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# 14ìž¥ - Homework 14.1 1. Evaluate Sol. C 1 Re z...

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Homework 1 § 14 . 1 1. Evaluate R C Re z dz, where C is the parabola y = x 2 from 0 to 1 + i . Sol. C : z ( t ) = t + t 2 i, 0 t 1. Z C Re z dz = Z 1 0 t (1 + 2 ti ) dt = [ t 2 2 + 2 t 3 3 i ] 1 0 = 1 2 + 2 3 i. 2. Evaluate R C ze z 2 / 2 dz, where C is the path from i along the axes to 1. Sol. C = C 1 C 2 where C 1 : z ( t ) = ti, 1 0 and C 2 : z ( t ) = t, 0 1. Z C ze z 2 / 2 dz = Z C 1 ze z 2 / 2 dz + Z C 2 ze z 2 / 2 dz = Z 0 1 ( ti ) e - t 2 / 2 idt + Z 1 0 te t 2 / 2 dt = Z 1 0 te - t 2 / 2 dt + Z 1 0 te t 2 / 2 dt = [ - e - t 2 / 2 ] 1 0 + [ e t 2 / 2 ] 1 0 = e 1 / 2 - e - 1 / 2 . 3. Consider R C Re z dz, where C is the shortest path from 0 to 1+ i . Find an upper bound of the absolute value of the integral. Sol. length of C : 2, | Re z | ≤ 1 . By ML-inequality, | R C ze z 2 / 2 dz | ≤ 2 . Thus an upper bound of the absolute value of the integral is 2. § 14 . 2 1. Find all contours C such that H C 1 1+ z 2 dz = 0. Sol. Notice that I C 1 1 + z 2 dz = I C 1 2 i ( 1 z - i - 1 z + i ) dz. (1). C : any contour for which z 0 = i, - i lie outside. By Cauchy’s integral theorem, the integral is zero. (2). C : any contour enclosing the points i, - i . Let C 1 and C 2 be contours such (CSE) C. K. Ko

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Homework 2 that C = C 1 C 2 and C 1 and C 2 are disjoint. Then I C 1 1 + z 2 dz = 1 2 i ( I C 1 1 z - i dz - I C 2 1 z + i dz ) = 1 2 i [ ± 2 πi - ( ± 2 πi )](according to direction) = 0 . 2. Evaluate (a) H C Re2 zdz, C as shown (b) H C 7 z - 6 z 2 - 2 z dz, C as shown (c) H C cos z z dz, C consists of | z | = 1 (counterclockwise) and | z | = 3 (clockwise). (d) H C Ln(2 + z ) dz, C the boundary of the square with vertices ± 1 , ± i (counter- clockwise). (e)
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## This note was uploaded on 11/08/2011 for the course EE 111 taught by Professor Kim during the Spring '11 term at Korea University.

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14ìž¥ - Homework 14.1 1. Evaluate Sol. C 1 Re z...

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