0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Jun 11, 2023 at 5:16

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .