(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find a Laurent Series of f(z)=[itex]\frac{1}{(2z-1)(z-3)}[/itex] about the point z=1 in the annular domain [itex]\frac{1}{2}<|z-1|<2[/itex].

2. Relevant equations

3. The attempt at a solution

By partial fraction decomposition, [itex]f(z)=\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\frac{1}{z-3}-\frac{2}{5}\frac{1}{2z-1}[/itex].

When [itex]\frac{1}{2}<|z-1|<2, \frac{(z-1)}{2}<1[/itex].

Hence: [itex]\frac{1}{z-3}=-\frac{1}{3-z}=-\frac{1}{3-z+1-1}=-\frac{1}{2-(z-1)}=-\frac{1}{2}\frac{1}{1-\frac{(z-1)}{2}}[/itex]

Therefore: [itex]\frac{1}{z-3}=-\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}[/itex]

And: [itex]f(z)=-\frac{1}{5}\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}-\frac{2}{5(2z-1)}.[/itex]

Is this complete? Is there something I could similarly do with the other term? Ive tried but havent been able to find anything that works.

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# Homework Help: Laurent Series Expansion Centered on z=1

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