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Thread: residue

  1. #1
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    residue

    Find the residue of the following.

    (a) $\displaystyle \text{Res}(f, z_0=0)$, where $\displaystyle f(z)=\frac{e^z}{\sin(2z^2)}$


    (b) $\displaystyle \text{Res}(f, z_0=0)$, where $\displaystyle f(z) = \frac{\cos(\frac{1}{z})}{e^z}
    $
    We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by pascal4542 View Post
    Find the residue of the following.

    (a) $\displaystyle \text{Res}(f, z_0=0)$, where $\displaystyle f(z)=\frac{e^z}{\sin(2z^2)}$


    (b) $\displaystyle \text{Res}(f, z_0=0)$, where $\displaystyle f(z) = \frac{\cos(\frac{1}{z})}{e^z}
    $
    We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.
    If $\displaystyle z_0$ is a pole of order $\displaystyle m$, then $\displaystyle \text{Res}\!\left(f;z_0\right)=\frac{1}{m-1}\lim_{z\to z_0}\frac{\,d^{m-1}}{\,dz^{m-1}}\left[(z-z_0)^mf\!\left(z\right)\right]$.

    In the case its a simple pole, we have $\displaystyle \text{Res}\!\left(f;z_0\right)=\lim_{z\to z_0}(z-z_0)f\!\left(z\right)$.

    Are you allowed to use these?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    If $\displaystyle z_0$ is a pole of order $\displaystyle m$, then $\displaystyle \text{Res}\!\left(f;z_0\right)=\frac{1}{m-1}\lim_{z\to z_0}\frac{\,d^{m-1}}{\,dz^{m-1}}\left[(z-z_0)^mf\!\left(z\right)\right]$.

    In the case its a simple pole, we have $\displaystyle \text{Res}\!\left(f;z_0\right)=\lim_{z\to z_0}(z-z_0)f\!\left(z\right)$.

    Are you allowed to use these?
    Yes I am. I will try that. Thank you.
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  4. #4
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    Quote Originally Posted by pascal4542 View Post
    Yes I am. I will try that. Thank you.

    I was wondering if you can check my work.

    For (a) $\displaystyle \text{Res}(f, z_0=0)$, where $\displaystyle f(z)=\frac{e^z}{\sin(2z^2)}$ we have

    $\displaystyle \frac{1+z+z^2/2!+z^3/3!+\cdots}{2z^2 -(2z^2)^3/3!+(2z^2)^5/5!- \cdots}$

    $\displaystyle = \frac{1}{2z^2}+\frac{1}{2z}+1/4+z/12 $


    So the residue is $\displaystyle 1/2$.


    For (b) $\displaystyle \text{Res}(f, z_0=0)$, where $\displaystyle f(z) = \frac{\cos(\frac{1}{z})}{e^z}$ we have


    $\displaystyle \cos(\frac{1}{z}) \cdot (1-z+z^2/2-z^3/6+z^4/24+\cdots).$


    I am stuck. I don't know how to get the residue. I know that $\displaystyle z_0$ is an essential singularity. I need help on part (b).
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  5. #5
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    You have a Taylor series multiplied by a non-terminating singular series so the residue is an infinite sum due to the term-wise multiplication of $\displaystyle z^n$ by $\displaystyle z^{-n}$ so then:

    $\displaystyle \begin{aligned}\frac{\cos(1/z)}{e^z}&=\sum_{n=0}^{\infty}\frac{(-1)^n}{1}\frac{1}{(2n)! z^{2n}} \sum_{n=0}^{\infty}\frac{(-1)^n z^n}{n!}\\
    &=\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{(-1)^k}{(2k)!z^{2k}}\frac{(-1)^{n-k}z^{n-k}}{(n-k)!}
    \end{aligned}
    $

    and note the coefficients on the $\displaystyle 1/z$ terms are when the exponent of z is $\displaystyle n=3k-1$. Now, can you extract out all those terms and come up with the expression:

    $\displaystyle \mathop\text{Res}\limits_{z=0} \frac{\cos(1/z)}{e^z}=\sum_{k=0}^{\infty} \frac{(-1)^{3k-1}}{(2k)!(2k-1)!}$
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