Laurent Series expansion in sigma notation

Question

I am trying to find the Laurent Series expansion in sigma notation of $$\frac{1}{z^3-2z^2+z}$$ where $0<|z-1|<1$. I've tried partial fractions and am still stuck on the approach necessary to achieve the answer of $(1)^n(z-1)^{n-2}$. Some guidance and instruction would be greatly appreciated. Thanks...

Answer 1

$f(z) = \frac{1}{z^3 - 2z^2 + z} = \frac{1}{z(z-1)(z+1)} = \frac{g(z)}{z-1}$

where $g(z) = \frac{1}{z(z+1)}$ is holomorphic at $z=1$

If you make the taylor series of $g$ at $z=1$ and then divide it by $z-1$ you should get the laurent series of $f$ at $z=1$

Answer 2

Rewrite

$$f(z) = \frac1{z (z^2-2 z+1)} = \frac1{(\zeta+1) \zeta^2} $$

where $\zeta = z-1$. Since $0 < |\zeta| \lt 1$, we may expand as follows:

$$f(\zeta) = \frac1{\zeta^2} \sum_{n=0}^{\infty} (-1)^n \zeta^n $$

Thus we have the Laurent series:

$$f(z) = \frac1{(z-1)^2} \sum_{n=0}^{\infty} (-1)^n (z-1)^{n} = \sum_{n=0}^{\infty} (-1)^n (z-1)^{n-2}$$