解答及评分标准 (30分) 1+-1=+y=2在除2=0外的扩充的全平面(包括∞点)上可导、解析 可导区域 解析区域 (2分) 导数公式 (2分) (6分) 3.2=1是f(x)=cos-的本性奇点,resf(1)=-sin1 (3分) 2=∞是f(2)=cs2-1的解析点,res(x)=sin1 (3分) z=0和z=∞是mnz的枝点,沿着正虚轴从z=0到z=∞作割线, 规定ll=1=0,则 作割线、规定单值分枝 (1分) z=1是可去奇点,resf(1)=0 (2分) z=k,k=-1,±2,±3,……是一阶极点 (1分) In k, ln|k-(-1)i,k=-2,-3,-4, 规定l2l11=2nmi,n=±1,土,±3,…,则 z=k,k=±1,±2,±3,…都是一阶极点, k=1 Ink+(-1 nk|+(-1)(2 0是方程 dz2 出-m=0的唯一正则奇点 (3分) 方程2d2+-=0在2=0点的指标为p2=p2=0 (3分) 25分) 解一般解为 u(x,t)=Cot+Do+∑ Do+>Dn cos cOS➪➶➹➘➴➷➬ ✱✲ ↕30 ✶ ➙ 1. x x 2 + y 2 − i y x 2 + y 2 = 1 z ❋➮ z = 0 ➱ P✃❐P➡òó ↕➡➢ ∞ ❲ ➙❸ ➭✦✲▼◆✛  1 z 0 = − 1 z 2 ➭✦ ❏❑ (2 ✶) ▼◆❏❑ (2 ✶) ✦ ✸ ✠✡ (2 ✶) 2. I |z|=2 cos z z 3 dz = −πi (6 ✶) 3. z = 1 ✺ f(z) = cos z z − 1 P ✔❒è❲✛ resf(1) = − sin 1 ❄ (3 ✶) z = ∞ ✺ f(z) = cos z z − 1 P▼◆❲✛ resf(∞) = sin 1 (3 ✶) 4. z = 0 ❳ z = ∞ ✺ ln z P ➬ ❲✛ û❮⑩❺❷❰ z = 0 ➾ z = ∞ ➘ ➴❯✛ ➆➇ ln z   z=1 = 0 ✛❨ ➘ ➴❯✲➆➇➷ ✼✶➬ (1 ✶) z = 1 ✺ ➭Ïè❲✛ resf(1) = 0 ✛ (2 ✶) z = k, k = −1, ±2, ±3, · · · ✺✱úå❲✛ (1 ✶) resf(k) =    (−1)k π ln k, k = 2, 3, 4, · · · i k = −1 (−1)k π ln |k| − (−1)k i, k = −2, −3, −4, · · · (1 ✶) (1 ✶) ➆➇ ln z   z=1 = 2nπi, n = ±1, ±2, ±3, · · · ✛❨ z = k, k = ±1, ±2, ±3, · · · ➨✺✱úå❲✛ resf(k) =    −2ni k = 1 (−1)k π ln k + (−1)k2ni, k = 2, 3, 4, · · · −(2n − 1)i k = −1 (−1)k π ln |k| + (−1)k (2n − 1)i, k = −2, −3, −4, · · · 5. z = 0 ✺ ➞➤ z d 2w dz 2 + dw dz − w = 0 PÐ✱⑩ ❨è❲✛ (3 ✶) ➞➤ z d 2w dz 2 + dw dz − w = 0 ❋ z = 0 ❲P➥➦❖ ρ1 = ρ2 = 0 (3 ✶) ⑦✲ ↕25 ✶ ➙ Ñ ✱Ò▼ ❖ u(x, t) = C0t + D0 + X∞ n=1  Cn sin nπ l at + Dn cos nπ l at cos nπ l x, u   t=0 = D0 + X∞ n=1 Dn cos nπ l at cos nπ l x = x, 19