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Math Help - [SOLVED] Poles, order of it and residue

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Poles, order of it and residue

    I must prove that the singularities of f are poles. I also must determine their order and the corresponding residue.

    f(z)=\frac{z}{\cos z}.

    My attempt :
    I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
    So I wrote \cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}.
    So if I can find the order of the zeros of \frac{1}{f(z)} I'm done.
    I've found out that \frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...
    The first term of the series really bothers me. If it wasn't there then \frac{1}{f(z)} would be equal to z\cdot h(z) with h(0) \neq 0 and so z=0 would be a 0 of order 1 of \frac{1}{f}, thus a pole of order 1 of f. But of course I should take in count the \frac{1}{z} term and look for z=k \frac{\pi}{2}, k \in \mathbb{Z}, but I'm stuck.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I must prove that the singularities of f are poles. I also must determine their order and the corresponding residue.

    f(z)=\frac{z}{\cos z}.

    My attempt :
    I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
    So I wrote \cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}.
    So if I can find the order of the zeros of \frac{1}{f(z)} I'm done.
    I've found out that \frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...
    The first term of the series really bothers me. If it wasn't there then \frac{1}{f(z)} would be equal to z\cdot h(z) with h(0) \neq 0 and so z=0 would be a 0 of order 1 of \frac{1}{f}, thus a pole of order 1 of f. But of course I should take in count the \frac{1}{z} term and look for z=k \frac{\pi}{2}, k \in \mathbb{Z}, but I'm stuck.
    f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

    The singularities are z = z_m = \frac{(2m+1)\pi}{2} where m is an integer.

    So you need to show that \lim_{z \rightarrow z_m} (z - z_m)^n f(z) is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

    eg. see - Wolfram|Alpha
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by mr fantastic View Post
    f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

    The singularities are z = z_m = \frac{(2m+1)\pi}{2} where m is an integer.

    So you need to show that \lim_{z \rightarrow z_m} (z - z_m)^n f(z) is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

    eg. see - Wolfram|Alpha
    I wasn't able to solve the problem using your tip.
    Here's what I've done :
    Considering the singularity at z=\frac{\pi}{2}, the series expansion of \cos z around this singularity (singularity for f) is 1+  \left (  z-   \frac{\pi}{2}  \right )^2 h(z) where h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-... so h\left ( \frac{\pi}{2} \right ) \neq 0.
    --------------------------------------------------------------------
    I've a side-note question : From this can I conclude that the order of the pole of r(z)=\frac{1}{\cos z} when z=\frac{\pi}{2} is 2?
    --------------------------------------------------------------------
    Now I wrote that the pole of f occurs when \frac{1}{f} has a zero. That is, when \frac{1}{z}\left [ 1+  \left (  z-   \frac{\pi}{2}  \right )^2 h(z) \right ]=0. I go nowhere.

    No let me try again you way : \lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left (  z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+  \left (  z-   \frac{\pi}{2}  \right )^2 h(z) \right ].
    For n=1, the limit is worth 0. Does that mean the order of the pole of f at z=\frac{\pi}{2} is 1, and that I've successfully solved the exercise?
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  4. #4
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    Quote Originally Posted by arbolis View Post
    I wasn't able to solve the problem using your tip.
    Here's what I've done :
    Considering the singularity at z=\frac{\pi}{2}, the series expansion of \cos z around this singularity (singularity for f) is 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) where h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-... so h\left ( \frac{\pi}{2} \right ) \neq 0.
    --------------------------------------------------------------------
    I've a side-note question : From this can I conclude that the order of the pole of r(z)=\frac{1}{\cos z} when z=\frac{\pi}{2} is 2?
    --------------------------------------------------------------------
    Now I wrote that the pole of f occurs when \frac{1}{f} has a zero. That is, when \frac{1}{z}\left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]=0. I go nowhere.

    No let me try again you way : \lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left ( z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ].
    For n=1, the limit is worth 0. Does that mean the order of the pole of f at z=\frac{\pi}{2} is 1, and that I've successfully solved the exercise?
    The link I posted shows that the limit exists and is finite. Therefore there is a simple pole. You should consider the general limit \lim_{z \rightarrow z_m} (z - z_m) f(z) to show that f(z) has a simple pole at z = z_m.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Ah I see, I should find that the limit is worth -\frac{\pi}{2} instead of 0.
    Thanks once again.
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