Complex Analysis - 4 Problems with Solutions - Midterm Exam II | MATH 412, Exams of Mathematics

Download Complex Analysis - 4 Problems with Solutions - Midterm Exam II | MATH 412 and more Exams Mathematics in PDF only on Docsity! May 4, 2009 Midterm II Name: Math 412 Problem 1. (6 points each) Evaluate, if possible. If an expression is not defined, explain why. (a): lim n→∞ [1 + (−1)nin] SOLUTION: The terms of the sequence oscillate among the four values 2, 1− i, 0, 1+ i repeatedly, hence never approaching a single value. Formally, you could quote a result that if a sequence contains more than one distinct subsequential limit, it cannot converge. 2 (b): lim n→∞ [ 3n n + 2 + i ( 1 + 3 n )n] SOLUTION: Note that lim n→∞ 3n n + 2 = lim n→∞ 3 1 + 2n = 3. I leave it for you to verify that lim n→∞ ( 1 + 3 n )n = e3 by taking the natural logarithm of the sequence and applying L’Hopital’s Rule. Therefore, the limit is 3 + ie3. 2 (c): ∞∑ n=1 ( 8− 6i 10 )n in SOLUTION: Since ∣∣∣∣8− 6i10 ∣∣∣∣ = 1, the terms of this series are not approaching zero. Hence, the series DIVERGES. 2 Problem 2. (8 points) Find the disk of convergence for the power series ∞∑ n=1 cn(z + 4i− 6)n, where cn = { (1− 3i− 4n ) n, if n is even (−1 + 2i + 2n ) n, if n is odd . Show all of your work. SOLUTION: By the Root Test, we know that the radius of convergence of this power series is ρ = 1 lim sup n→∞ |cn|1/n . The limit supremum is the largest accumulation point of the sequence |cn|1/n. The odd-numbered terms of the sequence |cn|1/n approach −1+2i and the even-numbered terms of the sequence |cn|1/n approach 1 − 3i. Hence, these are the two accumulation points. The latter one has the larger modulus, √ 32 + 12 = √ 10. Therefore ρ = 1√ 10 . The center of the disk is α = 6− 4i. Therefore, the disk in question is D1/ √ 10(6− 4i). 2 Problem 3. (a): (6 points) Evaluate ( 1 + i 2 )−i . SOLUTION: We have ( 1 + i 2 )−i = exp [ (−i) log ( 1 + i 2 )] = exp [ (−i) ( ln( √ 2/2) + i( π 4 + 2πn) )] = exp [ (−i) ( − ln √ 2 + i( π 4 + 2πn) )] = exp [π 4 + 2πn + i ln √ 2 ] = e( π 4 +2πn) [ cos(ln √ 2) + i sin(ln √ 2) ] . 2 (b): (8 points) Draw a sketch in the complex plane of the image under the function f(z) = logπ/3(z) of the set { z ∈ C : 1 e8 ≤ x2 + y2 ≤ e14 and y ≥ 0 } . SOLUTION: Recall the formula logπ/3(z) = ln |z|+ iθ, where θ ∈ ( π 3 , 7π 3 ] . 2 (c): ∫ C ez 2 (z + i)4 dz, where C is the positively-oriented boundary of a square with vertices (±9,±9). SOLUTION: We apply the higher-order Cauchy Integral Formula with n = 3, z0 = −i, and f(z) = ez 2 . We must compute three derivatives: f ′(z) = 2zez 2 f ′′(z) = (4z2 + 2)ez 2 , f ′′′(z) = (8z3 + 12z)ez 2 . Hence, f ′′′(−i) = 8i− 12i e = −4i e . Hence, the higher-order Cauchy Integral Formula gives the answer( 2πi 3! ) ( −4i e ) = 4π 3e . 2 (d): ∫ C−3 (0) 7z + 6 z2 − 4 dz SOLUTION: Let us first compute ∫ C+3 (0) 10z + 6 z2 − 4 dz, the corresponding positively-oriented contour integral. Both singular points z = ±2 lie inside the contour C. Therefore, using partial fractions decomposition, we have 7z + 6 z2 − 4 = 5 z − 2 + 2 z + 2 . Therefore, by the Cauchy Integral Formula, we have∫ C+3 (0) 10z + 6 z2 − 4 dz = ∫ C+3 (0) 5 z − 2 dz + ∫ C+3 (0) 2 z + 2 dz = 5(2πi) + 2(2πi) = 14πi. Therefore, correcting for orientation, we conclude that∫ C−3 (0) 7z + 6 z2 − 4 dz = −14πi. 2 Problem 6. (10 points) Prove that if f is analytic in and on a simple, closed contour C and z0 is not on C, then ∫ C f ′(z) z − z0 dz = ∫ C f(z) (z − z0)2 dz. SOLUTION: Recall that since f is analytic, so is f ′. Now, if z0 ∈ Exterior(C), then the integrands on both sides of the equation to be proved are analytic functions in and on C. Therefore, by Cauchy’s Theorem, both integrals evaluate to 0, and therefore, are obviously equal. Now suppose z0 ∈ Interior(C). Applying the Cauchy Integral Formula on the left side, we obtain 2πif ′(z0). On the right side, we must apply the higher-order version of Cauchy’s Integral Formula (with n = 1, z = z0) to get 2πif ′(z0). Once again, the answer is the same on both sides. 2