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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?

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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 $0<|z|<1$.

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  • $\begingroup$ Definitely the quickest way to do it, and the best, in my opinion. $\endgroup$ – Lubin Apr 8 '15 at 19:11

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