Expanding $f\left ( z \right )=\frac{e^{az}}{1+e^{z}}$ about $z= i\pi$

Question

$$f\left ( z \right )=\frac{e^{az}}{1+e^{z}} ,\left ( a\in\left ( 0,1 \right ) \right )$$ The point $z=i\pi$ is one of the nonremovable singularities of this function. In order to expand it about that point I introduced the change of variables $z=\omega +i\pi$ and expanded the function $f\left(\omega \right)=e^{ai\pi}\frac{e^{a\omega}}{1-e^{\omega}}$ about $\omega=0$. Here is what I've come up with. $$e^{a\omega}=1+a\omega+\frac{a^{2}\omega^{2}}{2}+\frac{a^{3}\omega^{3}}{6}+...$$ Then I tried to use this well-known expansion $$\frac{1}{1-x}= \sum_{k= 0}^{\infty}x^{n}$$ and just use $e^{\omega}$ instead of $x$, however $\left | e^{\omega} \right |$ need not be $<1$ here, so the series might not converge.