MA 513 HW 4

MA 513 HW 4 - MA 513 Homework 4 Solutions p.136: #11;...

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Unformatted text preview: MA 513 Homework 4 Solutions p.136: #11; p.140: #1,3,4; p.149: #2(a); p.160: #1(a)(e)(f); p.170: #1(a), 2,3. p.136: # 11: Let t = φ(τ ). Then dt = φ′ (τ )dτ. Therefore, with Z (τ ) = z (φ(τ )), we have Z ′ (τ )dτ = z ′ (φ(τ ))φ′ (τ )dτ = z ′ (t)dt. p.140: #1: Since |z 2 − 1| ≥ ||z |2 − 1| = 3 on C : |z | = 2, between 2 and 2i, and C has 1 length 4 (2πr ) = π, we have dz 1 ≤ π = π/3. 2−1 3 Cz z x #3: |e | = e ≤ 1 for x ≤ 0; |z | ≤ 4, and C has length 3 + 4 + 5 = 12. Therefore, (ez − z ) dz ≤ 12(1 + 4) = 60. C p. 149: #2(a): d 1 πz e dz π i/2 = e , so πz πz e i d 1 πz e dz = dz π i/2 = i 1 πi/2 1 πi 1 e − e = (i + 1). π π π p. 160: #1: Each of these integrands is analytic on and inside the unit circle, so the integral is zero. (a): z 2 /(z − 3) is analytic except at z = 3, which lies outside the unit circle. (e): tan z = sin z/ cos z is analytic except at the zeroes of cos z = cos x cosh y − i sin x sinh y. Thus cos z = 0 implies cos x = 0 and y = 0. Hence, z = π + kπ, k = 0, ±1, ±2, ... But all 2 these zeroes are outside the unit circle. (f): Log(z + 2) is analytic except along the line x + 2 ≤ 0, y = 0. But this lies outside the unit circle. e−z dz = 2πie−πi/2 = 2π. p. 170: #1(a): C z − (πi/2) #2: (a) dz dz 1 = π/2, = = 2πi +4 z + 2i z =2i C C (z + 2i)(z − 2i) since z = −2i lies outside the circle |z − i| = 2, but z = 2i lies inside the circle. z2 (b) C dz = (z 2 + 4)2 C d 1 dz = 2πi (z + 2i)2 (z − 2i)2 dz (z + 2i)2 = −4πi(4i)−3 = π/16. z =2i #3: C : |z | = 3. g (2) = C 2 s2 − s − 2 ds = 2πi(2s2 − s − 2) s−2 = 8πi. s=2 g (z ) = 0 for |z | > 3. ...
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This note was uploaded on 06/11/2010 for the course MA 513 taught by Professor Staff during the Spring '08 term at N.C. State.

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