Taylor vs. Laurent series.

Question

So I'm trying to figure out methods to get a Laurent series instead of a Taylor series. Here is the initial problem:

$$\frac{1}{(z-1)}$$

So if I'm trying to find the Laurent series and I know they need negative exponents, what's the strategy? I know the answer is to factor out a $\frac{1}{z}$ but how would I get there?

$$\frac{1}{(z-1)} = \frac{1}{z} \cdot \frac{1}{1 - \frac{1}{z}} = \frac{1}{z}(1 + \frac{1}{z} + \frac{1}{z^2} + ... )$$

The above Laurent series (it has negative exponents) converges when $|z| > 1$ right?

And what's the corresponding Taylor series representation? Is the takeaway here that you can represent the same function with two different series?

Answer 1

Assuming that you want to have your series centered at $0$, then it turns out that the Laurent series of $\frac1{z-1}$ in $D(0,1)$ is equal to its Taylor series there, which is$$-1-z-z^2-z^3-\cdots$$On the other hand, its Laurent series on $\Bbb C\setminus\overline{D(0,1)}$ is indeed what you got:$$z^{-1}+z^{-2}+z^{-3}+\cdots$$

Answer 2

We discussed this before, with regards to the function $1/(1-z)$.

You're apparently wondering about uniqueness. The Laurent (or Taylor) series around a given point can indeed be shown to be unique in a given region.