Find the Laurent series of $\frac{z+2}{z-1}$ where i) $0 < |z| < 1$ and ii) $|z-1| > 1$ I am struggling to find the algebraic manipulation to find the Laurent series of $\frac{z+2}{z-1}$ where i) $0 < |z| < 1$ and ii) $|z-1| > 1$
I tried putting the numerator as $(z-1) + 3$ but then my problem is that I cannot obtain an expression of the form $\frac{1}{1+w}$ to obtain the Laurent expansion since the expression becomes $ 1+\frac{3}{z-1}$.
Can someone help me find the algebraic manipulation to solve these types of problems please?
Thanks
 A: For i) you're on the right track. Just recall the formula of a geometric series $\sum_{n \ge0} x^n = \frac{1}{1-x}$ converges when $|x|<1$.
For ii) we see that the Laurent series for $|z-1| > 1$ is precisely the Laurent series for $|z-1|>0$, which you have already calculated to be
$$
\frac{z+2}{z-1} = 1 + 3 \frac{1}{z-1} \tag{1}
$$
which converges for all $|z-1|>0$.

To explain ii) need to recall a bit of Complex Analysis theory. Laurent's theorem states that given a function $f(z)$ we can express it as the series around some point $c$ as follows
$$
f(z) = \sum_{n \in \mathbb{Z}} a_n (z-c)^n
$$
where the coefficients $a_n$ are given by
$$
a_n = \frac{1}{2\pi i}\oint_{\gamma} \frac{f(z)}{(z-c)^{n+1}} \, dz \tag{2}
$$
where $\gamma$ is any positively oriented closed Jordan curve lying inside an annulus $A$ where $f(z)$ is holomorphic.
Now, the above is a mouthful, but the key thing to notice is that the curve $\gamma$ over which you integrate is arbitrary as long as it's in the region $A$ over which the function is holomorphic. This is a consequence of the homotopic version of Cauchy's integral theorem which states that given two smooth closed curves $\gamma_1$ and $\gamma_2$ in some region $U$ which are homotopic (which just means you can stretch one into the other smoothly), then for some function $f(z)$ which is holomorphic on $U$
$$
\int_{\gamma_1} f(z) dz = \int_{\gamma_2} f(z) dz \tag{3}
$$
are equal!
Because of the above, we see that if our function $f(z)$ happens to be holomorphic in a bigger region $U\supset U^*$, then if we have some $\gamma^* \in U^*$ over which we integrate $(2)$, then the integral will be the same as the one we would get by deforming $\gamma^* \to \gamma \in U \setminus U^*$.
In our case, we see that the function $f(z) = \frac{z+2}{z-1}$ is holomorphic for all $|z| \neq 1 \iff |z-1|>0$, but notice that this region encapsulates the sub-region we're interested in $|z-1|>1$. This means that if we choose some curve $\gamma \subset \{z \in \mathbb{C}:|z-1|>1\}$ over which to calculate $(2)$, since the function $\frac{f(z)}{(z-1)^{n+1}}$ will be holomorphic for all $|z-1| > 0$ we can deform our $\gamma \to \gamma_1 \subset \{z \in \mathbb{C}:|z-1| > 0\} \setminus \{z \in \mathbb{C}:|z-1|>1\}$ and we'll still get the same values of $a_n$, and hence, the same Laurent series.

The above discussion is the reason why these Laurent exercises are usually presented with regions that have extreme points at the singularities of the function $f(z)$, because if you have some other sub-region where the function is holomorphic then the function will continue to be holomorphic until you hit one of these singularities, so you might as well stretch the curve $\gamma$ to cover the whole thing.
Lastly, if you want some expression which only converges  on $|z-1| > 1$ you can do
$$
\frac{x+2}{x-1} = \frac{4}{x-1} + \frac{x-2}{x-1} = \frac{4}{x-1} + \left[\frac{1}{1- \left(\frac{1}{x-1} \right)}\right]^{-1} = \frac{4}{x-1}+\left[\sum_{n=0}^{\infty} (x-1)^{-n} \right]^{-1} \tag{4}
$$
As is, equation $(4)$ will only converge for $|z-1| > 1$. However, the above is not really the Laurent series, since if you try to write it in the form $\sum_{n \in \mathbb{Z}} a_n (z-1)^n$ you'll see that you just arrive at the expression in $(1)$ which we already argued has a bigger radius of convergence.
