show that function $f(z)=(z^2+3z)/(e^z-1)$ can be expressed as a power series Good day, 
I am looking for a help.
I should show that function $f(z)=(z^2+3z)/(e^z-1)$ can be expressed as a power series in some punctured disk around the origin. i.e. for some $r>0$ one has
$f(z)=\sum_{0}^{\infty}a_{k}z^{k}$ for $0<|z|<r$.
Thank you a lot for any help.
 A: This can always be done as long as the function is analytic in some punctured disc about the origin.  In this case the singularities occur when $e^z-1=0$.  So


*

*Solve this equation.

*Plot the solutions on a diagram (not essential but helpful).

*Find a disc about the origin which contains none of the solutions from step (1), except for the solution $z=0$.


Good luck!
A: Here is a proof that works for the current example:
Note that $\lim_{z \to 0} f(z) = 3$. Define $g(z) = f(z)$ for $z \neq 0$ close to zero and $g(0) = 3$. It is clear that $g$ is analytic for $z \neq 0$ close to zero
and only zero is in question. 
All we need to do is show that $g$ is differentiable at $z=0$.
We have ${g(z)-g(0) \over z} = { {z^2 + 3z \over e^z -1}  - 3 \over z} = {z^2 + 3z -3 e^z +3\over z(e^z -1)} = {z^2-3 {z^2 \over 2!}-3{z^3 \over 3!}-\cdots  \over z^2+{z^3 \over 2!} + \cdots} $, and we see that 
$\lim_{z \to 0} {g(z)-g(0) \over z}  =-{1 \over 2}$, hence
$g$ is differentiable at $z=0$ and $g'(0) = -{1 \over 2}$.
It follows that $g$ is analytic in a neighbourhood of $z=0$ hence it
has a power series expansion. Since $f(z) = g(z)$ for $z \neq 0$, it
follows that the same series is valid for $f$ in the corresponding
punctured neighbourhood.
