140 第3章解析函数的级数理论与留数定理 (3)f)=ea,o=1 5.求下列函数在指定点为中心的一切可能的Laurent级数，并指出各自的收敛 环域： (0f@=e-2-2万'0=0： 1 (②f8=k+2,m=-2: (8@=2-Dw=1 6.求下列函数在复平面上的所有奇点，并判断每个奇点的类型： 2 (4④)fa=cos oma=2a片9@--ae-山 1 7.求下列函数在无穷远点邻域内的Laurent级数，并由此判别无穷远点作为孤 立奇点的类型： @=6 afo=2sin} 僩f@=a-e-2雪④f@=e 8.判别无穷远点是否是下列函数的孤立奇点？若是，判别孤立奇点的类型. 1 1 ④f)=+ (1)=z (⑧)fa=ecosz, (()=sin 9.求下列函数在复平面上所有奇点的留数： 140 1 3 Ÿ )¤ºÍ?ÍnÿÜ3ͽn (3) f(z) = e 1 1−zßz0 = 1 © 5. ¶eºÍ3ç½:è•%òÉåU Laurent ?Íßøç—àg¬Ò Ççµ (1) f(z) = z (z − 1)(z − 2) ßz0 = 0 ¶ (2) f(z) = 1 z(z + 2)3ßz0 = −2 ¶ (3) f(z) = 1 z 2 (z − 1) ßz0 = 1 . 6. ¶eºÍ3E²°˛§k¤:ßø‰zá¤:a.µ (1) f(z) = z 2 (z + 1)3 , (2) f(z) = 1 sin z , (3) f(z) = 1 − cosz z 2 , (4) f(z) = cos 1 z , (5) f(z) = ze − 1 z 2 , (6) f(z) = cosz z 4 + 8z 2 + 16 , (7) f(z) = z 3 sin 1 z , (8) f(z) = 1 e z − 1 − 1 z , (9) f(z) = (1 − cosz)(ez 2 − 1) z 4 . 7. ¶eºÍ3ð:çS Laurent ?ÍßøddOð:äè ·¤:a.µ (1) f(z) = z 2 + 1 z − 6 , (2) f(z) = z 2 sin 1 z , (3) f(z) = 1 (z − 1) (z − 2) , (4) f(z) = 1 z 2 e z . 8. Oð:¥ƒ¥eºÍ·¤:º e¥ßO·¤:a.. (1) f(z) = 1 sin z , (2) f(z) = 1 e z − 1 − 1 z , (3) f(z) = 1 1 + cosz , (4) f(z) = z 2 + 4 z , (5) f(z) = 1 (z − 1)(z − 2) , (6) f(z) = z 2 sin 1 z , (7) f(z) = 1 z 2 e z (8) f(z) = e z cosz, (9) f(z) = e z sin 1 z . 9. ¶eºÍ3E²°˛§k¤:3͵