Contour integral of f z is independent of path in d

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contour integral of f ( z ) is independent of path in D . Proof: Refer to the following figure, and take the contour integral positive in counterclockwise in the domain D , Form theorem 1, we have We got the proof of theorem 2, that is the contour integral of f ( z ) is independent of path in D , if f ( z ) is analytic in the simply connected domain D . 1 2 C C dz z f dz z f ) ( ) ( A 2 C y x B D 1 C D C C enclosing contour closed a forms path 2 1 1 2 C C dz z f dz z f ) ( ) ( 2 1 C C C dz z f dz z f ) ( ) ( 0 C dz z f ) ( 0 1 2 C C dz z f dz z f ) ( ) ( . to from paths the are and where 2 1 B A C C
Theorem 3: In a doubly connected domain d with outer boundary C 1 and inner C 2 , those are simple closed paths ( in same direction ), as shown in the following figure, if f ( z ) is analytic in any domain D that contains d and its boundary curves (including C 1 and C 2 ), we have (57) 1 C dz z f ) ( 1 C 2 C d 2 C dz z f ) ( , ) ( , ) ( 0 0 0 at is but at is Supposing z analytic NOT z z z f z analytic z f . z z z f z 0 0 function the of pole) (or the y singularit the called is where ) ( closed for the function that of integral path closed the with deal we could How ? ) ( pole) (the y singularit the enclosing path 0 0 C dz z z z f z C
Let f ( z ) be analytic in a simply connected domain D , then for any point z 0 in D and any simple closed path C taken counterclockwise in D that encloses z 0 , we have Proof: we focus on the region bounded by simple closed path C , and make a circle C 1 of radius with center at z 0 as another closed path (taken counterclockwise ). From theorem 2, we have Then we just need to evaluate the integral along the circle C 1 Cauchy Integral Formulas (58) dz z z z f dz z z z f C C 1 0 0 ) ( ) ( C dz z z z f 0 ) ( C 0 z 1 C D C 0 z dz z z z f C 1 0 ) ( ) exp( , i z z 0 16 example refer to ) ( 0 z f π i 2 1
the integral (in parametric form ) now becomes Taking the limit , we obtain hence C 0 z 1 C d e i e e z f i i i 2 0 0 ) ( π C d d dz z z z f dz z z z f 2 0 0 0 1 ) ( ) ( ) ( d dz i z z and For 0 ), exp( 2 0 0 d e z f i i ) ( ) ( ) ( 0 0 2 1 z if dz z z z f C d e z f i i ) ( lim 2 0 0 0 0 . ) ( ) ( C dz z z z f i z f 0 0 2 1 dz z z z f C 0 ) ( ) ( 0 2 z π if 0 at integral the evaluate try to now we z z ) exp( i i
Example 20: Directly evaluate the integrals by the Cauchy Integral Formula, (a) , (b) . Solution : (a) For any contour C enclosing z 0 ( z 0 = 2 ), we have For any contour C for which z 0 = 2 lies outside , the function becomes analytic , the integral, from theorem 1, is equal to zero . dz z e C z 2 i ie ie z z 4268 46 2 2 2 2 . dz i z z C 2 6 3 dz z e C z 2 2 2 z z ie C 0 z 1 C
Solution : (continued) (b) For any contour C enclosing , the integral can be evaluated by (58) Otherwise (the point z 0 lies outside any contour , the function will be analytic

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