第二章 复变函数的积分.pdf

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Ҕ 
 b n n
f (x)dx= f (xi )∆xi f (z)dz= f (zi )∆zi
∫a lim∑∫C lim∑
∆xi →0 i=1 ∆zi →0 i=1
b b
cf(x)dx=c f x)( dx cf(z)dz=c f z)( dz
∫a ∫a ∫C ∫C
b b b

 [f +g]dx= fdx+ gdx [f +g]dz= fdz+ gdz
∫a ∫a ∫a ∫C ∫C ∫C
b a
f (x)dx=− f x)( dx f (z)dz=− f z)( dz
∫a ∫b ∫C ∫C
c b b
fdx+ fdx= fdx fdz+ fdz= fdz
∫a ∫c ∫a ∫C1 ∫C2 ∫C1∪C2
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Ҕ
ˤI
}C f(z) dz = }C(u +iv)(dx +idy)
= }C(udx -vdy)+i }C(udy+vdx)
ˤII
}C f(z) dz = }C(u +iv)(ei
dr +i r ei
d
)
= }C ei
[(udr -vrd
)+i(urd
+vdr)]

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၏}Czdz O̤B

 zdz = zdz + zdz
∫OAB ∫OA ∫AB
1 2
= iyd(iy) + (x i)d(x ++ i)
∫0 ∫0
10
1 1 3
= −+ ( × 22 + i × 2) = + 2i
2 2 2
2 x x
zdz = (x + i )d(x + i )
∫OB ∫0 2 2
2
= 1( + 1 i 2 xdx = 3 + 2) i
2 ∫0 2
zdz = zdz + zdz = 3 + 2i
∫OCB ∫OC ∫CB 2

2
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၏}C z2dz O̤B



z 2dz = z 2dz + z 2dz
∫OAB ∫OA ∫AB
1 2
= (iy)2 d(iy) + (x i)2 d(x ++ i)
∫0 ∫0
1
= (2 +11i)
3
2 x x
z 2dz = (x + i )2 d(x + i )
∫OB ∫0 2 2
2
= (1+ 1 i) 3 1 (2 += ixdx