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My attempt:

$$f(z)=\frac{1}{z^2-3z+2}$$ $$=\frac{1}{(z-2)(z-1)}=\frac{A}{(z-2)}+\frac{B}{(z-1)}$$

After finding common denominator, equating the numerators, and letting $z=0$, we get:

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

Letting $w=z-2$, we get:

$$f(w)=\frac{1}{1-(-w+1)}-\frac{1}{1-(-w)}$$

But this is the part where I get confused. I believe the next step is to find the sums about the new singularities, right? The singularity for the term on the left is $w=0$ and for the term on the right it is $w=-1$. The question is asking to find the series in the given region, where the $w=-1$ does not appear.

The rest of my attempt:

For $|w|<0$:

$$-\sum_{n=0}^\infty (-w+1)^n = -\sum_{n=0}^\infty (-z+3)^n$$

For $|w|>0$:

$$\sum_{n=0}^\infty \frac{1}{(-w+1)^{n+1}} = \sum_{n=0}^\infty \frac{1}{(-z+3)^{n+1}}$$

For $|w|<-1$:

$$-\sum_{n=0}^\infty (-w)^n = -\sum_{n=0}^\infty (-z+2)^n$$

For $|w|>-1$:

$$\sum_{n=0}^\infty \frac{1}{(-w)^{n+1}} = \sum_{n=0}^\infty \frac{1}{(-z+2)^{n+1}}$$

So what do I do about the $w=-1$ singularity?

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I'll continue from the step

$$\frac{1}{w}-\frac{1}{1+w}=\frac{1}{z-2}-\sum_{k=0}^{\infty} (-1)^k (z-2)^k, \quad 0<|z-2|<1 $$

which is the wanted Laurent series.

Note: we used the series expansion

$$\frac{1}{1+w}=\sum_{k=0}^{\infty}(-1)^k w^k, \quad |w|<1. $$

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  • $\begingroup$ Why couldn't anything be done with $\frac{1}{w}$? $\endgroup$ – whatwhatwhat Mar 16 '16 at 1:17
  • $\begingroup$ Just review the definition of Laurent series! $\endgroup$ – Mhenni Benghorbal Mar 16 '16 at 1:42
  • $\begingroup$ Oh ok so there's no point in trying to make it look like the standard $\frac{1}{1-z}$ form? I converted $\frac{1}{w}$ into $\frac{1}{1-(-w+1)}$, but maybe that was going the extra mile? $\endgroup$ – whatwhatwhat Mar 16 '16 at 1:47

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