I need to find the Laurent Series of $f(z) = \frac{1}{z-1}$, valid in the region $1<|z-2|<2$.
If I understand this, since $f(z)$ is analytic in $ A=\{z \in \mathbb{C}\ | 1 <|z-2|<2 \}$, we have that:
$$f(z)= \sum_{n=0}^\infty a_n(z-2)^n + \sum_{n=1}^\infty b_n(z-2)^{-n}, z \in A.$$
If $|z-2|>1$, then $\frac{1}{|z-2|}<1$, and with some manipulation we find that:
$$\frac{1}{z-1}=\frac{1}{1 + (z-2)}=\frac{1}{(z-2)(1 + 1/(z-2))}= \frac{1}{z-2}\cdot\frac{1}{1-(-1/(z-2))}.$$
Since $|\frac{-1}{z-2}| = |\frac{1}{z-2}|= \frac{1}{|z-2|}<1$, we have
$$\frac{1}{z-1}=\frac{1}{z-2}\cdot \sum_{n=0}^\infty \frac{(-1)^n}{(z-2)^n}= \sum_{n=0}^\infty \frac{(-1)^n}{(z-2)^{n+1}}= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(z-2)^n}.$$
If $|z-2|<2$, then $\frac{|z-2|}{2}<1$, but I can't seem to find a way to express $\frac{1}{z-1}$ in a suitable form $\frac{c}{1-(|z-2|/2)}$ to use the convergence of the geometric series.
Is this always possible? Or you just have to get lucky?
Since I also know that the coefficients of $a_n$ are given by: $$a_n = \frac{1}{2\pi i}\int_{\gamma}\frac{dz}{(z-1)(z-2)^{n+1}}$$
I guess I could find them that way, but it seems too hard to do
Is using the above formula the only way to do it? What if there is no pattern to $a_n$? I can't infinitely calculate them!
If anyone could help or find any mistake in my reasoning I would appreciate it!
If $1<|z-2|$, then\begin{align}\frac1{1+(z-2)}&=\frac1{1-\bigl(-(z-2)\bigr)}\\&=-\sum_{n=-\infty}^{-1}\bigl(-(z-2)\bigr)^n\\&=\sum_{n=-\infty}^{-1}(-1)^{n+1}(z-2)^n.\end{align}