ANSWER
We represent the function f as a sum: f(z)=2z−(z−2)(z−1)3=2z+3(z−11−z−21).
Using the geometric series we have z−11=−1−z1=−∑n=0∞zn,∣z∣<1 .
z−21=−21⋅ 1−2z1=−21∑n=0∞2nzn,∣z∣<2
. Thus, z−11−z−21=−∑n=0∞zn+21∑n=0∞2nzn=
=−1+21−z+4z−∑n=2∞(1−2n+11)zn, for ∣z∣<1 . Hence,
f(z)=−23−4z −3∑n=2∞(1−2n+11)zn,∣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 ∑n=1∞znan is missing)
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