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QUESTION


Show that $f(z)$ has a pole of order $m$ at $z_0$ if and only if
$1\over f(z)$ has a zero of order $m$ at $z_0$. Thus find the
order of the pole of $1/z(e^z-1)$ at $z=0$.



ANSWER


$f(z)$ has a pole of order $m$ at $z_o$ if and only if near $z_0$
we have $f(z)=(z-z_0)^{-m}g(z)$ where $g$ is analytic at $z_0$ and
$g(z_0)\not=0$. Thus ${1\over f(z)}=(z-z_0)^m/g(z)$ has a zero of
order $m$ at $z=z_0$.




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