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experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0man ... I stuck at finding the simple residue.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0if my memory serves me, it is the numerator divided by the derivative of the denominator evaluated at the pole

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0yeah you are correct ... had trouble fixing. \[ \huge \lim_{z \rightarrow e^{i \pi \over b}} {(z  e^{i{\pi }\over b}) z^{a1} \over 1 + z^b} = \lim_{z \rightarrow e^{i \pi \over b}} {z^{a1} \over bz^{b1}}\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1346858176216:dw somehow it got this fixed. I had this Q on my exam paper ... Man i couldn't do it.
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