Question #296323

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

1
Expert's answer
2022-02-16T15:22:18-0500

ANSWER

We represent the function ff as a sum: f(z)=2z3(z2)(z1)=2z+3(1z11z2).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 1z1=11z=n=0zn,z<1\frac{1}{z-1}=-\frac{1}{1-z}=-\sum _{ n=0 }^{ \infty }{ { z }^{ n } } , |z| <1 .

1z2=12 11z2=12n=0zn2n,z<2\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, 1z11z2=n=0zn+12n=0zn2n=\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+12z+z4n=2(112n+1)zn,=-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|z|<1 . Hence,


f(z)=32z4 3n=2(112n+1)zn,z<1f(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|z|<1 , since f(z)f(z) is analytic in this circle. In this case the series is the Taylor series of the function ff (a special case of the Laurent series, since the main part n=1anzn\sum _{ n=1 }^{ \infty }{ \frac{{a }^{n}}{z^{n}} } is missing)


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