<s>[INST] 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.
 [/INST] The geometric series $1-x+x^2-x^3+\cdots$ converges to $\frac{1}{x+1}$ for all $|x|<1$.
Substituting $x=\frac{y^4}{9}$ tells us that
$$1-\frac{1}{9}\,y^4 + \frac{1}{9^2} \, y^8 - \frac{1}{9^3}\, y^{12} + \cdots = \frac{1}{\frac{y^4}{9}+1} \equiv \frac{9}{y^4+9}$$
for all $\left|\frac{y^4}{9}\right|<1$, i.e. for all $|y| < \sqrt{3}$.
To get the power series we need, we just divide by $9$ and multiply by $y$:
$$\frac{1}{9}\,y-\frac{1}{9^2}\,y^5 + \frac{1}{9^3} \, y^9 - \frac{1}{9^4}\, y^{13} + \cdots = \frac{y}{y^4+9}$$
for all $|y| < \sqrt{3}$.
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