A f z is analytic in z 6 i since i x y 2 f z is

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Calculus
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Chapter 4 / Exercise 2
Calculus
Stewart
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(a) f ( z ) is analytic in { z 6 = ± i } . Since ± i ∈ {| x | + | y | < 2 } , f ( z ) is analytic in {| x | + | y | ≥ 2 , | z | ≤ 4 } . (b) f ( z ) is analytic in { z : sin ( z / 2 ) 6 = 0 } = { z 6 = 2 n π : n Z } . Since 2 n π ∈ {| x | + | y | < 2 } for n = 0 and | 2 n π | > 4 for n 6 = 0 and n Z , f ( z ) is analytic in {| x | + | y | ≥ 2 , | z | ≤ 4 } . (c) f ( z ) is analytic in { z 6 = - 1 , - 5 } . Since - 1 ∈ {| x | + | y | < 2 } for n = 0 and |- 5 | > 4, f ( z ) is analytic in {| x | + | y | ≥ 2 , | z | ≤ 4 } .
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Chapter 4 / Exercise 2
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Stewart
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60 Chapter 3. Complex Integrals 8. Let C denote the positively oriented boundary of the square whose sides lie along the lines x = ± 2 and y = ± 2. Evaluate each of these integrals (a) Z C zdz z + 1 ; (b) Z C cosh z z 2 + z dz ; (c) Z C tan ( z / 2 ) z - π / 2 dz . Solution. (a) By Cauchy Integral Formula, Z C zdz z + 1 = 2 π i ( - 1 ) = - 2 π i . (b) By Cauchy Integral Theorem, Z C cosh z z 2 + z dz = Z | z | = r cosh z z 2 + z dz + Z | z + 1 | = r cosh z z 2 + z dz for r = 1 / 2. By Cauchy Integral Formula, Z | z | = r cosh z z 2 + z dz = 2 π i cosh ( z ) z + 1 z = 0 = 2 π i and Z | z + 1 | = r cosh z z 2 + z dz = 2 π i cosh z z z = - 1 = - 2 π i cosh ( - 1 ) . Hence Z C cosh z z 2 + z dz = 2 π i ( 1 - cosh ( - 1 )) . (c) Note that tan ( z / 2 ) is analytic in { z 6 = ( 2 n + 1 ) π : n Z } and hence analytic inside C . Therefore, Z C tan ( z / 2 ) z - π / 2 dz = 2 π i tan ( π / 4 ) = 2 π i by Cauchy Integral Formula. 9. Find the value of the integral g ( z ) around the circle | z - i | = 2 oriented counterclock- wise when (a) g ( z ) = 1 z 2 + 4 ; (b) g ( z ) = 1 z ( z 2 + 4 ) .
3.2 Cauchy Integral Theorem and Cauchy Integral Formula 61 Solution. (a) Since - 2 i 6∈ {| z - i | ≤ 2 } and 2 i ∈ {| z - i | ≤ 2 } , Z | z - i | = 2 g ( z ) dz = Z | z - i | = 2 ( z + 2 i ) - 1 z - 2 i dz = 2 π i ( 2 i + 2 i ) - 1 = π 2 by Cauchy Integral Formula. (b) By Cauchy Integral Theorem, Z | z - i | = 2 g ( z ) dz = Z | z | = r g ( z ) dz + Z | z - 2 i | = r g ( z ) dz for r < 1 / 2. Since Z | z | = r g ( z ) dz = 2 π i 1 z 2 + 4 z = 0 = π i 2 and Z | z - 2 i | = r g ( z ) dz = 2 π i 1 z ( z + 2 i ) z = 2 i = - π i 4 by Cauchy Integral Formula, Z | z - i | = 2 g ( z ) dz = π i 4 10. Compute the integrals of the following functions along the curves C 1 = {| z | = 1 } and C 2 = {| z - 2 | = 1 } , both oriented counterclockwise: (a) 1 2 z - z 2 ; (b) sinh z ( 2 z - z 2 ) 2 . Solution. (a) Z | z | = 1 dz 2 z - z 2 = Z | z | = 1 ( 2 - z ) - 1 z dz = 2 π i ( 2 - 0 ) - 1 = π i (b) Z | z | = 1 sinh z ( 2 z - z 2 ) 2 dz = Z | z | = 1 ( sinh z )( 2 - z ) - 2 z 2 dz = 2 π i (( sinh z )( 2 - z ) - 2 ) 0 z = 0 = π i 2
62 Chapter 3. Complex Integrals 11. Show that if f is analytic inside and on a simple closed curve C and z 0 is not on C , then ( n - 1 ) ! Z C f ( m ) ( z ) ( z - z 0 ) n dz = ( m + n - 1 ) ! Z C f ( z ) ( z - z 0 ) m + n dz for all positive integers m and n . Proof. If z 0 lies outside C , then Z C f ( m ) ( z ) ( z - z 0 ) n dz = Z C f ( z ) ( z - z 0 ) m + n dz = 0 by Cauchy Integral Theorem, since f ( m ) z / ( z - z 0 ) n and f ( z ) / ( z - z 0 ) m + n are analytic on and inside C . If z 0 lies inside C , then ( n - 1 ) ! Z C f ( m ) ( z ) ( z - z 0 ) n dz = ( f ( m ) ( z )) ( n - 1 ) z = z 0 = f ( m + n - 1 ) ( z 0 ) and ( m + n - 1 ) ! Z C f ( z ) ( z - z 0 ) m + n dz = f ( m + n - 1 ) ( z 0 ) by Cauchy Integral Formula. Therefore, ( n - 1 ) ! Z C f ( m ) ( z ) ( z - z 0 ) n dz = ( m + n - 1 ) ! Z C f ( z ) ( z - z 0 ) m + n dz . 12. Let f ( z ) be an entire function. Show that f ( z ) is a constant if | f ( z ) | ≤ ln ( | z | + 1 ) for all z C . Proof. For every z 0 C , we have f 0 ( z 0 ) = 1 2 π i Z | z - z 0 | = R f ( z ) ( z - z 0 ) 2 dz for all R > 0. Since f ( z ) ( z - z 0 ) 2 ln ( | z | + 1 ) R 2 ln ( R + | z 0 | + 1 ) R 2 for | z - z 0 | = R , | f 0 ( z 0 ) | = 1 2 π i Z | z - z 0 | = R f ( z ) ( z - z 0 ) 2 dz ln ( R + | z 0 | )+ 1 R .

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