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Suppose $f$ is meromorphic on the Riemann sphere, and suppose also that $f(0) = 0$, $f(-1) = 2$, $f(3) = 3$, $f$ has a simple pole at $1$ with residue $1$, and $f$ has a triple pole at $2$ with residue $2$. The problem is to determine $f$ an a Laurent expansion on the annulus $\{z \in \mathbb{C}\ |\ 1 < |z| < 2\}$.

My attempt was to try to find $f$ of the form $\frac{1}{z - 1} + \frac{2}{z - 2} + \frac{A}{(z - 2)^2} + \frac{B}{(z - 2)^3}$; however we end up with only $2$ variables and $3$ equations. How do we proceed in this situation?

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You have a third variable: the value at $\infty$.

If we admit other poles that have not been mentioned, you have of course many more degrees of freedom, so to have a unique solution, we must assume that $f$ has no other poles than the two mentioned. Then we can write

$$f(z) = \frac{1}{z-1} + \frac{2}{z-2} + \frac{A}{(z-2)^2} + \frac{B}{(z-2)^3} + C,$$

and the three values given at the points $0, -1, 3$ produce a system of three linear equations in three unknowns.

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