Với mục đích trang bị cho sinh viên kỹ thuật ngành Điện tử Viễn thông các kiến thức cơ bản về toán áp dụng trong kỹ thuật, Khoa Điện tử Viễn thông Trường đại học Bách khoa Hà Nội đã xâydựng và đề xuất chương trình khung môn Toán kỹ thuật. | THE CAUCHY-GOURSAT THEOREM Tn the previous section we showed how to evaluate line integrations by brute-force reduction to real-valued integrals. In general this direct approach is quite difficult and we would like to apply some of the deeper properties of complex analysis to work smarter. In the remaining portions of this chapter we will introduce several theorems that will do just that. If wo scan over the examples worked in the previous section we see considerable differences when the function was analytic inside and on the contour and when it was not. We may formalize this anecdotal evidence into the following theorem Cauchy-Goursat theorem5 Lei f z be analytic in a domain D and let c be a simple Jordan curve3 inside D so that f z is analytic on and inside of c. Then fc f z dz 0. Proof Let c denote the contour around which we will integrate w f z . We divide the region within c into a series of infinitesimal rectangles. Sec Figure . The integration around each rectangle equals the product of the average value of w on each side and its length ớw fal . r . dw ỠỈL tZ w l . . r dĩ 2 Jdx ắè 2 d ty dwdx dw z t f dw x r di 2 flg r ôSĩ 2 È-ẫ 151 dx idy J Substituting w u 4- iv into dw dw _ du dv fdv du dx idy dy ch dyj Because the function is analytic the right side of 1 and equals zero. Thus the integration around each of these rectangles also equals zero. 28 We note next that in integrating around adjoining rectangles we transverse each side in opposite directions the net result being equivalent to integrating around the outer curve c. We therefore arrive at the result cf z dz 0 where f z IS analytic within and on the closed contour. The Cauchy-Goursat theorem has several useful implications. Suppose we have a domain where f z is analytic. Within this domain let us evaluate a line integral from point A to B along two different contours Cl and c . Then the integral around the closed contour formed by integrating along Cl and then back along C2 .
