考察下列函数在0<|z-x0内的洛朗级数 (1)f(z) 7,30=0(2)f(x) (1-z) (3)f(x)=e2,xo=0 解( (e-1)=-∑ z n=in. =(1-z)+1+(1-z)+…+(1-z) z(1-x) (3)e2=1+-+ 2!z23!z3 n. z0 考察下列函数在 < 内的洛朗级数 0 | | . z z −   0 1 (1) ( ) , 0 z e f z z z − = = 0 1 (2) ( ) , 1 (1 ) f z z z z = = − 1 0 (3) ( ) , 0 z f z e z = = 1 1 1 1 1 (1) ( 1) ! z z n n e e z z z z n  = − 解 = − =  1 1 1 (2) (1 ) 1 (1 ) (1 ) (1 ) n z z z z z − − = − + + − + + − + − 1 2 3 1 1 1 1 (3) 1 2! 3! ! z n e z z z n z = + + + + + +