If $\displaystyle f(z) = \frac{e^{iz}}{z^2-1}$ then we can set $g(z)=e^{iz}$ and $h(z)=z^2-1$. The Maclaurin expansion for $e^{iz}$ is $$\sum\limits_{n=0}^\infty \frac{(iz)^n}{n!}$$ so $\displaystyle f(z)=\frac{1}{z^2-1} \sum\limits_{n=0}^\infty \frac{(iz)^n}{n!}$.
The geometric series for $-\frac{1}{1-z^2}$ is $$-\sum\limits_{n=0}^\infty z^{2n}$$
So we have the quotient $$-\frac{\sum\limits_{n=0}^\infty \frac{(iz)^n}{n!}}{\sum\limits_{n=0}^\infty z^{2n}}$$
Am I on the right track? I'd appreciate a hint on what to do next.
You get $\left(-\sum\limits_{n=0}^\infty \frac{(iz)^n}{n!}\right)\cdot\left(\sum\limits_{n=0}^\infty z^{2n}\right)$, not $-\frac{\sum\limits_{n=0}^\infty \frac{(iz)^n}{n!}}{\sum\limits_{n=0}^\infty z^{2n}}$. You can now just multiply this out term by term.