# Computing a Laurent series

Let $$f(z) = \frac{1}{(2z-1)(z-3)}$$. Compute the Laurent series about the point z = 1 in the annular domain $$\frac{1}{2} < |z-1| < 2$$

My attempt: I broke f(z) up into the partial fraction decomposition:

$$-\frac{2}{5(2z-1)} + \frac{1}{5(z-3)} = -\frac{2}{5}*\frac{1}{(1-\frac{(z+\frac{1}{2})}{2})} +\frac{1}{5}*\frac{1}{1-(z-2)} =$$

$$-\frac{2}{5}\sum_{n=0}^\infty(-1)^{n}\frac{(z+1)^{n}}{2^n}-\frac{1}{5}\sum_{n=0}^\infty(z-2)^n$$

And that was my answer. But I was told I was wrong, and I'm not sure where I went wrong in there. So if someone could point out where I went wrong, it would be greatly appreciated!

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1
You have a Laurent series around $z=-1$ plus a Laurent series around $z=2$. – anon Jul 17 '12 at 3:40
A) You were asked for a series with powers of z-1. B) The geometric series formula requires the ratio to be less than 1 in absolute value. – user31373 Jul 17 '12 at 3:41
So for A) I should manipulate my partial fractions so that I have z-1, but what should I do for B)? I'm not sure how I could make the ratio less than 1 in absolute value. – Rand Jul 17 '12 at 3:50

## 2 Answers

Writing $w=z-1$, we have by partial fractions

$$\begin{array}{c l}\frac{1}{(2z-1)(z-3)} & =\frac{1}{(2w+1)(w-2)} \\ & =\frac{1}{5}\left(-\frac{2}{2w+1}+\frac{1}{w-2}\right)\\ & =-\frac{2/5}{1+2w}-\frac{1/10}{1-w/2}.\end{array}$$

And $|-2w|,|w/2|<1\iff \frac{1}{2}<|w|<2$. Do you see how this works out?

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I think so. So we would have:  $$-\frac{2}{5}*\frac{1}{1-(-2(z-1))} - \frac{1}{10}*\frac{1}{1-\frac{z-1}{2}} = -\frac{2}{5}*\sum_{n=0}^\infty [(z-1)^{n}(-2)^{n}] - \frac{1}{10}*\sum_{n=0}^\infty[\frac{(z-1)^{n}}{2^{n}}]$$and then I just combine the two series. – Rand Jul 17 '12 at 4:05
Yes. | Edit: Now no. | Edit: Now yes. – anon Jul 17 '12 at 4:08

$$\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\left(\frac{1}{z-3}-\frac{1}{z-\frac{1}{2}}\right)=-\frac{1}{10}\frac{1}{1-\frac{z-1}{2}}-\frac{1}{5(z-1)}\frac{1}{\left(1+\frac{1}{2(z-1)}\right)}$$

Now only check that $$\left|\frac{z-1}{2}\right|<1\,\,,\,\,|2(z-1)|^{-1}<1$$ and you'll be able to use the developments $$\frac{1}{1-z}=1+z+z^2+...=\sum_{n=0}^\infty z^n\,\,,\,\,\frac{1}{1+z}=1-z+z^2-...=\sum_{n=0}^\infty(-1)^{n}z^n$$

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