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Math Help - [SOLVED] Pole/essential singularity

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    MHF Contributor arbolis's Avatar
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    [SOLVED] Pole/essential singularity

    I must find the singularities and classify them (poles, essential or removable singularities) and find their principal part.
    The function is f(z)=\frac{\sin z}{z^2(z-\pi)}.

    My attempt :
    By a first look, I notice that the function has a pole or singularity at z=0 and z=\pi.
    Let's examinate what happens when z=0.
    The Taylor series of \sin z in z=0 is \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}.
    Thus f(z)=\frac{1}{z(z-\pi)}-\frac{z}{z-\pi}+\frac{z^3}{z-\pi}-\frac{z^5}{z-\pi}+... .
    Now I'm unsure of myself. I see that the first term is undetermined when z=0 and z=\pi while all the other terms are undetermined only when z=\pi. As a consequence I'm tempted to say that z=0 is a pole of order 1. (Isn't a removable singularity also?).

    \pi would be an essential singularity since there are infinite terms where z=\pi is not defined, but I don't know how to justify formally.

    Plus, another doubt I have is why did I find the Taylor series of f in z=0? I could have done it for an arbitrary z_0, or not?
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    Quote Originally Posted by arbolis View Post
    I must find the singularities and classify them (poles, essential or removable singularities) and find their principal part.
    The function is f(z)=\frac{\sin z}{z^2(z-\pi)}.

    My attempt :
    By a first look, I notice that the function has a pole or singularity at z=0 and z=\pi.
    Let's examinate what happens when z=0.
    The Taylor series of \sin z in z=0 is \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}.
    Thus f(z)=\frac{1}{z(z-\pi)}-\frac{z}{z-\pi}+\frac{z^3}{z-\pi}-\frac{z^5}{z-\pi}+... .
    Now I'm unsure of myself. I see that the first term is undetermined when z=0 and z=\pi while all the other terms are undetermined only when z=\pi. As a consequence I'm tempted to say that z=0 is a pole of order 1. (Isn't a removable singularity also?).

    \pi would be an essential singularity since there are infinite terms where z=\pi is not defined, but I don't know how to justify formally.

    Plus, another doubt I have is why did I find the Taylor series of f in z=0? I could have done it for an arbitrary z_0, or not?
    \lim_{z \rightarrow \pi} f(z) = - \frac{1}{\pi^2} using l'Hopital's Rule so z = \pi is clearly a removable singularity.

    Since \lim_{z \rightarrow 0} z f(z) = - \frac{1}{\pi} it can be said that z = 0 is a simple pole. You should calculate the Laurent series for f(z) in powers of z and confirm that the principle part is - \frac{1}{\pi} \frac{1}{z}.
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    MHF Contributor arbolis's Avatar
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    Thanks for the help!

    I have a question, you wrote
    Since it can be said that is a simple pole.
    , but why did you multiplied f(z) by z? (I understand the rest assuming that I understand the reason of the product)
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    Quote Originally Posted by arbolis View Post
    Thanks for the help!

    I have a question, you wrote , but why did you multiplied f(z) by z? (I understand the rest assuming that I understand the reason of the product)
    Think about what the principle part of the Laurent series is for a function g(z) that has a pole of order n at z = z_0. What must happen if g(z) is then multiplied by (z - z_0)^n and what does that mean about \lim_{z \rightarrow z_0} (z - z_0)^n g(z) ....?
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    Quote Originally Posted by arbolis View Post
    Thanks for the help!

    I have a question, you wrote , but why did you multiplied f(z) by z? (I understand the rest assuming that I understand the reason of the product)

    This is a definition: the point a is a simple pole of a function f(z) if there exists an analytic function g(z) s.t f(z)=\frac{g(z)}{z-a}\, \forall{z\neq{a}}

    Tonio
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Think about what the principle part of the Laurent series is for a function g(z) that has a pole of order n at z = z_0. What must happen if g(z) is then multiplied by (z - z_0)^n and what does that mean about \lim_{z \rightarrow z_0} (z - z_0)^n g(z) ....?
    Here is what I think : if g(z) has a pole of order n then the principal part of the Laurent series has ... n terms?
    Or has a term to the -n th power?
    So that the singularity of g(z) must disappear when multiplied by (z-z_0)^n and hence \lim_{z \rightarrow z_0} (z - z_0)^n g(z) exists and is different from \pm \infty.
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    Quote Originally Posted by arbolis View Post
    Here is what I think : if g(z) has a pole of order n then the principal part of the Laurent series has ... n terms?
    Or has a term to the -n th power?
    So that the singularity of g(z) must disappear when multiplied by (z-z_0)^n and hence \lim_{z \rightarrow z_0} (z - z_0)^n g(z) exists and is different from \pm \infty.
    Correct.
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