Res[(z^2)/(z^2+1),i]求留数

carol_lu2022-10-04 11:39:540条回答

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求留数Res tanπz,z=1/2
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tanπz= tan{π*(z-1/2)+π/2}=-cotπ*(z-1/2)
=-cosπ*(z-1/2)/sinπ*(z-1/2)
z=1/2是分母的一级零点,是tanπz的一级极点,分母单独求导后求极限得:-1/π