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I had to find the Maclaurin series expansion of the function: $f(z) = \frac{z}{z^4 + 9}$. I did find the correct expansion: $\sum_{i=0}^{\infty}\frac{(-1)^n}{3^{2n + 2}}z^{4n+1}$.

However, the answer in my solutions manual is:

$\sum_{i=0}^{\infty}\frac{(-1)^n}{3^{2n + 2}}z^{4n+1}$ . ($|z|<\sqrt3$)

Q: Why does |z| needs to be smaller than $\sqrt3$? I know it probably has to do with the singularities of f, but I can't really figure out why.

Thanks in advance!

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Hopefully you know that a power series always converges exactly in a disk (and possibly points on its boundary, with the extreme cases of $\{0\}$ and all of $\mathbb C$ also being possible), where the disk is as large as it can be without including a singularity.

The singularities of your function are the poles at points where $z^4=-9$, and all those points have a modulus of $|-9|^{1/4}=\sqrt 3$.

So $\sqrt 3$ must be the radius of convergence.

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