Expand $\dfrac{z}{(z-1)(2-z)}$ in the Laurent series valid for $|z-1|>1$. The question is to expand
$$f(z)=\frac{z}{(z-1)(2-z)}=\frac{1}{z-1}+\frac{2}{2-z}$$
in the Laurent series valid for
(i)  $|z|<1$
(ii) $|z-1|>1$
I solved for (i) using the geometric series but for the (ii) I'm absolutely confused. Any help would be appreciated!
 A: It's been a while since I've done this, but:
$|z-1| > 1 \Leftrightarrow \frac{1}{|z-1|} < 1$, so you can use the geometric series method for $q = \frac{1}{|z-1|}$.
$$ \frac{1}{z-1} + \frac{2}{2-z} = \frac{1}{z-1} + \frac{2}{1 - (z - 1) } = \frac{1}{z-1} + \frac{2}{z-1}\cdot\frac{1}{ \frac{1}{z - 1} - 1}$$
And $\frac{1}{q - 1} = - \sum_{n=0}^{\infty} q^n$ iff $|q| < 1$. So...
$$\frac{1}{z-1} + \frac{2}{z-1}\cdot\frac{1}{ \frac{1}{z - 1} - 1} = \frac{1}{z-1} - \frac{2}{z-1}\cdot\sum_{n=0}^{\infty}\left( \frac{1}{z-1} \right)^n \\ = (1-2)(z-1)^{-1} - 2\sum_{n=2}^{\infty}\left(z-1\right)^{-n} $$
Ok, so here's a shot at an explanation.
Here's the function plotted (as an $\mathbb R \to \mathbb R$ function).
$f$." />
As it is apparent from the definition of $f(z) = \frac{1}{z-1} + \frac{2}{2-z}$, it has simple poles at $1$, and $2$. This means roughly that it tends to infinity with approximately the same speed as $\frac{1}{x}$ does at $0$.
On the plot this is quite noticeable.
A insert-name-here series of a function is essentially an infinite sum.
Power series are of the form $\sum_{n=0}^{\infty}a_n(z-a)^n$. Where we say that the specific series is "around the point $a$".
For example, $f$ has a Taylor series around $0$, which is a special kind of power series. Specifically it is:
$$ T(z) = \sum_{n=0}^{\infty}\frac{1 - 2^{n+1}}{2^{n+1}}z^n$$
Now these series are most useful if they converge to the function at least on some 'nice' set.
For example, the Taylor-series of a function around $a$ will converge on a disk around $a$. The radius of this disk can be any nonnegative real number or positive infinity. On this disk the convergence is really nice: it is uniform on every compact subset. This means that, for example you are allowed to do things like exchange the order of summation and differentiation.
One important, but easy example is the exponential function's Taylor series around $0$, which has a convergence radius of $\infty$. Exchanging the order of operations, you can prove that $\exp' = \exp$ like so:
$$ \left(\sum_{k=0}^\infty \frac{x^k}{k!}\right)' = \sum_{k=0}^\infty \left(\frac{x^{k}}{k!}\right)' = \sum_{k=1}^\infty \frac{kx^{k-1}}{k!} = \sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!} = \sum_{k=0}^\infty \frac{x^{k}}{k!}$$
Anyway, now you might have a better idea as to why do we bother with power series. Let's look at $f$ again.
As you can see, it's Taylor series at $0$ has no chance of converging at $1$, (because the function itself goes to infinity there), so the radius of convergence must be at most $1$. As it happens, it is exactly $1$.
As you may know, in the complex case, differentiability implies a very strong property, analyticity. Meaning that if a function is complex-differentiable on some set, then it has a power-series (namely its Taylor-series), and that series converges to the function on the set. So power series have a very special place in investigating holomorphic functions.
Ok, but how do Laurent series come to the picture?
They are what happen when you investigate meromorphic functions - These are holomorphic functions with a few poles. - using power series around its poles.
Namely: try to construct a power series for $f$ around $1$. The usual power series won't work, because of the pole right in the middle (meaning that the radius of convergence will be $0$).
So as it turns out, Laurent series have a lot of nice properties too.
They'll converge not on a disk, but on the next best thing, a "disk with a hole in the center", an annulus.
You can ask what's the Laurent series of a function around a point if that function is holomorphic on an annulus around that point.
E.g. In the case of $f$, you can ask about any point and annulus that doesn't contain $1$ or $2$.
Let me list a few:

*

*You can ask (and you did with question ii) what is the Laurent series of $f$ around $1$, such that it converges on the (somewhat special) annulus described by $|z - 1| > 1$


*You can also ask what's the Laurent series of $f$ around $1$ such that it converges on the annulus $0 < |z-1| < 1$.


*Or the same question for $|z| < 1$. In this case, you'll get back the usual Taylor-series, since the function is holomorphic on the whole disk.
Hope this wall of text helps a bit. If you have further questions, ask away.
