Find the Laurent expansion $f(z)=\frac{1}{z(1-z)^2}$.

Question

Find the Laurent Series expansion of

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

at $z=1$.

How do I do this when I am not given any region?

Answer 1

Hint: $$f(z)=\frac{1}{z(1-z)^2}=\frac{1}{1-(1-z)}\cdot\dfrac{1}{(1-z)^2}\\=\dfrac{1}{(z-1)^2}\sum_{n=0}^\infty(-1)^n(z-1)^n$$

Answer 2

You do not need a region. The function is defined in a pointed neighbourhood of $1$, that is all you need. If it helps, shift the function so that the pole is at $0$: $f(1+z)=g(z)$, then the shifted function is $g(z)=\frac{1}{z^2(1+z)}$. The Laurent expansion of this function is $\sum\limits_{n=-2}^{\infty} (-1)^n z^n$, as you can easily obtain from the power series of $\frac{1}{1+z}$.

Can you finish from here?