z=1泰勒展开 解:结果:∑a1(-1,z-1k2 首先应分项分式 (+11=21.2 (z+1)2(z+1 + z+12-1+22 ∑(-1)(三 又 (z+1) 2+ 2 dz ∑(-1) 2∑(1)k(2 k=0 n+l n0 2 2 2 2 2 0 2 2 2 ( -1) ,| -1| 2 ( 1-1) 2 1 1- ( 1 3. 1) ( 1) ( 1) 1 1 1 1 1 -1 (-1) ( , 1, ) 1 -1 2 2 1 -1 2 2 2 1 1 - ( 1 1) 1 ( ) k k k k k k a z z z z z z z z z z z z z z z z z ¥ = ¥ = å < + = = + + + + + = = = å + + + ¢ é ù = ê ú ë + + = + û Q 解: 果: 首先 泰勒展 分 。 分式: 又 开 结 应 项 å å å ¥ = + ¥ = - ¥ = - - = - - - = - - × - - - = 0 1 0 1 0 ) 2 1 ( 2 1 ( 1 ) 2 1 ) 2 1 ( 2 1 ( 1 ) 2 1 ) 2 1 ( 1 ) ( 2 1 n n n k k k k k k n z z k z dz d