Expand $\frac{z}{z^4+9}$ To Taylor Series 
expand $$\frac{z}{z^4+9}$$ to taylor series 

$$\frac{z}{z^4+9}=\frac{z}{9}\frac{1}{1--\frac{z^4}{9}}$$
Can we write $$\frac{z}{9}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z^4}{9}\right)^n=\sum_{n=0}^{\infty}(-1)^n\frac{z^{4n+1}}{9^{n+1}}$$?

Yes, your solution is a good solution.
In fact, you are expanding around $0,$ but one can choose different points. Also, note that the radius of convergence is $\sqrt{3}.$
