6 Questions on Complex Analysis with Solution - Midterm Exam | MATH 417, Exams of Mathematics

Download 6 Questions on Complex Analysis with Solution - Midterm Exam | MATH 417 and more Exams Mathematics in PDF only on Docsity! Math 417, Complex Analysis, Midterm Exam Friday, July 11, 2008 YOU MUST SHOW ALL OF YOUR COMPUTATIONS IN THE EXAM BOOKLET TO RECEIVE FULL CREDIT 1. Find all values of: (a) log(3 − 4i) (b) (2 + 2i)i 2. Complete each of the following: (a) Is |ez| = e|z|? Explain. (b) Explain why the following reasoning is incorrect: |eiz| = | cos z + i sin z| = √ cos2 z + sin2 z = 1 for all z 3. Determine the values of z for which the function f(z) = xez is analytic. If f is analytic at z = 0, then compute f ′(0). 4. Consider the function u(x, y) = e2x sin(2y) + 2x. (a) Show that u(x, y) is harmonic in the entire z plane. (b) Find a harmonic conjugate v(x, y) of u(x, y). Then express f = u + iv as a function of z. 5. Let C be a contour consisting of the two straight-line segments: (1) from z = i to z = 1 + i and (2) from z = 1 + i to z = 1 − 2i. Compute the integral: I = ∫ C ez dz (a) by finding a parametric representation z(t) = x(t) + iy(t), a ≤ t ≤ b for each line segment and computing: ∫ b a f(z(t))z′(t) dt over each arc of the contour and (b) verifying the result above by using an antiderivative F (z) of f(z) = ez. 6. Consider the integral: I = ∫ C dz z(z + 5) where C is the rectangle with corners at z = 3 + 3i, z = −3 + 3i, z = −3 − 3i, and z = 3 − 3i, oriented counterclockwise. (a) Find an upper bound on |I|. Justify your answer. (b) Compute the exact value of |I|. 5. Let C be a contour consisting of the two straight-line segments: (1) from z = i to z = 1 + i and (2) from z = 1 + i to z = 1 − 2i. Compute the integral: I = ∫ C ez dz (a) by finding a parametric representation z(t) = x(t)+iy(t), a ≤ t ≤ b for each line segment and computing: ∫ b a f(z(t))z′(t) dt over each arc of the contour and (b) verifying the result above by using an antiderivative F (z) of f(z) = ez. Solution: (a) See HW 4 solutions. (b) The function f(z) = ez is entire so it has an antiderivative F (z) = ez everywhere in the complex plane. The value of the integral is then ∫ C ez dz = F (1 − 2i) − F (i) = e1−2i = ei 6. Consider the integral: I = ∫ C dz z(z + 5) where C is the rectangle with corners at z = 3 +3i, z = −3 +3i, z = −3− 3i, and z = 3− 3i, oriented counterclockwise. (a) Find an upper bound on |I|. Justify your answer. (b) Compute the exact value of |I|. Solution: (a) We use the ML-Bound formula to find an upper bound on |I|. First, the length of C is the perimeter of the square which is L = 24. Next, we find an upper bound on |f(z)| by using some properties of moduli as follows: |f(z)| = ∣ ∣ ∣ ∣ 1 z(z + 5) ∣ ∣ ∣ ∣ = 1 |z||z + 5| To find an upper bound on |f(z)| we look for the smallest possible values of |z| and |z + 5| for all z on the contour. The value of |z| is smallest when z = 3,−3, 3i,−3i since these points are the ones on C closest to the origin. The value of |z +5| is smallest when z = −3 since this point is the point on C closest to z = −5. Therefore, |f(z)| = 1 |z||z + 5| ≤ 1 |3|| − 3 + 5| = 1 6 = M and an upper bound on |I| is |I| ≤ ML = 1 6 · 24 = 4 (b) We find the exact value of |I| by first using the Method of Partial Fractions to rewrite the integral as I = ∫ C dz z(z + 5) = 1 5 ∫ C dz z − 1 5 ∫ C dz z + 5 The function 1 z + 5 is analytic everywhere on and inside the simple closed contour C. Therefore, by the Cauchy-Goursat Theorem we have ∫ C dz z + 5 = 0 To evaluate the first integral we notice that 1 z is analytic everywhere except z = 0, so we can deform the path into a circle of radius 1 centered at the origin. Therefore, ∫ C dz z = 2πi The value if I is then I = ∫ C dz z(z + 5) = 1 5 (2πi) − 1 5 (0) = 2πi 5 and its modulus is |I| = 2π 5 which agrees with the upper bound we found in part (a).