I need to calculate the principal part of the Laurent expansion of $f$ around a given $z_0$ in an annulus of the form $\{z\in \mathbb{C}:0<|z-z_0|<r$} and then use this to find $Res(f,z_0)$

i)the first case is $$ f(z)=\frac{z^2}{(z-1)^2} $$ At a first , I see that $f(z)$ can be written as $$ \frac{A}{(z-1)^2}+\frac{B}{(z-1)}+\frac{C}{(z-1)^0} $$ and it's easy to see that $A=1 ,B=2 , C=1$ therefore $$f(z)= \frac{1}{(z-1)^2}+\frac{2}{(z-1)}+1$$ which is in the form of a Laurent expansion at $z_0=1$ , so along with the uniqueness of the Laurent expansion , I conclude that this is it and also $Res(f,1)=2$

Is my approach and justification valid ?

If I'd choose an approach where I'd try to construct a geometric series and end up with a series , would I have to take both cases for $1<|z-1|<2$ and $|z-1|>2 $ ?

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    $\begingroup$ Yes, it's valid. Here, the simplest way is to expand $z^2 = ((z-1)+1)^2 = (z-1)^2 + 2(z-1) + 1$ [probably that's how you did it]. I don't understand what you mean in the last paragraph, the function here has only one singularity, so the Laurent series with centre $1$ converges in $\mathbb{C}\setminus\{1\}$, and the Laurent series with other centres are Taylor series converging in the disk $\{z : \lvert z-z_0\rvert < \lvert z_0-1\rvert\}$. $\endgroup$ – Daniel Fischer Aug 29 '14 at 20:50
  • $\begingroup$ @DanielFischer That's how I did it( seeing it first at your other answer correcting my errors ). Convergence of the Laurent series still confuses me. If the function had more singularities, say like $f(z)=\frac{1}{(z^2+1)^3}$ which has 2 poles ( of 3rd order ) ? Thanks for the quick response! $\endgroup$ – helplessKirk Aug 29 '14 at 21:12
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    $\begingroup$ A Laurent series (of a holomorphic function $f$ with only isolated singularities) always converges [locally uniformly] in an annulus $r < \lvert z-z_0\rvert < R$ [where $r = 0$ or $R = +\infty$ are possible], and there are always non-removable singularities on both bounding circles if $R < +\infty$ [if $r = 0$, then the inner bounding circle is degenerate, and $z_0$ is an isolated singularity of $f$]. So in your example, the possibilities are 1. both poles in the outside of the annulus, then it's a Taylor series and the radius of convergence is the distance from the centre to the closer pole; $\endgroup$ – Daniel Fischer Aug 29 '14 at 21:22
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    $\begingroup$ 2. one pole inside the smaller circle, and the other outside the larger circle, then you have a real annulus [or a punctured disk] of convergence, and $r$ is the distance from the centre to the nearer pole, $R$ the distance to the farther pole; 3. both poles inside the smaller circle, then $r$ is the distance to the farther pole, and $R = +\infty$. $\endgroup$ – Daniel Fischer Aug 29 '14 at 21:22
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    $\begingroup$ @helplessKirk: See here. $\endgroup$ – Mhenni Benghorbal Aug 29 '14 at 22:48

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