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So i understand everything with regards to how to find the laurent series as was detailed in this post already:

Finding the Laurent series of $f(z)=1/((z-1)(z-2))$

My question extends on this in how would i find the laurent series not about the origin. For example how about around $$z_0 = 2$$ ? Would that change any of my calculations?

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D.C. the III
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1 Answers1

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

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

With the latter we use that $\frac{1}{1-x}=1+x+x^2+...$

We must add the series for $\frac{-1}{z-1}$ with the term $\frac{1}{z-2}$. We get a Laurent series. It can't be a regular Taylor since $z=2$ is a singularity.

The series we obtain this way converges for $0<|z-2|<1$. It is an annulus (when viewed in the complex plane), in the reals it would be a pair of intervals (the intersection of an annulus with the real line).

By a small change in the work we could also obtain a Laurent series that converges instead for $|z-2|>1$. The idea is to write

$$\frac{-1}{z-1}=\frac{-1}{(z-2)+1}=\frac{-1}{z-2}\frac{1}{1-\frac{1}{z-2}}$$ and use with the second factor that $\frac{1}{1-x}=1+x+x^2+...$. Thisgives us a Laurent series that have infinitely many terms with negative powers of $(z-2)$.

Karanko
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  • So my series would be $$\sum (-(z-2))^n $$ ? Would my conditions for convergence change? I.e: for the annulus aroulnd zero it was $$ 1 < |z| < 2 $$ would it change to $$ 1 < |z-2| < 2 $$? I ask because i am having trouble getting the coeffcients to work – D.C. the III Apr 05 '15 at 23:34
  • @dc3rd I will answer above. – Karanko Apr 05 '15 at 23:41
  • i am still a little confused on the coeffcients. So the series will be representsed with the sum of the decomposition of $$\frac {-1}{(z-1)} \ added\ to\ \frac {1}{(z-2)} $$ correct? But im having issues with pulling out coefficients, especially the residue. Sorry for the reptetiveness – D.C. the III Apr 05 '15 at 23:57
  • @dc3rd Yes, we need to add those two. If you need the residue we need the first series we computed that converges in $0<|z-2|<1$. For that one we only need the term in which $(z-2)$ appears to power $-1$. That is $\frac{1}{z-2}$. The series of $\frac{-1}{z-1}$ gives only non-negative powers. – Karanko Apr 06 '15 at 00:01