Answer to Question #235405 in Complex Analysis for Fozia Sayda

Question

1/z(e^z-1) at its poles

Answer 1

$f(z)=\dfrac{1}{z(e^z-1)}$

The singularities are $z = 0$ and $z = 2nπi, n = ±1, ±2, . . . .$

The singularity at $z = 0$ is a pole of order 2 since $z = 0$ is a zero of order 2 of $z(e^z-1)$.

$\dfrac{1}{z(e^z-1)}=\dfrac{1}{z(z+z^2/2!+z^3/3!+...)}$

$=\dfrac{1}{z^2(1+z/2!+z^2/3!+...)}=\dfrac{1}{z^2}g(z), g(0)\not=0$

The singularities $z = 2nπi, n = ±1, ±2, . . . .$ are simple poles since they are simple zeros of $\dfrac{1}{z(e^z-1)}.$