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Thread: [SOLVED] Pole/essential singularity

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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 $\displaystyle f(z)=\frac{\sin z}{z^2(z-\pi)}$.

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

    $\displaystyle \pi$ would be an essential singularity since there are infinite terms where $\displaystyle z=\pi$ 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 $\displaystyle f$ in $\displaystyle z=0$? I could have done it for an arbitrary $\displaystyle z_0$, 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 $\displaystyle f(z)=\frac{\sin z}{z^2(z-\pi)}$.

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

    $\displaystyle \pi$ would be an essential singularity since there are infinite terms where $\displaystyle z=\pi$ 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 $\displaystyle f$ in $\displaystyle z=0$? I could have done it for an arbitrary $\displaystyle z_0$, or not?
    $\displaystyle \lim_{z \rightarrow \pi} f(z) = - \frac{1}{\pi^2}$ using l'Hopital's Rule so $\displaystyle z = \pi$ is clearly a removable singularity.

    Since $\displaystyle \lim_{z \rightarrow 0} z f(z) = - \frac{1}{\pi}$ it can be said that $\displaystyle z = 0$ is a simple pole. You should calculate the Laurent series for $\displaystyle f(z)$ in powers of z and confirm that the principle part is $\displaystyle - \frac{1}{\pi} \frac{1}{z}$.
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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 $\displaystyle f(z)$ by $\displaystyle z$? (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 $\displaystyle f(z)$ by $\displaystyle z$? (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 $\displaystyle g(z)$ that has a pole of order n at $\displaystyle z = z_0$. What must happen if $\displaystyle g(z)$ is then multiplied by $\displaystyle (z - z_0)^n$ and what does that mean about $\displaystyle \lim_{z \rightarrow z_0} (z - z_0)^n g(z)$ ....?
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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 $\displaystyle f(z)$ by $\displaystyle z$? (I understand the rest assuming that I understand the reason of the product)

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

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