I'm asked to evaluate the following integral: [itex]\int_{c} \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}dz[/itex] where c is the unit circle. This function has a simple pole at [itex]z=\frac{1}{3}[/itex] and a second order pole at [itex]z=\frac{1}{2}[/itex], both of which are within my region of integration. I then went about computing the residues using the following formula (which I, mistakenly perhaps, thought was general and applied to any mth-order pole): [itex]Res f(z) = \frac{1}{(1-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[2z-1]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}[/itex] for which I got 2 (this was just the residue for the first pole). The residue I got for the other singularity was 6. They are supposed to be 1/2 and 2, respectively. What am I doing wrong, exactly? Does this residue formula not apply to this case? If so, why not? Thanks in advance.(adsbygoogle = window.adsbygoogle || []).push({});

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# Complex analysis: residue integration question

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