# Laurent series expansion in a region

Consider the function $$f(z) = \frac{1}{z(z-1)(z-2)}$$ in the region $$2 < |z| < \infty$$. I would like to find the partial fraction decomposition and the Laurent series expansion of $$f(z)$$ in this region.

Partial Fraction Decomposition: To begin, we need to express $$f(z)$$ in partial fraction form. We seek constants $$A, B, C$$ such that:

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

And the Laurent series expansion.

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– Community Bot
Jun 11, 2023 at 5:16

Note that $${B\over z-1}={Bz^{-1}\over1-z^{-1}}=Bz^{-1}+Bz^{-2}+Bz^{-3}+\dotsb$$ and $${C\over z-2}={Cz^{-1}\over1-2z^{-1}}=Cz^{-1}+2Cz^{-2}+4Cz^{-3}+\dotsb$$ so the Laurent series is $$(A+B+C)z^{-1}+\sum_2^{\infty}(B+2^{n-1}C)z^{-n}$$.
Now, we just have to do the partial fractions part to get $$A,B,C$$.
Clearing fractions, $$1=A(z-1)(z-2)+Bz(z-2)+Cz(z-1)$$ Substituting in turn $$z=0$$, $$z-1$$, $$z=2$$, we get simple equations for $$A,B,C$$. The reader is encouraged to carry out the arithmetic.