复变函数 Res f(z)=Res f(z)=-l 2=00 (3)f(z)=z"sin-,z=0 f(z)仅有z0及无穷远点两个孤立奇点,相对而 ,z=0处的残数较无穷远点处的残数好求,故 m=2k+1 Res f()=-Resf(=)=(1) m=2k (2k+1)1 ( ) ( ) 1 z z Res f z Res f z = = = − = − 1 (3) ( ) sin , m f z z z z = =  f(z)仅有z=0及无穷远点两个孤立奇点，相对而 言，z=0处的残数较无穷远点处的残数好求，故 1 0 0, 2 1 ( ) ( ) ( 1) , 2 (2 1)! k z z m k Res f z Res f z m k k + = =  = +  = − =  − =   +