<s>[INST] Find the Laurent Expansion of $f(z)$ Find the Laurent Expansion for
$$f(z)=\frac{1}{z^4+z^2}$$
about $z=0$.
I have found the partial fraction decomposition
$$f(z)=\frac{1}{z^4+z^2}=\frac{1}{z^2}-\frac{1}{2i(z-i)}+\frac{1}{2i(z+i)}.$$
Next I wanted to expand each of the three terms separately.  I have
$$\frac{1}{z^2}=\frac{1}{z^2},$$
$$\frac{1}{2i(z-i)}=-\frac{1}{2z}i\sum_{n=0}^{\infty}\left(\frac{i}{z}\right)^n,\quad |z|>1$$
$$\frac{1}{2i(z+i)}=-\frac{1}{2z}i\sum_{n=0}^{\infty}\left(-\frac{i}{z}\right)^n,\quad |z
|>1.$$
Therefore, I believe that my Laurent expansion should be
$$\frac{1}{z^4+z^2}=\frac{1}{z^2}+\frac{1}{2z}i\sum_{n=0}^{\infty}\left(\frac{i}{z}\right)^n-\frac{1}{2z}i\sum_{n=0}^{\infty}\left(-\frac{i}{z}\right)^n,\quad |z|>1.$$
I had a few questions, though.
1) What about the $z$ in the denominators outside the sums?  What's that all about?
2) Does the same radius of convergence $|z|>1$ apply for $\frac{1}{z^2}$ as did for the other two series?  What does it mean to expand about $z=0$, and yet the radius of convergence for those two expansions above are $|z|>0$?
3) Can I do anything to clean up this answer?
 [/INST] You may just write, as $z \to 0$:
$$\begin{align}f(z)=\frac{1}{z^4+z^2}&=\frac{1}{z^2(1+z^2)}\\\\&=\frac{1}{z^2}(1-z^2+z^4-z^6+z^8...)
\\\\&=\frac{1}{z^2}-1+z^2-z^4+z^6-...
\end{align}
$$ and this gives the Laurent expansion of $f$ near $z=0$, on $0<|z|<1$.
</s>