Final Exam Solutions - Complex Analysis - Spring 2009 | MATH 412, Exams of Mathematics

Download Final Exam Solutions - Complex Analysis - Spring 2009 | MATH 412 and more Exams Mathematics in PDF only on Docsity! May 22, 2009 Final Exam Name: Math 412 SOLUTIONS Problem 1. (7 points each) Define f(z) = z − 3 + 4i z2 + 4 + 4i . (a): Compute f(−i). Express your answer in the form a + bi, where a, b ∈ R. SOLUTION: We have f(−i) = (−i)− 3 + 4i (−i)2 + 4 + 4i = −3 + 3i 3 + 4i = (−3 + 3i)(3− 4i) (3 + 4i)(3− 4i) = 3 + 21i 25 = 3 25 + 21 25 i.  (b): Find all singularities of f . Express any such points in polar or exponential form. SOLUTION: We are looking for the values of z such that z2 = −4− 4i. That is, we must find the two square roots of −4− 4i = √ 32ei( 5π 4 +2πn). Thus, (−4− 4i)1/2 = 4 √ 32ei( 5π 8 +πn), where we may take n = 0 and n = 1. Thus, the two points excluded from the domain of f are 4 √ 32ei( 5π 8 ) and 4 √ 32ei( 5π 8 +π) = 4 √ 32ei( 13π 8 ).  Problem 2. Draw a sketch in the complex plane of each set below. (a): (7 points) {z ∈ C : |z − 1| < |z − 2|} SOLUTION: This is the set of complex numbers that are closer to 1 than they are to 2. This is precisely the set of complex numbers whose real part is less than 32 . That is, {z ∈ C : Re(z) < 3 2}.  (b): (7 points) The image of the set S = { z ∈ C : 1 e22 < x2 + y2 ≤ 1 e4 and x < 0 and y < 0 } under the function f(z) = log4π/3(z). SOLUTION: Recall that f(z) = ln |z|+ iθ, where θ ∈ ( 4π 3 , 10π 3 ] . We are given that e−11 < |z| ≤ e−2. Thus, −11 < ln |z| ≤ −2. Therefore, the image of S consists of horizontal strips extending from the vertical lines x = −11 to x = −2. The first strip exists in the range 4π/3 < y ≤ 3π/2 and the second strip exists in the range 3π ≤ y ≤ 10π/3.  Follow-Up Question: Circle all words below that apply to the given set: (c): (4 points) the set in part (a) above: SOLUTION: OPEN and CONNECTED.  (d): (4 points) the set in part (b) above: SOLUTION: BOUNDED.  Problem 3. (6 points each) Evaluate, if possible, or explain why the given expression is meaningless. (a): the derivative of the function f(z) = f(x + iy) = (x2 − y2 − 3x + 1) + (2xy − 3y)i. SOLUTION: Using the theorem for the Cauchy-Riemann Equations, we have f ′(x, y) = ux(x, y) + ivx(x, y) = (2x− 3) + i(2y).  (b): ∫ C (y − x− 3x2i)dz, where C consists of a straight line segment from z = 0 to z = i. SOLUTION: The function f(x + iy) = y − x − 3x2i is not analytic (Cauchy-Riemann equations fail everywhere), so we have no shortcuts. We must parametrize and integrate the curve. We use z(t) = it, (0 ≤ t ≤ 1). Then dz = idt. We have x = 0 and y = t, so the integral becomes∫ 1 0 (ti)dt = i 2 . SOLUTION: The singularity z = 0 is a pole of order 6, while the singularity at z = −2 is a simple pole.  (d): (7 points) Find Res[f, z0] for each singularity z0 in (b). SOLUTION: We begin with z0 = 0, a pole of order 6. The formula we need (p. 294) is Res[f, 0] = 1 5! lim z→0 [( 1 z + 2 )′′′′′] = lim z→0 −120(z + 2)−6 5! = 1 120 · (−120)(2−6) = − 1 64 . Next, for z0 = −2, a simple pole, we have Res[f,−2] = lim z→−2 1 z6 = 1 (−2)6 = 1 64 .  (e): (5 points) Evaluate ∫ C−7 (i) f(z)dz. SOLUTION: By the Cauchy Residue Theorem, we have∫ C−7 (i) f(z)dz = Res[f, 0] + Res[f,−2] = 2πi [ − 1 64 + 1 64 ] = 0.  Problem 6. (6 points each) In (a) and (b), evaluate, if it exists. In (c), decide whether or not the series converges absolutely, converges conditionally, or diverges. (a): lim n→∞ ( 1 2 + i 4 )n SOLUTION: Note that ∣∣∣∣12 + i4 ∣∣∣∣ = √( 1 2 )2 + ( 1 4 )2 < 1, so that the terms of this sequence tend to zero. Thus, the limit is 0.  (b): lim n→∞ (3ni + i− 2)(4n + 2i)(n− i) (2 + i)n3 − in2 + 1 SOLUTION: Note that (3ni + i− 2)(4n + 2i)(n− i) (2 + i)n3 − in2 + 1 = (3i)(4)(1)n3 + “lower terms” (2 + i)n3 + “lower terms” , so as n →∞, this tends to 12i 2 + i .  (c): ∞∑ n=1 3n 4n−1 in SOLUTION: Apply the Ratio Test: L = lim n→∞ ∣∣∣∣∣ 3(n+1) 4n 3n 4n−1 ∣∣∣∣∣ = limn→∞ ∣∣∣∣ (n + 1)4n−1n4n ∣∣∣∣ = limn→∞ ∣∣∣∣n + 14n ∣∣∣∣ = 14 < 1. Thus, by the Ratio Test, the series converges absolutely.  Problem 7. (10 points) Prove that whenever 0 < |z| < 4, we have 1 4z − z2 = ∞∑ n=0 zn−1 4n+1 . SOLUTION: Write 1 4z − z2 = A z + B 4− z . Therefore, A(4− z) + Bz = 1. Thus, we conclude that 4A = 1 and B −A = 0. Thus, A = B = 1 4 . We have 1 4z − z2 = 1 4z + 1 4(4− z) = 1 4z + 1 16(1− z4 ) = 1 4z + 1 16 ∞∑ n=0 (z 4 )n = 1 4z + ∞∑ n=0 ( zn 4n+2 ) = ∞∑ n=−1 ( zn 4n+2 ) = ∞∑ n=0 ( zn−1 4n+1 ) . Problem 8. (8 points each) You may write “FREE POINTS” next to any one of the three parts below. Only do the other two parts of your choice for grading. (a): Let u(x, y) and v(x, y) be real-valued harmonic functions in a domain D ⊆ R2. Prove that the function F : C → C defined by F (x + iy) = (uy − vx) + i(ux + vy) is analytic in D. SOLUTION: Since u and v are harmonic, we know that uxx + uyy = 0 and vxx + vyy = 0. Therefore, if we write F (x + iy) = U(x, y) + iV (x, y), where U(x, y) = uy − vx and V (x, y) = ux + vy, we can verify the Cauchy-Riemann equations for F : Vy = (ux + vy)y = uxy + vyy = uyx − vxx = Ux and Vx = (ux + vy)x = uxx + vyx = −(uyy + vxy) = −Uy. Therefore, F is analytic in D.  (b): Let f be analytic in a simply connected domain D and let z1 and z2 be two distinct complex numbers that lie in the interior of the simple closed positively-oriented contour C in D. Show that 1 2πi ∫ C f(z)dz (z − z1)(z − z2) = f(z2)− f(z1) z2 − z1 . SOLUTION: We have 1 (z − z1)(z − z2) = A z − z1 + B z − z2 = A(z − z2) + B(z − z1) (z − z1)(z − z2) . Therefore, A + B = 0 and Az2 + Bz1 = −1. Hence, A = −1 z2 − z1 and B = 1 z2 − z1 . Therefore, 1 2πi ∫ C f(z)dz (z − z1)(z − z2) = − 1 2πi ∫ C f(z)dz (z − z1)(z2 − z1) + 1 2πi ∫ C f(z)dz (z − z2)(z2 − z1) = 1 z2 − z1 (−f(z1) + f(z2)) , where we have used Cauchy’s Integral Formula in the last step.  (c): Show that for all z ∈ D1(0), we have Log(1 + z) = ∞∑ n=1 (−1)n+1 z n n .