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Thread: Quick question on order of poles

  1. #1
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    Quick question on order of poles

    For a complex function $\displaystyle f(z)$ which has an isolated singularity at $\displaystyle z=z_0$

    1. How do I find out if $\displaystyle z=z_0$ is a pole or not?
    2. If it is a pole what is the order of the pole

    Consider for e.g. $\displaystyle f(z)=\frac{1}{zsin(z)}$. Is$\displaystyle z=0 $pole? If yes what is the order of the pole?

    Is there a way to find answer to the above question without really writing down the Laurent Series about $\displaystyle z=z_0$

    A related question I would have is - do we have similar method to comment on essential and removable singularities?

    Thanks
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    For a complex function $\displaystyle f(z)$ which has an isolated singularity at $\displaystyle z=z_0$

    1. How do I find out if $\displaystyle z=z_0$ is a pole or not?
    2. If it is a pole what is the order of the pole

    Consider for e.g. $\displaystyle f(z)=\frac{1}{zsin(z)}$. Is$\displaystyle z=0 $pole? If yes what is the order of the pole?

    Is there a way to find answer to the above question without really writing down the Laurent Series about $\displaystyle z=z_0$
    You don't need the whole Laurent Series, only the first non-zero term. If you know that $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ and that $\displaystyle (z-z_0)^k f(z)\to c\neq 0$ when $\displaystyle z\to z_0$, then $\displaystyle z_0$ is a pole of order $\displaystyle k$ for $\displaystyle f$. And the residue is $\displaystyle c$.

    In your example, you may know that $\displaystyle \frac{\sin z}{z}\to 1$ when $\displaystyle z\to 0$, hence $\displaystyle z^2\frac{1}{z\sin z}\to_{z\to 0} 1$, hence 0 is a pole of order 2, with residue 1.
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    Quote Originally Posted by Laurent View Post
    You don't need the whole Laurent Series, only the first non-zero term. If you know that $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ and that $\displaystyle (z-z_0)^k f(z)\to c\neq 0$ when $\displaystyle z\to z_0$, then $\displaystyle z_0$ is a pole of order $\displaystyle k$ for $\displaystyle f$. And the residue is $\displaystyle c$.

    In your example, you may know that $\displaystyle \frac{\sin z}{z}\to 1$ when $\displaystyle z\to 0$, hence $\displaystyle z^2\frac{1}{z\sin z}\to_{z\to 0} 1$, hence 0 is a pole of order 2, with residue 1.
    Let me just re-phrase (so that I get it)
    1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $\displaystyle (z-z_0)^k f(z)$ when $\displaystyle z\to z_0$
    Then 'k' is the order of the pole at$\displaystyle z_0$. Have I got it correct?

    Few questions (I tried to work out a logic for this)
    1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
    2. Why is the assumption $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ important?
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    Thinking about this I got a related question (which might help me understand why the limit method to find the order works)

    Is is correct to say

    if $\displaystyle f(z) = \frac{a_1}{z}+\frac{a_2}{z^2}+\frac{a_3}{z^3}+...+ \frac{a_k}{z^k}$

    and Limit $\displaystyle z \to 0:f(z)$ exists then
    1. each of $\displaystyle {a_1},{a_2},...,{a_k} = 0$
    2. Above is true in both cases:
    a) $\displaystyle k$ is finite
    b) $\displaystyle k$ in infinite


    Quote Originally Posted by aman_cc View Post
    Let me just re-phrase (so that I get it)
    1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $\displaystyle (z-z_0)^k f(z)$ when $\displaystyle z\to z_0$
    Then 'k' is the order of the pole at$\displaystyle z_0$. Have I got it correct?

    Few questions (I tried to work out a logic for this)
    1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
    2. Why is the assumption $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ important?
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  5. #5
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    Quote Originally Posted by Laurent View Post
    You don't need the whole Laurent Series, only the first non-zero term. If you know that $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ and that $\displaystyle (z-z_0)^k f(z)\to c\neq 0$ when $\displaystyle z\to z_0$, then $\displaystyle z_0$ is a pole of order $\displaystyle k$ for $\displaystyle f$. And the residue is $\displaystyle c$.
    $\displaystyle c$ then is the coefficient on the $\displaystyle \frac{1}{(z-z_0)^k}$ term and the residue is:

    $\displaystyle \lim_{z\to z_0} \frac{1}{(k-1)!} \frac{d^{k-1}}{dz} (z-z_0)^k f(z)$
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  6. #6
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    Quote Originally Posted by aman_cc View Post
    Let me just re-phrase (so that I get it)
    1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $\displaystyle (z-z_0)^k f(z)$ when $\displaystyle z\to z_0$
    Then 'k' is the order of the pole at$\displaystyle z_0$. Have I got it correct?
    You don't really have to check every $\displaystyle k$ one by one: if $\displaystyle k$ is more than the order, the limit is 0, and if it is less than the order, the limit doesn't exist. Therefore, if the limit is finite and non-zero, $\displaystyle k$ is the right order. In your example, I could have written "$\displaystyle \sin z\sim_{z\to 0} z$, hence $\displaystyle \frac{1}{z\sin z}\sim\frac{1}{z^2}$ hence the order is 2. "

    Few questions (I tried to work out a logic for this)
    1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
    2. Why is the assumption $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ important?
    Your questions are related to each other. With your definition of a pole (using Laurent series), you have: "for all $\displaystyle z$ in a small disc centered at $\displaystyle z_0$, $\displaystyle f(z)=\sum_{n=-k}^\infty a_n (z-z_0)^n$ and $\displaystyle a_{-k}\neq 0$". This definition concerns $\displaystyle f$ not only "at $\displaystyle z_0$" but also on a neighbourhood of $\displaystyle f$. And it implies that $\displaystyle f$ is analytic on a disc around $\displaystyle z_0$.

    The condition $\displaystyle (z-z_0)^k f(z)\to c\neq 0$, on the other hand, does not say anything about the regularity of $\displaystyle f$ outside $\displaystyle z_0$. If however you know (in addition) that $\displaystyle f$ is analytic on a subset (like a disc) around $\displaystyle z_0$ (but deprived of $\displaystyle z_0$), then you can see that the function $\displaystyle g:z\mapsto (z-z_0)^k f(z)$ is analytic on the same subset, and has a finite limit $\displaystyle c$ at $\displaystyle z_0$, so that $\displaystyle z_0$ is a removable singularity of $\displaystyle g$ (this is a property you probably know). In other words, $\displaystyle g$ (with $\displaystyle g(z_0)=c$) is analytic in the previous subset plus the point $\displaystyle z_0$. Hence $\displaystyle g(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$ for some sequence $\displaystyle a_n$ with $\displaystyle a_0=g(z_0)=c\neq 0$. And from the relation $\displaystyle g(z)=(z-z_0)^k f(z)$, one gets $\displaystyle f(z)=\sum_{n=0}^\infty a_n (z-z_0)^{n-k}$ $\displaystyle =\sum_{n=-k}^\infty a_{n+k} (z-z_0)^n$. Thus we find a Laurent series and again the initial definition of pole and order. This justifies my previous post.
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