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Consider the function: $$f(z)=\frac{1}{z^3(z+3)}.$$ Suppose we want to compute the integral of this function on the positively oriented contour defined by $|z|=2$. We note that only the singularity at $z_0=0$ is within the contour. So the residue theorem states that, $$\oint_{C:|z|=2}\frac{1}{z^3(z+3)}=2\pi i\cdot Res(f(z),0).$$ So, it remains to compute the Laurent expansion of $f(z)$ about $z_0=0$ for the function. My question arises during this step. We write the Laurent series as follows: $$\frac{1}{z^3}\frac{1}{3-(-z)}=\frac{1}{z^3}\frac{1}{3}\frac{1}{1-(-\frac{z}{3})}=\frac{1}{3z^3}\sum_0^\infty \frac{(-1)^nz^n}{3^n}=\sum_0^{\infty}\frac{(-1)^nz^{n-3}}{3^{n+1}}.$$ This Laurent series is valid the annular domain given by, $$\{z:\,0<|z|<3\}.$$ However, there is another Laurent expansion for this function. Namely, we have the following:

$$\frac{1}{z^3}\frac{1}{z}\frac{1}{1-(-\frac{3}{z})}=\frac{1}{z^3}\frac{1}{z}\sum_1^{\infty}\frac{(-1)^n3^n}{z^n}=\sum_1^{\infty}\frac{(-1)^n3^n}{z^{n-4}}$$ This is valid the annular domain given by, $$\{z:\,3<|z|<\infty\}.$$

Of these two expressions (both valid in two different regions), which do we use when computing the residue?

My guess is for the former (because it is valid in a neighborhood about $z_0=0$), but I'm not too sure. Any help with clarifying this would be appreciated.

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Yes, it's the former. It's the one which is defined in $D(z_0,r)\setminus\{z_0\}$ for some $r>0$ (in your case, $z_0=0$).

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