I see that $ z / (e^z - 1) $ has a singularity at $ z = 0 $.
Is $ z / (e^z - 1) $ analytic everywhere in $ |z| < \pi $? If yes, how is it determined to be analytic everywhere despite a singularity existing at $ z = 0 $?
I want to know how to show $z/(e^z - 1)$ is analytic or not starting from the definition of analytic function.
$f(z) = z/(e^z-1)$ is holomorphic in $0 < |z| < \pi$ as the quotient of two holomorphic functions with a non-zero denominator. $f$ has an “isolated singularity” at $z=0$.
This singularity is removable because $\lim_{z \to 0} f(z) = 1$ exists, see Riemann's theorem on removable singularities. This means that the function $$ F(z) = \begin{cases} 1 & \text{ if } z = 0 \\ f(z) & \text{ if } 0 < |z| < 1 \end{cases} $$ is holomorphic in $|z| < \pi$. $F$ is called the “holomorphic extension of $f$ over $a$.”
In a slight abuse of language, the extension is sometimes also denoted as $f$. In that sense, $f$ is analytic in the disk.