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Math Help - The order of the poles of cos(z)/z^2 and sin(z)/z^2

  1. #1
    CSM
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    The order of the poles of cos(z)/z^2 and sin(z)/z^2

    g(z)=\frac{sin(z)}{z^2}and f(z)=\frac{cos(z)}{z^2} both have a pole at z=0

    From the Laurent series expansion I deduced that for f(z) this pole is of order one, while for g(z) the pole is order two.

    Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: The order of the poles of cos(z)/z^2 and sin(z)/z^2

    Quote Originally Posted by CSM View Post
    Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)

    The series expansion of \sin z and \cos z around z=0 are well known, so we find immediately the order of the pole: \sin z/z^2=1/z+\ldots (simple) and \cos z/z^2=1/z^2+\ldots (double) .
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    MHF Contributor chisigma's Avatar
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    Re: The order of the poles of cos(z)/z^2 and sin(z)/z^2

    Quote Originally Posted by CSM View Post
    g(z)=\frac{sin(z)}{z^2}and f(z)=\frac{cos(z)}{z^2} both have a pole at z=0

    From the Laurent series expansion I deduced that for f(z) this pole is of order one, while for g(z) the pole is order two.

    Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)
    If You know that f(z) has a pole in z=a but You don't know the Laurent expansion of f(z) around z=a , then the order of the pole in the value of n for which...

    \lim_{z \rightarrow a} (z-a)^{n}\ f(z)= \lambda (1)

    ... where is \lambda \ne 0. An interesting 'counterexample' is the function \ln z , for which z=0 is a 'singularity'. If we try to apply (1) in order to determine the 'order of the pole' we discover that...

    \lim_{z \rightarrow 0} z^{n}\ \ln z =  \begin{cases} -\infty &\phantom{|}\textrm{if }n=0\\ 0 &\phantom{|}\textrm{if } n>0 \end{cases} (2)

    ... so that no value on n satisfies (1). The reason is that z=0 is a singularity of the function \ln z but isn't a 'pole'...

    Kind regards

    \chi \sigma
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