May I ask for a little help in solving a problem about Bernoulli number generating function?
Bernoulli number generating function is given by: $$f(z):=\begin{cases} \frac{z}{e^{z}-1} & z \in D_{2\pi}(0)\\ 1& z=0 \end{cases}$$
I have to show that $$f(z)+\frac{z}{2}=\frac{z}{2i} \cot\left( \frac{z}{2i} \right).$$
Good, I have shown that this function is even and it holds $$f(z)+\frac{z}{2}=\left ( \frac{z}{2} \right )\frac{e^{z}+1}{e^{z}-1}=\left(\frac{z}{2}\right )\frac{{e^\frac{z}{2}}+e^{\frac{-z}{2}} }{e^{\frac{z}{2}}-e^{\frac{-z}{2}}},$$ but what else I can do, I don't know...
Thank you in advance!