Answer to Question #296323 in Calculus for pavani

Question

Find all possible Taylor's series and Laurent Series expansions of f(z)= 2z-3/(z-2)(z-1) about z=0

Answer 1

ANSWER

We represent the function $f$ as a sum: $f(z) =2z-\frac{3}{(z-2)(z-1)}=2z+ 3\left ( \frac{1}{ z-1 }- \frac{1}{z-2}\right ).$

Using the geometric series we have $\frac{1}{z-1}=-\frac{1}{1-z}=-\sum _{ n=0 }^{ \infty }{ { z }^{ n } } , |z| <1$ .

$\frac{1}{z-2}=-\frac{1}{2} \cdot\ {\frac{1}{1-\frac{z}{2}}}= -\frac{1}{2}\sum _{ n=0 }^{ \infty }{ \frac{{ z }^{n}}{2^{n}} } , |z| <2$

. Thus, $\frac{1}{z-1}-\frac{1}{z-2}=-\sum_{n=0}^{\infty}{ { z } ^{ n } } +\frac{1}{2}\sum _{ n=0 }^{ \infty }{ \frac{{ z }^{n}}{2^{n}} } =$

$=-1+\frac{1}{2} -z+\frac{z}{4} -\sum _{ n=2 }^{ \infty }{ \left ( 1-\frac{1}{2^{n+1}} \right )z^{n} } ,$ for $|z|<1$ . Hence,

$f(z)= -\frac{3}{2}- \frac{ z}{4} \ -3\sum _{ n=2 }^{ \infty }{ \left ( 1-\frac{1}{2^{n+1}} \right )z^{n} } , |z| <1\\$ .

The decomposition obtained takes place inside the circle $|z|<1$ , since $f(z)$ is analytic in this circle. In this case the series is the Taylor series of the function $f$ (a special case of the Laurent series, since the main part $\sum _{ n=1 }^{ \infty }{ \frac{{a }^{n}}{z^{n}} }$ is missing)