Laurent series and regions Some help to understand why this is true would be amazing;
For the function
$$\frac{1}{z-1}\;\;,$$ the laurent series is different depending on the region.
1) for the region $|z|<1$, you rewrite the function as $\frac{-1}{1-z}$, and the laurent series is
$$
-\sum_{n=0}^\infty z^n
$$
2) for the region $1<|z|<2$, you rewrite the function as $\frac{1}{z}$ $\frac{1}{1-\frac{1}{z}}$, and the laurent series is 
$$
\sum_{n=0}^\infty\frac{1}{z^{n+1}}
$$
What I dont understand is, why can't I rewrite the functions in the same way, regardless of the region?
 A: In both cases you use the geometric series expansion
$$
\frac{1}{1-z}=\sum_{k=0}^\infty z^n
$$
which is only valid for $|z|<1$, so you cannot apply the same rule for $|z|>1$, unless you rewrite the expression as
$$
\frac{1}{z}\frac{1}{1-\frac{1}{z}}
$$
in which case you can apply the geometric series expansion to the second factor because $|z|>1 \implies |1/z|<1$.
A: You can't write the function in a unique way for both regions because althought the function is the same, the object we are dealing with is the Laurent development of the function, NOT the function itself; the Laurent development is a representation of the function, it's "a way to see the function" and it's natural to expect this representation changes when "the point of view" (i.e. the domain) changes.
Moreover look at the radius of convergence: the first series you wrote converges on the unit disk, the second converges over the whole complex plane minus the unit disk: thus these two series converge on disjoint domain.
A: You can. The function is $\frac{1}{z-1}$. The point is that a Laurent series centered at some point (here $z=0$) is a power series expansion with both positive and negative integer powers. Such a series can only be convergent on some annuli bounded by singularities of the function. In the present case there is a singularity for $|z|=1$ which will then separate the plane into two annuli of convergence: $|z|<1$ (actually a disk) and $|z|>1$ (in fact, also a disk).
