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Unformatted text preview: Examples: • γ ( t ) := t +3 it , t ∈ [0 , 2], parametrizes a segment of the line Im z = 3Re z . • γ ( t ) := 4 cos t + 4 i sin t , t ∈ [0 , 2 π ), parametrizes the circle  z  = 4. Notice that the parametrization for a given curve C is not unique. We denote by − C the curve oriented in the opposite direction: − C := { γ ( a + b − t )  t ∈ [ a, b ] } . (675) The arc length of a curve C is defined by L ( C ) := sup a = t <t 1 < ··· <t n 1 <tn = b n ∈ N n summationdisplay m =1  γ ( t m ) − γ ( t m − 1 )  . (676) The arc length is an intrinsic property of the curve C , i. e. it does not depend on the choice of the parametrization γ . The curve C is rectifiable if L ( C ) < ∞ . This is true for Lipschitzcontinuous functions γ , for example. Definition of the line integral For a given n ∈ N , we choose n +1 points in the interval [ a, b ], a = t < t 1 < ··· < t n = b. (677) With the mapping γ we obtain n + 1 points z m := γ ( t m ) ∈ C , m = 0 , . . . , n . Now we can divide the curve C into n segments: C = n uniondisplay m =1 C m , C m = { γ ( t )  t ∈ [ t m − 1 , t m ] } , m = 1 , . . . , n. (678) Consider a rectifiabe curve C . For a continuous complex function f which is defined (at least) at each point z ∈ C , and for any choice of points ζ m ∈ C m , m = 1 , . . . , n , we define the sums S n := n summationdisplay m =1 f ( ζ m ) ( z m − z m − 1 ) . (679) 113 Remark: S n depends on the partition of the interval [ a, b ] and on the choices of ζ 1 , . . . , ζ n . The line integral of f over the path of integration C is defined by integraldisplay C f ( z ) dz := lim n →∞ S n . (680) Under the assumptions that f is continuous and C is rectifiable, the limit on the righthand side of (680) exists. It turns out that it is independent of the choice of points ζ m , m = 1 , . . . , n , and also that the line integral is independent of the choice of the parametrization γ of C . If the curve C is closed, γ ( b ) = γ ( a ), then we also write contintegraldisplay C f ( z ) dz. (681) Basic Properties 1. Linearity: integraldisplay C ( k 1 f 1 ( z ) + k 2 f 2 ( z )) dz = k 1 integraldisplay C f 1 ( z ) dz + k 2 integraldisplay C f 2 ( z ) dz. (682) 2. Sense reversal: integraldisplay − C f ( z ) dz = − integraldisplay C f ( z ) dz. (683) 3. Partitioning of path: integraldisplay C f ( z ) dz = integraldisplay C 1 f ( z ) dz + integraldisplay C 2 f ( z ) dz (684) Evaluation via antiderivative This evaluation method requires a com plex function which is holomorphic on a simply connected domain....
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This note was uploaded on 11/17/2011 for the course MATH 529 taught by Professor Staff during the Spring '08 term at UNC.
 Spring '08
 Staff
 Math

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