auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING Engineering Math EEE 3640 1
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING Statement of the formula (without proof) For f(z) analytic in a domain D then it has a derivative in D which is also analytic in D. The values of the derivative at a point z 0 in D is given by the formula: f n z 0 = ර here is any simple closed path in D that encloses z 0 and whose full interior belongs to D and the integration is taken counterclockwise around. Engineering Math EEE 3640 2
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING Although the formula is stated as a method for finding the derivative, it is actually very useful for finding the integral of many analytic functions. When rearranged as follows, you can see that finding the integral of an analytic function is transformed into an exercise of finding the derivative: ර n+1 dz = f n z 0 The challenge is to manipulate a given function until it has the form of the integral on the left, determine values for, z 0, and n and then find the derivative(s) of. Engineering Math EEE 3640 3
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING ර n+1 dz = f n z 0 The expression on the right is ර n+1 dz = f n z 0 Engineering Math EEE 3640 4
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING ර n+1 dz = f n z 0 The expression on the right is interpreted as follows: f n z 0 = dn f(z) dz n ቤ z = z 0 Engineering Math EEE 3640 5
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING Example: Evaluate ර sin 7z+e 1.5z z 2 4 dz, z 3i = 5 (W) First, we observe that the root of the denominator (z = 2) does lie within the contour of integration, so we can not just say this integral equals zero. Further observation shows that this integrand fits the form for auchy s Integral Formula by substituting: n + 1 = 4 n = 3 = sin 7z + e 1.5z z 0 = 2 Engineering Math EEE 3640 6
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING n = 3 = sin 7z + e 1.5z z 0 = 2 Put these values into the formula to get: sin 7z + e 1.5z z 2 4 dz = ቤ 3! f 3 z dz To proceed, we must find the third derivative of z=2 Engineering Math EEE 3640 7
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING = sin 7z + e 1.5z f 1 f 2 f 3 z = 7 cos 7z + 1.5e 1.5z z = 49 sin 7z + 2.25e 1.5z z = 343 cos 7z + 3.375e 1.5z Engineering Math EEE 3640 8
DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING So, the final result is: ර sin 7z + e 1.5z z 2 4 auchy s Integral Formula = 3! 343 cos 7z + 3.375e1.5z อ z = 2 = 3! 343 cos 7 2 + 3.375e 1.5 2 = 6.28318i 6 343 0.13674 + 3.375 20.08554 = 1.04720i 46.90087 + 67.78869 = 21.87367i Engineering Math EEE 3640 9
DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING Example: Evaluate ර auchy s Integral Formula z 6 2z 1 6 dz, z = 1 Get it in the right format: z 6 6 dz = z 6 6 dz = z 6 64 6 dz = 1 64 z 6 6 dz 2 z 1 2 2 6 z 1 2 z 1 2 z 1 2 This fits the form for auchy s Integral Formula for: = z 6, z 0 = 1 2, n + 1 = 6 n = 5 Engineering Math EEE 3640 10
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING n = 5 = z 6 z 0 = 1 2 Put these values into the formula to get: z 6 2z 1 6 dz = 1 64 5! z 6 5 ቮ z= 1 2 To proceed, we must find the fifth derivative of Engineering Math EEE 3640 11
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING = z 6 f 1 z = 6z 5 f 2 z = 30z 4 f 3 z = 120z 3 f 4 z = 360z 2 f 5 z = 720z Engineering Math EEE 3640 12
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING So, the final result is: z 6 2z 1 6 dz = 1 64 5! 720zቚ z= 1 2 = 6.28318 120 720 64 1 2 i = 0.29452i Engineering Math EEE 3640 13
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING Example: Evaluate cosh 2z z i 2 4 dz, z = 1 This fits the form for auchy s Integral Formula for: n + 1 = 4 n = 3 = cosh 2z z 0 = i 2 Engineering Math EEE 3640 14
DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING n =3 = cosh 2z auchy s Integral Formula z 0 = i 2 Put these values into the formula to get: cosh 2z z i 4 dz = cosh 2z 3! ቤ z= i 2 2 To proceed, we must find the third derivative of cosh 2z Engineering Math EEE 3640 15
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING = cosh 2z f 1 f 2 f 3 z = 2sinh 2z z = 4cosh 2z z = 8sinh 2z Engineering Math EEE 3640 16
auchy s Integral Formula DEPARTMENT OF ELETRIAL AND OMPUTER ENGINEERING So, the final result is: cosh 2z z i 2 4 dz = 8 sinh 2zቚ 3! i z= 2 = 6.28318i 6 8 sinh i = 1.04720i 8 0.84147i = 7.04949 Engineering Math EEE 3640 17