Δ z k 0 when n the above definition is not very

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Δ z k | → 0 when n → ∞ . The above definition is not very helpful for purposes of evaluating a contour integral. We turn to the following result.,, whose proof is omitted as it involves concepts beyond the scope of this course. Theorem 5.1. Let C be a contour in the complex plane parametrized by z = f ( t ) , t [ a, b ] , and let G ( z ) be a complex-valued function which is continuous on the contour C . Then (a) G ( f ( t )) is continuous on [ a, b ] (b) dz = dz dt dt = f 0 ( t ) dt The integral of G along the contour C is defined by Z C G ( z ) dz = Z b a G ( f ( t ) f 0 ( t ) dt Example 5.8. Evaluate the integral Z C dz z 2 where C is the circle | z | = 2 traversed in the counterclockwise direction and parametrized by z = f ( t ) = 2 e it , t [0 , 2 π ]. 62
Solution: We have Z C dz z 2 = Z 2 π 0 2 ie it dt 2 e it = 2 πi Example 5.9. Evaluate the integral Z C z - 1 dz where C is the polygonal path that joins 0 to 1 + i to 3 + i . The contour C may be viewed as the sum of the contour C 1 joining the origin to the point 1 + i and the contour C 2 which is the segment that joins 1 + i t0 3 + i . The following theorem states that the properties of the definite integral have analogues for contour integrals: Theorem 5.2. Let F ( z ) and G ( z ) be continuous complex valued functions defined on a contour C , and let C be the inverse of C . Also let α be a complex constant. Then (a) Z C F ( z ) ± G ( z ) dz = Z C F ( z ) dz ± Z C G ( z ) dz (b) Z C α F ( z ) dz = α Z C F ( z ) dz (c) Z C F ( z ) dz = - Z C F ( z ) dz (d) If C = C 1 + C 2 , then Z C F ( z ) dz = Z C 1 F ( z ) dz + Z C 2 F ( z ) dz The following theorems give additional properties of contour integrals: Theorem 5.3. Let L be the length of a contour C which is parametrized by z = f ( t ) , t [ a, b ] . Then L = Z C | dz | = Z b a | f 0 ( t ) | dt Theorem 5.4. Let G be a complex-valued function which is continuous on a contour C parametrized by z = f ( t ) , t [ a, b ] , and let L be the length of this contour. Let G be bounded on C , so there exists M > 0 such that | G ( z ) | ≤ M for every z C . Then Z C G ( z ) dz ML 63
Example 5.10. Without evaluating the integral, show that Z C dz z 2 - 1 4 π 3 where C is the circle | z | = 2 traversed in the counterclockwise direction. Solution: We have L = 2 π (2) = 4 π , while | z 2 - 1 | ≥ | z | 2 - 1 = 3, so that 1 z 2 - 1 1 3 = M . Applying Theorem 5.4, we have Z C dz z 2 - 1 ML = 1 3 (4 π ) = 4 π 3 Exercise 5.2.1. 1. Evaluate each of the following integrals: (a) Z 1 0 (2 t + it 2 ) dt (b) Z 0 - 2 (1 + i ) cos( it ) dt (c) Z 1 0 (1 + 2 it ) 5 dt (d) Z 2 0 t ( t 2 + i ) 2 dt 2. Evaluate the following contour integrals: (a) Z Γ (2 z + 1) dz where Γ is the line segment from z = - i to z = 1. (b) Z Γ (2 z + 1) dz where Γ is the circular arc z ( t ) = e it , t [ - π/ 2 , 0]. (c) Z Γ ( | z - 1 + i | 2 - z ) dz where Γ is the semicircle z ( t ) = 1 - i + e it , t [0 , π ]. (d) Z Γ z dz , where Γ is the circle | z | = 2 traversed once counterclockwise. 3. Prove the following upper bounds: (a) If C is the circle traversed counterclockwise, then Z C dz z 2 - i 3 π 4 .

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