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Math Help - f(z)=(z-3)sein(1/(2+z))

  1. #1
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    f(z)=(z-3)sein(1/(2+z))

    For the Laurent series for f(z)=(z-3)\sin{\frac{1}{z+2}} about z=-2.
    The convergence region is the disc with center in z=-2 and 0<radius<5, since we have singularities in z=3 and z=-2.
    Is this correct?
    The textbook unswer is that the series converges for \forall z\neq -2
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  2. #2
    Behold, the power of SARDINES!
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    The function \sin\left( \frac{1}{z+2}\right)only has only one singularity and it is at  z=-2. (z-3) is an entire function with a zero at 3.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by hurz View Post
    For the Laurent series for f(z)=(z-3)\sin{\frac{1}{z+2}} about z=-2.
    The convergence region is the disc with center in z=-2 and 0<radius<5, since we have singularities in z=3 and z=-2.
    Is this correct?
    The textbook unswer is that the series converges for \forall z\neq -2
    Setting...

    (1)

    ... You obtain...

    (2)

    ... so that is...

    (3)

    The (3) converges everywhere is and has at z=-2 an essential singularity...

    Kind regards

    \chi \sigma
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