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Math Help - what type of singularity f has and what is the residue of f

  1. #1
    Junior Member sorv1986's Avatar
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    Cool what type of singularity f has and what is the residue of f

    Suppose f(z)=sin(1/cos(1/z))be a complex function.

    Then what type of singularity f has at z=0 and what will be the residue of f at z=0?

    thanks. regards
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  2. #2
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    Re: what type of singularity f has and what is the residue of f

    Quote Originally Posted by sorv1986 View Post
    Suppose f(z)=sin(1/cos(1/z))be a complex function.

    Then what type of singularity f has at z=0 and what will be the residue of f at z=0?

    thanks. regards
    Where did you get this question from? It seems like it isn't even "well-posed": f has a bunch of essential singularities at the points \left( \pm \frac{\pi}{2} \pm k\pi\right)^{-1} for all k\in \mathbb{N}, and these singularities accumulate at 0, so for starters 0 is not even an isolated singularity, and every closed curve that contains the origin in its interior will contain an infinite number of other singularities.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: what type of singularity f has and what is the residue of f

    The function has in z=0 an essential singularity. Setting \frac{1}{z}=s the function becomes...

    f(s)=\sin \frac{1}{\cos s} (1)

    ... and the residue in z=0 is the term in s of its Taylor expansion, that is...

    \lim_{s \rightarrow 0} f'(s) (2)

    The computation of the derivative is left to You and the result is that the limit (2) is zero...

    Kind regards

    \chi \sigma
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  4. #4
    Junior Member sorv1986's Avatar
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    Re: what type of singularity f has and what is the residue of f

    Quote Originally Posted by chisigma View Post
    The function has in z=0 an essential singularity. Setting \frac{1}{z}=s the function becomes...

    f(s)=\sin \frac{1}{\cos s} (1)

    ... and the residue in z=0 is the term in s of its Taylor expansion, that is...

    \lim_{s \rightarrow 0} f'(s) (2)

    The computation of the derivative is left to You and the result is that the limit (2) is zero...

    Kind regards

    \chi \sigma
    i am a little bit confused. Why r u calculating the limit (2). Besides this, if s-> 0, then z->∞. You are calculating the residue of f(z) at z=∞ if i am right.How did you get z=0 is an essential singularity? sorry if i misunderstood. Thanks for kind reply.
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  5. #5
    Junior Member sorv1986's Avatar
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    Re: what type of singularity f has and what is the residue of f

    Quote Originally Posted by Jose27 View Post
    Where did you get this question from? It seems like it isn't even "well-posed": f has a bunch of essential singularities at the points \left( \pm \frac{\pi}{2} \pm k\pi\right)^{-1} for all k\in \mathbb{N}, and these singularities accumulate at 0, so for starters 0 is not even an isolated singularity, and every closed curve that contains the origin in its interior will contain an infinite number of other singularities.
    i got it from an exam paper. thanks for your kind reply.appreciated .
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: what type of singularity f has and what is the residue of f

    Originally Posted by sorv1986...

    ... I am a little bit confused. Why r u calculating the limit (2). Besides this, if s-> 0, then z->∞. You are calculating the residue of f(z) at z=∞ if i am right.How did you get z=0 is an essential singularity?... sorry if i misunderstood...

    I do hope that the following example will clarify the procedure to obtain an easy solution of Your question. Let's consider the function...

    e^{\frac{1}{z}}= 1 + \frac{1}{z} + \frac{1}{2\ z^{2}} + ... + \frac{1}{n!\ z^{n}} + ... (1)

    The (1) is the Laurent expansion of the function e^{\frac{1}{z}} around z=0 and, because the infinite number of term in z^{-n}, it has in z=0 an essential singularity. By definition the residue in z=0 is the coefficient of the tem in z^{-1} of the Laurent expansion and in this case in (1) You see that is r=1. Setting z^{-1}=s the (1) becomes...

    e^{s}= 1 + s + \frac{s^{2}}{2} + ... + \frac{s^{n}}{n!}} + ... (2)

    Now the (2) is the Taylor expansion of the analytic function e^{s} around s=0 and, of course, the coefficient of the term in z^{-1} in (1) and the coefficient of the term in s in (2) are the same. But the coefficient of the Taylor expansion of an analytic function f(s) around s=0 is f'(0) and that justifies the formula...

    r_{z=0} f(\frac{1}{z}) = \lim_{s \rightarrow 0} f^{'} (s) (3)

    Kind regards

    \chi \sigma
    Last edited by chisigma; December 11th 2011 at 02:07 PM.
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