Why does $\frac{z+2}{z-1}$ not have a series expansion around $|z-1|>1$?

Question

Determine the Laurent series of $$z \mapsto \frac{z+2}{z-1}$$ over the region $$C := \left\{ z \in \mathbb{C} \mid |z-1| > 1 \right\} $$

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With the region being simplified to

$$ \frac{1}{|z-1|} < 1$$

the solution would be

$$ 1+\frac{3}{z-1} $$

However, I am confused as to how this is the conclusion. Why is there no expansion of further terms of the Laurent series? Because if you change the inequality you get a actual expansion like so:

In the annulus: $$C:=\{z \in \mathbb{C} \mid 0<|z|<1 \} \\$$ $$|z| < 1$$ Gives: $$f(z)=\frac{z+2}{z-1}=1-3\sum_{n=0}^{\infty}z^n$$

So why does that give a series solution but the first way does not?

Answer 1

I believe the first series you got is infact the Laurent series of $f(z)$ about z=1 and the second the Taylor series about z=0. What you've shown is that generally they don't agree.

Page-442:Laurent's theorem: If $f(z)$ is analytic at an annlus centered at $a$, then $f(z)$ can be expressed as a Laurent series centered at $a$.

Remark of theorem on page-443: If there are no singularities in $D$, then the inner edge of the annulus maybe completly collapsed, thereby transforming the annulus into a disc In this case, the laurent series does not contain negative powers, and, we recover Taylor series as a special case of Laurent series (paraphrased, not same as original)

Picture:

Source: Visual Complex Analysis