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The question is about complex analysis and Calculus of Residues. One of the 2 pictures is an example of a solution. Please solve parts a, b and c.

The question is about complex analysis and Calculus of Residues. One of the 2 pictures is an example of a solution. Please solve parts a, b and c.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter7: Distance And Approximation
Section7.2: Norms And Distance Functions
Problem 41EQ
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The question is about complex analysis and Calculus of Residues.

One of the 2 pictures is an example of a solution.

Please solve parts a, b and c.

Example 21.1.2. The integral f cos z dz/z², where C is the circle |z| = 1, is zero because 22n (-1)". (2n)! COS Z 1 1 22 - z2 2 2 2 4! n=0 0 (no 1/z term in the Laurent expansion). Therefore, by Equation (21.1) yields bị the integral must vanish. When C is the circle |2| 2, fc e dz/(z – 1)³ = ine because e = ee (z – 1)" = e |1+ (z – 1) + 2! (z – 1)? = e n! n=0 and ez 1 1 1 = e (z – 1)3 1)3 Thus, b1 = e/2, and the integral is 2ribi = ine. (z – 1)2 2 -
Transcribed Image Text:Example 21.1.2. The integral f cos z dz/z², where C is the circle |z| = 1, is zero because 22n (-1)". (2n)! COS Z 1 1 22 - z2 2 2 2 4! n=0 0 (no 1/z term in the Laurent expansion). Therefore, by Equation (21.1) yields bị the integral must vanish. When C is the circle |2| 2, fc e dz/(z – 1)³ = ine because e = ee (z – 1)" = e |1+ (z – 1) + 2! (z – 1)? = e n! n=0 and ez 1 1 1 = e (z – 1)3 1)3 Thus, b1 = e/2, and the integral is 2ribi = ine. (z – 1)2 2 -
21.1. Evaluate each of the following integrals, for all of which C is the circle |2| = 3: ez dz. (b) P z(2 – in) COS Z 4z – 3 dz. (c) P - dz. (a) . 2(2 - c z(z – 2) 22 +1 dz. (d) P 2 – 1) cosh z dz. + 72 1 (f) - COS Z dz. (e) 22 (:). of dz dz. (i) P 3(2+5) sinh z dz. 4 (g) (h) Z COS dz. dz (1) dz. 22 dz. (j) (k) Jc sinh 2z tan z dz. dz ez dz (m) dz. 22 sin z (n) (z – 1)(2 – 2)*
Transcribed Image Text:21.1. Evaluate each of the following integrals, for all of which C is the circle |2| = 3: ez dz. (b) P z(2 – in) COS Z 4z – 3 dz. (c) P - dz. (a) . 2(2 - c z(z – 2) 22 +1 dz. (d) P 2 – 1) cosh z dz. + 72 1 (f) - COS Z dz. (e) 22 (:). of dz dz. (i) P 3(2+5) sinh z dz. 4 (g) (h) Z COS dz. dz (1) dz. 22 dz. (j) (k) Jc sinh 2z tan z dz. dz ez dz (m) dz. 22 sin z (n) (z – 1)(2 – 2)*
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