=另1) 02+1(3-2) 0z-21k2) f)=}离展开点z=2最近的孤立奇点为z=0,它们 之间的距离为2,故收敛半径R=2. 5、把f(z)=n(1+z)在z=0邻域内展成泰勒级数. 解:-1t2-:0水k川 两边对2积分,得:f=2"+c=2-+ k0k+11 k=1 k 当z=0时,c=f(0)=ln1=lne2mr=i2nr,n=0,±1,±2,. 21 4P2121 1 0 ( 1) ( 2) 2 k k k k z + = − = − (| 2 | 2) z − 离展开点 最近的孤立奇点为 ,它们 1 f z( ) z = z = 2 z = 0 之间的距离为2,故收敛半径R=2. 5、把 在 邻域内展成泰勒级数. f z z ( ) ln(1 ) = + z = 0 解: 0 1 '( ) ( 1) 1 k k k f z z z = = = − + (| | 1) z 两边对z积分,得: 1 1 0 1 ( 1) ( 1) ( ) 1 k k k k k k f z z c z c k k − + = = − − = + = + + 当z=0时, 2 (0) ln1 ln 2 , i n c f e i n = = = = n = 0, 1, 2