《数学物理方程》课程PPT教学课件（讲稿）总复习－V 例题

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  = − = − − 

∑ 的收敛半径 R=lim --=lim -=0 n→)00 n→)0 n+1,n+1 R=lim一/ (n+1) -lim nn”(n+1)”1m→0n+1n

5. n n  n z 的收敛半径 1 1 lim lim 0 n n n n R → → n n = = = ! n n n z n  1 ! ( 1)! 1 1 lim / lim ( ) ( 1) 1 n n n n n n n n n R e n n n n → → + + + + = = = + +

在除z=1的区域的罗朗级数 (=-1e=(=-1)e=(-2∑s(c-1 n=0!(-1)n+2 ∑ ∑ 1=-00 (-n+2)!

6. 1 2 1 ( 1) z z e − − 在除z=1的区域的罗朗级数 1 1 2 2 2 1 1 0 ( 1) ( 1) ( 1) ( 1) ( 1) ! n z z n n z e z e z z n − − − − = − − = − = − −  2 2 0 ( 1) ( 1) ( 1) ( 1) ! ( 2)! n n n n n n z z n n − + = =− − − = − = − − +  

7.计算 +16 +162J∞x4+242J∞(x2+4)x2-4) d x 2∞(x2-4 /2 )(x x/2 2J∞(x-2e4)x+2em4)(x-2e/)(x+2eH 上半平面极点 /4 /4 2riRe sf(2e)+Re sf (2e//) aril 4(2etn/4)3+ 丌i e 4(2 i3/4 32 32

7. 计算 4 0 16 dx x  +  4 4 4 2 2 0 2 / 2 2 / 2 / 4 / 4 / 4 / 4 1 1 16 2 2 2 ( 4 )( 4 ) 1 2 ( 4 )( 4 ) 1 2 ( 2 )( 2 )( 2 )( 2 ) i i i i i i dx dx dx x x x i x i dx x e x e dx x e x e x e x e          − −  − −  − − − = = = + + + − = = − − = − + − +      上半平面极点 / 4 3 / 4 2 [Re (2 ) Re (2 )] i i i sf e sf e   = +  / 4 / 4 / 4 3 3 / 4 3 1 1 1 1 2 2 [ ] [ ] 2 4(2 ) 4(2 ) 32 32 i i i i i i e e e e        − = + = − + =

7.已知p(p)=L(),求 e ar e"o(n)的像函数 e“(z)dlz a(t-T O(rdx]=le a jo(p) o(p) 0 p+ 卷积 8求解asim=0)- So sin a(t=)ndr L[asin ot]=L[p(D)]-LlL sino(t-tp(r)dr p2+o2(以、00()a=[(p2+o2)-op(p) O O O ao 0-O (p) P(t) sintvo-a P+-√o2-0p2+(y2-) √-O o(p) P(t=at

7. 已知   ( ) [ ( )] p L t = ，求 0 ( ) t at a e e d     −  的像函数 0 [ ( ) ] t at a L e e d     −  卷积 8. 求解 0 sin ( ) sin ( ) ( ) t a t t t d        = − −  0 [ sin ] [ ( )] [ sin ( ) ( ) ] t L a t L t L t d        = − −  2 2 2 2 ( ) ( ) a p p p p       = − + + 2 2 a p p     = + − [( ) ] ( ) 2 2 2 2 2 2 2 ( ) ( ) a a p p p            − = = + − − + − 2 2 ( ) sin a t t       = − − ( ) 0 ( ) [ ( ) ] [ ] ( ) t a t at p L e d L e p p a       − − − = = = +   =1 2 ( ) a p p  = ( )t at =

9.长/细干两端恒温零度,初始时温度分布为 0<x< <x≤l 求温度。 -nat n元x L.已 SIn 0<x in <x<l

9. 长 l 细干两端恒温零度，初始时温度分布为 0 0 0 0 2 2 t x l T x l u l x l T x l l =     =  −     求温度。 2 2 0 sin n a t n n n x u a e l −  = = 0 0 0 0 0 2 sin 2 n n t x l T x n x l a l l x l T x l l  = =     =  −     