$f(z) = \frac{1}{z(1-z)},\qquad 0 < |z| < 1$
at first the function is analytic.
then, $f(z) = \frac{1}{z(1-z)}$ due to partial fraction decomposition it becomes $f(z) = \frac{1}{z} + \frac{1}{1-z}$
this is where I become confused.
When doing $0 <|z|$:
there is no Taylor series, is that right?
Now for $|z| < 1$:
$f(z) = \frac{1}{z}$ stays the same, and for $\frac{1}{1-z}$ is turned into geometric series which is $\sum_\limits{n=0}^{\infty} z^n$
so the final result is $f(z) = \frac{1}{z} + \sum\limits_{n=0}^{\infty} z^n$? So is it correctly done right...or I am doing it fully wrong.