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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    Jun 11, 2023 at 5:16

1 Answer 1


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.


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