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Math Help - 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:

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

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

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

    Multiply it out, take derivatives, equate coefficients and solve for 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; June 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:


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

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

    then using the product rule

     = \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:

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

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

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

    Multiply it out, take derivatives, equate coefficients and solve for 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 h(z) conflicts with your h(z). Mine was just general terms. You have:
    \frac{1}{f(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+a_0+\cdots
    and:
    f(z)=\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f'''(z_0)}{6}(z-z0)^3+\cdots
    and therefore
    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
    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:
    1=\frac{b_2 f''(z_0)}{2}
    and:
    0=\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}
    Solving for b_1
    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 h(z) conflicts with your h(z). Mine was just general terms. You have:
    \frac{1}{f(z)}=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+a_0+\cdots
    and:
    f(z)=\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f'''(z_0)}{6}(z-z0)^3+\cdots
    and therefore
    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
    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:
    1=\frac{b_2 f''(z_0)}{2}
    and:
    0=\frac{b_2 f'''(z_0)}{6}+\frac{b_1 f''(z_0)}{2}
    Solving for b_1
    b_1=-2/3 \frac{f'''(z_0)}{(f''(z_0))^2}
    This might be a stupid question, but how do we know that  f'(z_{0}) = 0 ?
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  7. #7
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    If f(z) has a zero of order 2 at z_0, then f(z) can be written as:

    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:

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

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

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

    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:

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

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

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