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Math Help - Quick question on order of poles

  1. #1
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    Quick question on order of poles

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

    1. How do I find out if 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. f(z)=\frac{1}{zsin(z)}. Is  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 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 f(z) which has an isolated singularity at z=z_0

    1. How do I find out if 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. f(z)=\frac{1}{zsin(z)}. Is  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 z=z_0
    You don't need the whole Laurent Series, only the first non-zero term. If you know that f is analytic in U\setminus\{z_0\} and that (z-z_0)^k f(z)\to c\neq 0 when z\to z_0, then z_0 is a pole of order k for f. And the residue is c.

    In your example, you may know that \frac{\sin z}{z}\to 1 when z\to 0, hence 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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  3. #3
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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 f is analytic in U\setminus\{z_0\} and that (z-z_0)^k f(z)\to c\neq 0 when z\to z_0, then z_0 is a pole of order k for f. And the residue is c.

    In your example, you may know that \frac{\sin z}{z}\to 1 when z\to 0, hence 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 (z-z_0)^k f(z) when z\to z_0
    Then 'k' is the order of the pole at  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 f is analytic in U\setminus\{z_0\} important?
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  4. #4
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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 f(z) = \frac{a_1}{z}+\frac{a_2}{z^2}+\frac{a_3}{z^3}+...+  \frac{a_k}{z^k}

    and Limit z \to 0:f(z) exists then
    1. each of {a_1},{a_2},...,{a_k} = 0
    2. Above is true in both cases:
    a) k is finite
    b) 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 (z-z_0)^k f(z) when z\to z_0
    Then 'k' is the order of the pole at  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 f is analytic in 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 f is analytic in U\setminus\{z_0\} and that (z-z_0)^k f(z)\to c\neq 0 when z\to z_0, then z_0 is a pole of order k for f. And the residue is c.
    c then is the coefficient on the \frac{1}{(z-z_0)^k} term and the residue is:

    \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 (z-z_0)^k f(z) when z\to z_0
    Then 'k' is the order of the pole at  z_0. Have I got it correct?
    You don't really have to check every k one by one: if 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, k is the right order. In your example, I could have written " \sin z\sim_{z\to 0} z, hence \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 f is analytic in 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 z in a small disc centered at z_0, f(z)=\sum_{n=-k}^\infty a_n (z-z_0)^n and a_{-k}\neq 0". This definition concerns f not only "at z_0" but also on a neighbourhood of f. And it implies that f is analytic on a disc around z_0.

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