We have I 1 1 2 Z \u03b3 1 dz z 1 1 2 Z \u03b3 1 dz z 1 2 1 4 Z \u03b3 1 dz z i 1 4 Z \u03b3 1 dz z

We have i 1 1 2 z γ 1 dz z 1 1 2 z γ 1 dz z 1 2 1 4

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We have I 1 = 1 2 Z γ 1 dz z + 1 + 1 2 Z γ 1 dz ( z + 1) 2 - 1 4 Z γ 1 dz z + i - 1 4 Z γ 1 dz z - i = 1 2 · 0 + 1 2 · 0 - 1 4 · 0 - 1 4 · 2 πi = - π 2 i,
Section 3.7 Cauchy’s Theorem for Multiply-Connected Regions 79 where the first 3 integrals are 0 because the integrands are analytic inside an on γ 1 , and the fourth integral follows from Example 4. Similarly, I 2 = 1 2 Z γ 2 dz z + 1 + 1 2 Z γ 2 dz ( z + 1) 2 - 1 4 Z γ 2 dz z + i - 1 4 Z γ 2 dz z - i = 1 2 · 2 πi + 1 2 · 0 - 1 4 · 0 - 1 4 · 0 = πi, where the first, third, and fourth integrals follow from Example 4, and the second integral follows from Example 4, Sec. 3.2. Thus, the desired integral is equal to I 1 + I 2 = π 2 i. 33. Let p ( z ) = a n z n + · · · + a 1 z + a 0 . Then we have 1 2 πi Z C p ( z ) z - z 0 dz = a n 1 2 πi Z C z n z - z 0 dz + · · · + a 1 1 2 πi Z C z z - z 0 dz + a 0 1 2 πi Z C 1 z - z 0 dz = a n z n 0 + · · · + a 1 z 0 + a 0 [by Exercise 32] = p ( z 0 ) . 37. The integrand factors as 1 z 4 + 1 = 1 z - e i π 4 z - e i 3 π 4 z - e i 5 π 4 z - e i 7 π 4 , and all of these roots are distinct and inside C 2 (0). By Exercise 36, we have Z C 2 (0) 1 z 4 + 1 = Z C 2 (0) 1 z - e i π 4 z - e i 3 π 4 z - e i 5 π 4 z - e i 7 π 4 = 0 .
80 Chapter 3 Complex Integration Solutions to Exercises 3.8 1. Apply Cauchy’s formula with f ( z ) = cos z at z = 0. Then Z C 1 (0) cos z z dz = Z C 1 (0) cos z z - 0 dz = 2 πif (0) = 2 πi. 5. Apply Cauchy’s formula with f ( z ) = - Log z at z = i . Then Z C 1 2 ( i ) Log z - z + i dz = Z C 1 2 ( i ) - Log z z - i dz = - 2 πi Log i = - 2 πi ln 1 + i π 2 = π 2 . 9. We apply the generalized Cauchy formula with f ( z ) = sin z at z = π with n = 2. Then Z γ sin z ( z - π ) 3 dz = 2 πi 2! f (2) ( π ) = πi ( - sin π ) = 0 . 13. Follow the solution in Example 2. Draw small nointersecting negatively oriented circles inside γ , γ 1 centered at 0 and γ 2 centered at i . Then Z γ z + cos( πz ) z ( z 2 + 1) dz = Z γ 1 z + cos( πz ) z ( z 2 + 1) dz + Z γ 2 z + cos( πz ) z ( z 2 + 1) dz = I 1 + I 2 . Apply Cauchy’s formula with f ( z ) = z +cos( πz ) z 2 +1 at z = 0. Then (recall γ 1 is negatively oriented) I 1 = Z γ 1 z + cos( πz ) z ( z 2 + 1) dz = - 2 πif (0) = - 2 πi 0 + cos 0 0 2 + 1 = - 2 πi. Apply Cauchy’s formula with f ( z ) = z +cos( πz ) z ( z + i ) at z = i . Then I 2 = Z γ 2 z + cos( πz ) z ( z 2 + 1) dz = Z γ 2 z + cos( πz ) z ( z + i )( z - i ) dz = - 2 πif ( i ) = - 2 πi i + cos πi i (2 i ) = + πi ( i + cosh π ) . So I 1 + I 2 = - π - (2 - cosh π ). 17. Factor the denominator as z 3 - 3 z +2 = ( z +2)( z - 1) 2 . Apply the generalized Cauchy formula (6), with f ( z ) = 1 z +2 at z = 1, with n = 1. Then Z C 3 2 (0) dz ( z + 2)( z - 1) 2 = 2 πif 0 (1) = 2 πi - 1 3 2 = - 2 πi 9 . 21. Let ζ = e it , 0 t 2 π . Then = ie it dt , and we can reexpress the integral as F ( z ) = 1 2 π Z 2 π 0 e it e it - z dt = 1 2 π Z C 1 (0) 1 ζ - z i .
Section 3.8 Cauchy’s Integral Formula 81 Since | z | < 1, z is within C 1 (0), by Cauchy’s Theorem, we have F ( z ) = 1 2 πi Z C 1 (0) 1 ζ - z = 1 . 25. Define g ( z ) = R 1 0 cos( zt ) dt and let ζ = t for 0 t 1. Then g ( z ) becomes g ( z ) = R [0 , 1] cos( ) . Let φ ( z, ζ ) = cos( ). Then φ ( z, ζ ) is continuous in ζ [0 , 1] and analytic in z C . Furthermore, the derivative d φ d z = ζ cos( ) is continuous in ζ [0 , 1]. By Theorem 3.8.5, g ( z ) is analytic in C , i.e., entire.

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