MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

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∞∑ 4. Let n=0 n n n! (z + 2i)n . First of all, center is −2i since Let z n = nn n! (z + 2i)n .As lim n→∞ ∣ n n n! (n+1) ( n+1) (n+1)! radius of convergence is 1 e . | = lim n→∞ z + 2i = 0 ⇒ z = −2i. n n (n + 1) = lim (n + 1) n+1 n→∞ n n (n + 1) n = lim ( n ) n 1 = n→∞ n + 1 e , 5. Let ∞∑ n=0 n! n n (z + 1)n . The center point is −1 since Let z n = n! n n (z + 1) n . Then lim n→∞ we say that radius of convergence is e. ∣ z + 1 = 0 ⇒ z = −1. n! n n (n+1)! (n+1) ( n+1) | = lim n→∞ (n + 1) n = e, n 15.3 1-3 Radius of convergence by differentiation or integration : 1. Let ∞∑ n=2 n(n − 1) 3 n (z − 2i) n . Since lim n→∞ we say that radius of convergence is 1 4 . ∣ 4 n n(n+1) 4 n+1 (n+1)(n+2) n + 2 | = lim n→∞ 4n = 1 4 , 2. Let ∞∑ n=1 4 n n(n + 1) zn . Because radius of convergence is found that 3. 3. Let ∞∑ n=1 lim ∣ n(n−1) 3 n n→∞ | = lim n(n+1) n→∞ 3 n+1 3n − 3 n + 1 = 3, n 2 n (z + i)2n . Radius of convergence for this serie is 2 in that lim n→∞ ∣ n 2 n n+1 n→∞ 2 n+1 | = lim 47 2n n + 1 = 2.
15.4 1-3 Taylor and Maclaurin series : 1. Let center be 0 for e −2z . We know that e z = ∞∑ n=0 z n n! = 1 + z + z2 2! + . . . at z = 0. But here the function is e −2z , then Maclaurin serie for this function is ∞∑ e −2z (−2z) n = n! n=0 Radius of convergence for this serie is ∞ in that = 1 + (−2z) + 4z2 2! + . . . lim R = lim n→∞ n→∞ ∣ ∣ (−2)n n! (−2) n+1 (n+1)! n=0 −(n + 1) | = lim = ∞. n→∞ 2 1 2. Let center be 0 for (1−z 3 ) . 1 ∞ 1 − z = ∑ z n = 1 + z + z 2 + z 3 + . . . is known at 0. Then Maclaurin serie for 1 (1−z 3 ) is 1 ∞ 1 − z 3 = ∑ z 3n = 1 + z 3 + z 6 + z 9 + . . . n=0 Radius of convergence for this serie is 1 since 3. Let center be −2i for e z .We know that e z = ∞∑ n=0 lim R = 1 n→∞ z n n! = 1 + z + z2 2! + . . . at z = 0. But here the center is −2i, then Taylor serie for this function is e z = ∞∑ n=0 z n n! Radius of convergence for this serie is ∞ in that = 1 + (z + 2i) + (z + 2i)2 2! + . . . lim n→∞ ∣ 1 n! 1 (n+1)! | = lim n→∞ n + 1 = ∞. 15.5 48
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 and 42: Let D be the union of C and the int
Page 43 and 44: Let g(z) = ln(z−1) z−5 . Singul
Page 45 and 46: Hence we see that when k ∈ {0, 1,
Page 47: Because lim n→∞ 1 √ n 2 + 1 =
Page 51: ∞∑ 11. Let n=1 (−1) n 2 n n z

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