Results 1 to 3 of 3

Thread: f(z)=(z-3)sein(1/(2+z))

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    45

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

    For the Laurent series for $\displaystyle f(z)=(z-3)\sin{\frac{1}{z+2}}$ about $\displaystyle z=-2$.
    The convergence region is the disc with center in z=-2 and $\displaystyle 0<radius<5$, since we have singularities in z=3 and z=-2.
    Is this correct?
    The textbook unswer is that the series converges for $\displaystyle \forall z\neq -2$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    The function $\displaystyle \sin\left( \frac{1}{z+2}\right)$only has only one singularity and it is at $\displaystyle z=-2$. $\displaystyle (z-3)$ is an entire function with a zero at 3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    Quote Originally Posted by hurz View Post
    For the Laurent series for $\displaystyle f(z)=(z-3)\sin{\frac{1}{z+2}}$ about $\displaystyle z=-2$.
    The convergence region is the disc with center in z=-2 and $\displaystyle 0<radius<5$, since we have singularities in z=3 and z=-2.
    Is this correct?
    The textbook unswer is that the series converges for $\displaystyle \forall z\neq -2$
    Setting...

    (1)

    ... You obtain...

    (2)

    ... so that is...

    (3)

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum