MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

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g(z) = sinh πz z 2 − 3z = sinh πz z(z − 3) Singular points of g are z 0 = 0 and z 1 = 3. Let D be the union of C and the interior part of C. Then z 1 = 3 /∈ D but z 0 = 0 ∈ D. f(z) = we get 12. ∮ C ∮ g(z)dz = C sinh πz z−3 is analytic in D. Hence by use of Cauchy integral formula, sinh πz z − 3 .1 0 dz = 2πi.f(0) = 2πisinh z −3 = 2π sin(i.0) = 0 −3 Let g(z) = tan z z−i . Singular point z 0 = i is enclosed by C. f(z) = tan z is analytic in D when D is the union of C and the interior part of C. So by the use of Cauchy integral formula, we get ∮ tan z dz = 2πi tan i z − i 13. C g(z) = e−3πz 1 2z + i = 2 e−3πz z + i 2 Singular point z 0 = − i 2 is enclosed by C. f(z) = 1 2 e−3πz is analytic in D when D is the union of C and the interior part of C. So by use of Cauchy integral formula, we get ∮ e −3πz 1 2z + i ∮C dz = dz = 2πi. 1 2 e− 3 2 πi = 2 e−3πz C z + i 2 15. We sketch C : |z − 4| = 2 as follows: 41 πi 3 2 (cos π + i sin π) = −2 3 πi
Let g(z) = ln(z−1) z−5 . Singular point z 0 = 5 is enclosed by C. Let D be the union of C and the interior part of C, f(z) = ln(z − 1). Then f(z) is analytic in D. Therefore, by use of Cauchy integral formula, we get ∮ C ln(z − 1) dz = 2πif(5) = 2πi ln 4 z − 5 16. Let C 1 : |z| = 3 and C 2 : |z| = 1. We sketch C = C 1 ∪ C 2 = 2 as follows: g(z) = sin z z 2 − 2iz = sin z z(z − 2i) Singular points of g are 0 and 2i. Only z 0 = 2i is contained in the ring-shaped domain bounded by C 1 and C 2 . f(z) = sin z z is analytic on that domain. Hence by use of Cauchy integral formula, we get ∫ sin z 2i z 2 dz = 2πi.f(2i) = 2πi.sin = π sin 2i − 2iz 2i 18.Let C a simple closed path enclosing z 1 and z 2 . C By use of Cauchy integral formula, we obtain ∮ C (z − z 1) −1 (z − z 2 ) −1 dz = ∮ 1 C (z−z 1 )(z−z 2 ) dz = ∮ 1 1 C 1 z−z 2 . z−z 1 dz + ∮ 1 1 C 2 z−z 1 . z−z 2 dz = 2πi[ 1 z−z 2 ] z=z1 + 2πi[ 1 z−z 1 ] z=z2 = 2πi 1 z 1 −z 2 + 2πi 1 z 2 −z 1 = 2πi( 1 z 1 −z 2 − 1 z 1 −z 2 ) = 0 14.4 1-8: Contour Integration Let C : |z| = 2 and D be the union of C and the interior part of C. 42
Page 1 and 2: MCS 351 ENGINEERING MATHEMATICS SOL
Page 3 and 4: a + b = 90 ◦ 3. z 1 = x 1 + iy 1
Page 5 and 6: the fact that tan θ = π 2 , θ =
Page 7 and 8: For this reason, tan α = 0 and so
Page 9 and 10: k = 2 =⇒ z 2 = 3√ 5(cos(θ + 4
Page 11 and 12: 4. Let’s now determine the region
Page 13 and 14: 12-15 Function values: 12. For z =
Page 15 and 16: 13.4 1-10 Cauchy-Riemann equations:
Page 17 and 18: =⇒ f(z) = x 3 − 3xy 2 + i(3x 2
Page 19 and 20: 13-17 Polar form: 13. z = √ i ⇒
Page 21 and 22: then, cosh(z 1 + z 2 ) = cos(i(z 1
Page 23 and 24: Then we obtain that e 2z + 1 = 0
Page 25 and 26: ⇒ e x . cos y = − cos(1) and e
Page 27 and 28: 4. z(t) = 1 + i + e −πit , 0 ≤
Page 29 and 30: • t = 0 ⇒ z(0) = 0 • t = 2
Page 31 and 32: • x = −1 ⇒ y = 4 − 4 = 0
Page 33 and 34: f(z) = e 2z is analytic in C. So we
Page 35 and 36: 27. z(t) = ( π 4 − t) + ti, (0
Page 37 and 38: z 2 + 4 = 0 =⇒ z 2 = −4 =⇒ z
Page 39 and 40: İf z = x + iy, dz = dx + idy = idy
Page 41: Let D be the union of C and the int
Page 45 and 46: Hence we see that when k ∈ {0, 1,
Page 47 and 48: Because lim n→∞ 1 √ n 2 + 1 =
Page 49 and 50: 15.4 1-3 Taylor and Maclaurin serie
Page 51: ∞∑ 11. Let n=1 (−1) n 2 n n z

MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

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