<s>[INST] 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 $$\frac{z}{9}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z^4}{9}\right)^n=\sum_{n=0}^{\infty}(-1)^n\frac{z^{4n+1}}{9^{n+1}}$$?
 [/INST] Your  expansion as  Taylor series around $0$ (i.e. as  Maclaurin series)  is  fine.  But  we also  have to state  the  validity of the series representation of $f(z)$ 
\begin{align*}
f(z)=\frac{z}{z^4+9}=\frac{z}{9}\cdot\frac{1}{1-\left(-\frac{z^4}{9}\right)}=\sum_{n=0}^\infty  (-1)^n\frac{z^{4n+1}}{9^{n+1}}
\end{align*}
The range of validity is  $\left|-\frac{z^4}{9}\right|<1$ or equivalently $|z|<\sqrt{3}$.

We  now consider the Taylor expansion  around other points $z_0\in\mathbb{C}$. The function
  \begin{align*}
f(z)&=\frac{z}{z^4+9}\\
&=-\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(1+i\right)}
-\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(-1-i\right)}\\
&\qquad+\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(-1+i\right)}
+\frac{i}{12}\cdot\frac{1}{z-\sqrt{\frac{3}{2}}\left(1-i\right)}\tag{1}
\end{align*}
  has  four simple poles at $z_1=\sqrt{\frac{3}{2}}\left(1+i\right),z_2=\sqrt{\frac{3}{2}}\left(1-i\right),z_3=\sqrt{\frac{3}{2}}\left(-1+i\right)$ and $z_4=\sqrt{\frac{3}{2}}\left(-1-i\right)$, one in each quadrant residing on the diagonals.

According to (1) we derive an expansion at $z=z_0$ via
\begin{align*}
\frac{1}{z-a}&=\
\frac{1}{(z-z_0)-(a-z_0)}\\
&=-\frac{1}{a-z_0}\cdot\frac{1}{1-\frac{z-z_0}{a-z_0}}\\
&=-\frac{1}{a-z_0}\sum_{n=0}^\infty\left(\frac{z-z_0}{a-z_0}\right)^n\\
&=-\sum_{n=0}^\infty\frac{1}{(a-z_0)^{n+1}}(z-z_0)^n\tag{2}
\end{align*}

We obtain from (1) and (2) for $z,z_0\in\mathbb{C}\setminus\{z_1,z_2,z_3,z_4\}$:
  \begin{align*}
f(z)&=\frac{z}{z^4+9}\\
&=\frac{i}{12}\sum_{n=0}^\infty\left[
\frac{1}{\left(\sqrt{\frac{3}{2}}\left(1+i\right)-z_0\right)^{n+1}}
+\frac{1}{\left(\sqrt{\frac{3}{2}}\left(-1-i\right)-z_0\right)^{n+1}}\right.\\
&\qquad\qquad\qquad\left.-\frac{1}{\left(\sqrt{\frac{3}{2}}\left(-1+i\right)-z_0\right)^{n+1}}
-\frac{1}{\left(\sqrt{\frac{3}{2}}\left(1-i\right)-z_0\right)^{n+1}}\right](z-z_0)^n
\end{align*}
  The radius of convergence of the expansion around $z=z_0$ is the distance to the pole in the same quadrant of $z_0$ or the distance to the two nearest poles if $z_0$ resides on an axis.

</s>