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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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    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.2012-07-17

2 Answers 2

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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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    Yes. | Edit: Now no. | Edit: Now yes.2012-07-17
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$\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\left(\frac{1}{z-3}-\frac{1}{z-\frac{1}{2}}\right)=-\frac{1}{10}\frac{1}{1-\frac{z-1}{2}}-\frac{1}{5(z-1)}\frac{1}{\left(1+\frac{1}{2(z-1)}\right)}$

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