Maclaurin Series Representation for $f(z)=\frac{z}{z^4+9}$ I need help finding the Maclaurin series representation for $$f(z)=\frac{z}{z^4+9}$$ 
I first tried to factorize $z^4+9$, but am I missing something? I could not figure out how to factorize this. Is there another approach to this? I am open to any approach. Just help me figure this out through factorizing if possible.

Hint 1:
$$f(z)=\frac{z}{z^4+9}=\frac{1}{1+\left(\frac{z^4}{9}\right)}\cdot \frac{z}{9}$$
Hint 2:
$$\frac{1}{1-z}=\sum_{n=0}^\infty{z^n}$$

Using both hints,
$$\frac{1}{1+\left(\frac{z^4}{9}\right)}=\sum_{n=0}^\infty{\frac{(-1)^n}{3^{2n}}}z^{4n}\implies f(z)=\sum_{n=0}^\infty \frac{(-1)^n}{3^{2n+2}}z^{4n+1}\,\,\, ,|z| < \sqrt{3}$$
