复变函数 Res f(=)=lim[(z-1f(=I 4 Res()=m[(=+1)(2) 4 (2)f(z)=e2-,z=1 f(=)=1+++…=1+ 12 z-12(z-1) →Resf(=)=1  1 1 2 1 1 1 ( ) lim ( 1) ( ) 4 1 ( ) lim ( 1) ( ) 4 z z z z Res f z z f z Res f z z f z = → = − → − = − =  = + = −     1 1 (2) ( ) , 1 z f z e z − = = 2 2 1 1 1 ( ) 1 1 1 2! 1 2!( 1) ( ) 1 z u u f z z z Res f z = = + + + = + + + − −  =