Calculating the Laurent Series for complicated functions I would appreciate a nudge in the right direction for the following questions, as I haven't really had any exposure to finding Laurent Series expansions for more complicated functions 
Find the Laurent Series expansion for:
$$1)\frac{z-7}{z^2+z-2}$$    $ 1<|z|<2$ , $|z|>2$ ,  $0<|z-1|<1$
For this question I tried to follow the standard route for this question by using partial fraction decomposition and then attempting to put each fraction in the form of $\frac{1}{1-z}$ but I feel like I'm having an issue when I come to this stage:
By partial fraction decomposition we have: $$\frac{z-7}{(z-1)(z+2)}=\frac{3}{z+2} - \frac{2}{z-1}$$ And attempting to put $\frac{3}{z+2}$ and $-\frac{2}{z-1}$ in the form $\frac{1}{1-z}$ has so far yielded an embrassing $$\frac{3}{z+2}=\frac{3}{2+z}=\frac{3}{2}\frac{1}{1+\frac{z}{2}}$$ At this point I think it's as simple as doing some algebraic manipulation but I can't even seem to do this? And once I've done that is this still just a simple "plug and play" type scenario with Taylor Series?
$$2)\frac{1-cosz}{(z-2\pi)^3}$$ I don't even know where to start for this. I thought perhaps partial fraction decomposition again but I don't see how this would help (if even possible) since we would still struggle to get the form of $\frac{1}{1-z}$
$3)$Let $f(z)= \sqrt {z^2-3z+2}$, $1<|z|<2$. Does the Laurent series expansion for $f(z)$ exist? Justify your answer.
This is completely beyond me as I can't imagine where I would begin to check if the expansion exists. I'm only used to (as you can probably tell) calculating Laurent Series of the form $\frac{1}{1-z}$
Additionally I'm not entirely sure how the conditions i.e $1<|z|<2$ etc affect the question or how the Laurent Series expansion is calculated, so if someone could briefly explain that I'd appreciate it.
Thanks
 A: For context, the correct expansion is
$$\frac{1}{1-w} = \sum_{n=0}^\infty w^n\qquad\text{when $|w|<1$.}$$
Which means it's critical to manipulate the expression so that $|w|$ is in fact smaller than 1.
1) The partial fraction expansion is the correct first step:
$$\frac{z-7}{z^2+z-2} = \frac{3}{z+2} - \frac{2}{z-1}.$$
Where to go from here depends on the domain you're interested in.  On the annulus $1<|z|<2$, we want
$$\frac{3}{z+2} = \frac{3}{2}\cdot\frac{1}{1+\frac{z}{2}} = \frac{3}{2}\cdot\frac{1}{1-\left(-\frac{z}{2}\right)}
=\sum_{n\ge0}(-1)^n\frac{3}{2^{n+1}}z^n\qquad
\text{because $\displaystyle{\left|-\frac{z}{2}\right|<1}$,}$$
but
$$-\frac{2}{z-1} = \frac{2}{z}\cdot\frac{1}{1-\frac{1}{z}}
  = \sum_{n<0} 2z^n\qquad\text{because $\displaystyle{\left|\frac{1}{z}\right|<1}$.}$$
The Laurent series for the original expression on the annulus is just the sum of these.
On the complement of the disk, you want to do something similar, but keep in mind that now the first fraction has to be expanded in powers of $2/z$ rather than $z/2$.
On the punctured disk $0<\left|z-1\right|<1$, first rewrite the partial fractions in terms of $t=z-1$, expand about $t$, then re-express in terms of $z$.  For the purposes of the first fraction, $t$ is small. The second is just $2/t$; not much else to be done there.  (Note this expansion will actually work out to $0<\left|z-1\right|<3$.)
2) Here the question is to find (a?) Laurent series for
$$\frac{1-\cos z}{(z-2\pi)^3}.$$
I can only give general advice here since you don't specify a domain, but:


*

*There are well-know Taylor expansions (and therefore Laurent expansions) for $\cos z$ and $\sin z$ centered at $z=0$.  If the center is somewhere else, there are always sum-of-angle formulas.

*There is a general rule for multiplying Laurent series:
$$\left(\sum_n a_nz^n\right)\cdot\left(\sum_n b_nz^n\right)
  = \sum_n c_nz^n\qquad\text{where $\displaystyle{c_n = \sum_{i+j=n}a_ib_j}$.}$$
(Note that this only necessarily works if both series terminate in the negative direction or both terminate in the positive direction; we want that sum to be finite or there's a whole bunch of issues that come up.)


3) Is $\sqrt{z^2-3z+2}$ even well-defined on the annulus? If not, it definitely doesn't have a Laurent series.
