总复习一V 例题
总复习－V 例题
1.指出|-a+-b=c在复平面上轨迹 a-b=c线段 a-b>c不存在
1. 指出 z a z b c − + − = 在复平面上轨迹； a b c − a b a b c − a b c − = 线段 不存在
2.∞()=x4-y2+(x2+x3y)在|<5是不是解析函数 02u=12( )≠0 Ou 4x 2(xy+x2) 3计算积分g2 丌l 2 d 丌l l 丌l 2∈l|=2e解析 2ie2=2丌
2. ( ) ( ) z x y i xy x y = − + + 4 4 2 2 在 是不是解析函数 4 4 a. u x y = − 2 2 2 2 2 2 12( ) 0 u u x y x y + = − z 5 b. 3 2 4 , 2( ) u v x xy x x y = = + 3. 计算积分 2 2 z z e dz i z = + 2 2 ( ) 2 2 z z z z e e dz dz i i z z = = = + − − 2 2 i z − = 2 2 2 i ie − = = z e 解析
2z-1 dz z+z-1 dz dz 2(2 2(2 2-(z dz tIrEs(o)+res(l) =3z( Re(0) Re(1) (-2) dz z=1 dz 4丌 2(2
4. 2 3 3 2 1 ( 1) z z dz z z = − − 2 3 3 2 2 3 3 3 1 ( 1) ( 1) ( 1) z z z z z dz dz dz z z z z z z = = = + − = = + − − − 3 3 2 [Re (0) Re (1)] ( 1) z dz i s s z z = = + − 3 0 1 Re (0) 1 ( 1) z s z = = − = − 2 3 2 1 1 1 1 1 Re (1) ( ) ( 2) 1 2 2 z z d s z dz z − = = − = − = = 3 3 4 ( 1) z dz i z z = = − −
dz tIrEs(o)+res()l Re(0) Re(1)=2(x2)21=(-2)=3 3 2z-1 dz=-4Ti
2 2 3 2 [Re (0) Re (1)] ( 1) z dz i s s z z = = + − 2 3 0 0 1 1 Re (0) 2 2 ( 1) ( 1) z z d s dz z z = = − = = = − − 2 3 Re (1) ( ) ( 2) 2 z z 1 1 d s z z dz − − = = − = − = = 2 2 3 0 ( 1) z dz z z = = − 2 3 3 2 1 4 ( 1) z z dz i z z = − = − −