylor series must be zero or $\displaystyle f(z_0)=0$ and $\displaystyle f'(z_0)=0$.
    But basically what you're using is the fact that if $\displaystyle z_{0} $ is a zero of $\displaystyle f(z)$ of order k, then $\displaystyle f(z{o}), f'(z_{o}), ... , f^{(k-1)}(z_{0}) = 0 $, right?

    I don't why I didn't realize that before.
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