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I just wanna see if I have some of the fundamentals nailed in terms of understanding.

Say we have some complex valued function defined as

$$f(z)=\tfrac{1}{z(z-1)}$$

and we want to evaluate it's laurent series centred at $z_0=0$ on the annulus $0<|z|<1$.

Or take the function

$$f(z)=\tfrac{1}{(z-1)(z+1)}$$

this time taking it's Laurent series centred at $z_0=1$ on the annulus $0<|z-1|<2$

I have a few questions regarding said laurent series:

i) my first question is just on the general situation weve got here . I believe in the first example I mentioned that we have a function which has two singularities, a simple pole at zero and another at 1. Then we have defined a region on which to calculate the laurent series that stops us from hitting the singularities by only calculating at a distance greater to one and less than the other . similarly we have in my example a complex valued function defined on some region but this time it's singularities are +-1 . and we want to centre our expansion at 1 so no we define our region in such a way that the distance between 1( which we'll never hit because we require the distance between it and any point to be greater than zero ) and any other point is less than 2 so we know have a circle that is centred at one and has one side touching 3 and the other -1 (it's singularity) but never touches either.Is this correct ?

ii)a. when actually calculating the series mentioned in my first series I know we evaluate in a region valid for $|z|<1$ in the following way :

$$\tfrac{1}{z(z-1)}=\tfrac{-1}{z}\tfrac{1}{1-z}=\tfrac{-1}{z} \Sigma_{n=0}z^n$$ which is valid for $|z|<1/|c|$ c in this case being the coeffiecient of z in in the secon equality above.

My question however is this, evaluating for $|z|<1$ is the same as evaluating for $0<|z|<1$ in this case right ? , because the series i found is valid below 1 up until 0 which it is undefined, at right ?

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  • $\begingroup$ Your typography is a little difficult for me to read, but at least for your first question, I think you have the right idea. The region you define the Laurent series on won't contain a singularity, but the singularities can be on the boundary of the region (in fact, regions of convergence for Laurent series are either disks or annuli, depending on the singularities). $\endgroup$ – Clayton Nov 30 '18 at 18:55
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    $\begingroup$ We actually don't choose the region. The Laurent series itself does that. If you take the Laurent series around a pole, it's radius of convergence will be the distance to the next closest place the function is not analytic.Your final example is true everywhere that the expressions are defined. $$\frac 1{1-z} = \sum z^n$$ is true where the series converges, which includes $|z| < 1$ and excludes $|z| > 1$. But $$\frac{-1}z\frac 1{1-z}=\frac{-1}z \sum z^n$$ also requires $\frac 1z$ to exist, so excludes $z = 0$. $\endgroup$ – Paul Sinclair Dec 1 '18 at 5:50

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