Find the Laurent expansion of
$$\frac{1}{z(z-1)}=\frac{1}{z-1}-\frac{1}{z}$$
in $0<|z|<1$ and $|z|>1$
I think I'm getting confused about which expansion corresponds to which domain. I've looked at a few examples, but have been unable to make a solid connection.
For the $|z|>1$, I think it makes sense to
$$-\frac{1}{z}+\frac{1}{z}\frac{1}{1-\frac{1}{z}}=-\frac{1}{z}+\frac{1}{z}\sum_{n=0}^\infty\frac{1}{z^n}=-\frac{1}{z}+\sum_{n=0}^\infty\frac{1}{z^{n+1}}$$
but further expanding this,
$$-\frac{1}{z}+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}\cdots=\frac{1}{z^2}+\frac{1}{z^3}+\cdots=\sum_{n=2}^\infty\frac{1}{z^n}$$
However this doesn't deal with the $z=0$ situation, so I know it won't work for $0<|z|<1$. I have a hard time distinguishing how to expand functions into the Laurent series.
Best Answer
The Laurent series for a given function can vary as one varies the annulus. Your exercise is such an example.
You have worked out the case when $|z|>1$ correctly.
In the case when $0<|z|<1$, you simply use $$ \frac{1}{1-z}=\sum_{n=0}^\infty z^n\;, $$ so that $$ \frac{1}{z-1}-\frac1z=-\frac1z+\sum_{n=0}^\infty (-1)z^n $$