Midterm Exam Solutions - Complex Analysis with Applications | MATH 417, Exams of Mathematics

Download Midterm Exam Solutions - Complex Analysis with Applications | MATH 417 and more Exams Mathematics in PDF only on Docsity! Math 417 – Midterm Exam Solutions Friday, July 9, 2010 Solve any 4 of Problems 1–6 and 1 of Problems 7–8. Write your solutions in the booklet provided. If you attempt more than 5 problems, you must clearly indicate which problems should be graded. Answers without justification will receive little to no credit. 1. (a) Evaluate (−1 + i)50 and write your answer in the form a + bi. (b) Find all values of log(−2i). (c) Find all solutions to the equation z3 = −8. Write your answers in the form a+ bi. Solution: (a) The modulus of z = −1 + i is |z| = √ 2 and the principal argument is Θ = 3π 4 . Using DeMoivre’s Theorem we have (−1 + i)50 = r50 (cos 50Θ + i sin 50Θ) (−1 + i)50 = (√ 2 )50 ( cos 150π 4 + i sin 150π 4 ) (−1 + i)50 = 225 ( cos 75π 2 + i sin 75π 2 ) We note that the angle 75π 2 is equivalent to 3π 2 since 75π 2 −18(2π) = 3π 2 . There- fore, (−1 + i)50 = 225 ( cos 3π 2 + i sin 3π 2 ) = −225i (b) The modulus of z = −2i is |z| = 2 and the principal argument is Θ = −π 2 . Using the definition of log z we have log z = ln r + i (Θ + 2kπ) log(−2i) = ln 2 + i ( −π 2 + 2kπ ) where k = 0,±1,±2, . . .. (c) The solutions to the equation are the cube roots of −8. We use the formula: z1/3 = r1/3 [ cos ( Θ + 2kπ 3 ) + i sin ( Θ + 2kπ 3 )] , k = 0, 1, 2 The modulus of z = −8 is |z| = r = 8 and the principal argument is Θ = π. Therefore, the solutions are z1/3 = 81/3 [ cos ( π + 2(0)π 3 ) + i sin ( π + 2(0)π 3 )] = 2 ( cos π 3 + i sin π 3 ) = 1 + i √ 3 z1/3 = 81/3 [ cos ( π + 2(1)π 3 ) + i sin ( π + 2(1)π 3 )] = 2 (cosπ + i sin π) = −2 z1/3 = 81/3 [ cos ( π + 2(2)π 3 ) + i sin ( π + 2(2)π 3 )] = 2 ( cos 5π 3 + i sin 5π 3 ) = 1 − i √ 3 2. (a) Sketch the set of points defined by the inequality |z + 2| ≤ |z|. (b) Sketch the image of the set of points in the z-plane defined by −π 2 ≤ x ≤ π 2 , 0 ≤ y < ∞ under the transformation w = sin z. Solution: (a) Letting z = x + iy we have |z + 2| ≤ |z| |x + iy + 2| ≤ |x + iy| |(x + 2) + iy| ≤ |x + iy| √ (x + 2)2 + y2 ≤ √ x2 + y2 (x + 2)2 + y2 ≤ x2 + y2 x2 + 4x + 4 + y2 ≤ x2 + y2 4x + 4 ≤ 0 x ≤ −1 Part (a) -1 1 2 x -2 -1 1 2 y -2-3 Part (b) -1 1 2 u -2 -1 1 2 v -2 Solution: (a) Let z = x + iy. Then z = x − iy and we have f(z) = z2z̄ f(z) = (x + iy)2(x − iy) f(z) = [ (x2 − y2) + i(2xy) ] (x − iy) f(z) = x(x2 − y2) + 2xy2 + i [ −y(x2 − y2) + 2x2y ] f(z) = x3 + xy2 + i(y3 + x2y) (b) Let u(x, y) = x3 + xy2 and v(x, y) = y3 + x2y. Both u(x, y) and v(x, y) have continuous derivatives of all orders everywhere in the complex plane. The first partial derivatives are ux = 3x 2 + y2, vy = 3y 2 + x2 uy = 2xy, vx = 2xy In order for the Cauchy-Riemann equations to be satisfied we need ux = vy uy = −vx 3x2 + y2 = 3y2 + x2 2xy = −2xy 2x2 = 2y2 4xy = 0 x = ±y xy = 0 The second equation says that either x = 0 or y = 0. If x = 0 then the first equation says that = 0. If y = 0 then the first equation says that x = 0. Thus, the C-R equations are only satisfied when z = 0 and f ′(z) exists only when z = 0. 5. Determine the values of z for which the function f(z) = z̄ex is differentiable and evaluate f ′(z) at each point. At what points, if any, is f(z) analytic? Solution: Let z = x + iy. Then f(z) = z̄ex f(z) = (x − iy)ex f(z) = xex + i(−yex) We have u(x, y) = xex and v(x, y) = −yex. These functions have continous derivatives of all orders everywhere in the complex plane. The first partial derivatives are ux = xe x + ex, vy = −ex uy = 0, vx = −yex In order for the Cauchy-Riemann equations to be satisfied we need ux = vy uy = −vx xex + ex = −ex 0 = −yex xex + 2ex = 0 yex = 0 ex(x + 2) = 0 Since ex > 0 for all x, the first equation tells us that x = −2 and the second equation tells us that y = 0. Therefore, the C-R equations are only satisfied when z = −2 and f ′(z) exists only when z = −2. There is no neighborhood of z = −2 throughout which f ′(z) exists. Thus, f(z) is analytic nowhere. 6. Consider the function u(x, y) = xy3 − x3y + 2x − 6y. (a) Show that u(x, y) is harmonic in the entire complex plane. (b) Find a harmonic conjugate v(x, y) of u(x, y). Solution: (a) The function u(x, y) has continuous derivatives of all orders everywhere in the complex plane. The first and second partial derivatives are ux = y 3 − 3x2y + 2, uxx = −6xy uy = 3xy 2 − x3 − 6, uyy = 6xy We can see that uxx + uyy = −6xy + 6xy = 0 for all x, y. Therefore, u(x, y) is harmonic in the entire complex plane. (b) A harmonic conjugate v(x, y) of u(x, y) must satisfy the Cauchy-Riemann equa- tions. vy = ux vx = −uy vy = y 3 − 3x2y + 2 vx = −3xy2 + x3 + 6 Integrating the first equation with respect to y we have ∫ vy dy = ∫ (y3 − 3x2y + 2) dy v(x, y) = 1 4 y4 − 3 2 x2y2 + 2y + φ(x) Differentiating this equation with respect to x and setting the result equation to the equation for vx above we get ∂ ∂x v(x, y) = vx ∂ ∂x ( 1 4 y4 − 3 2 x2y2 + 2y + φ(x) ) = −3xy2 + x3 + 6 −3xy2 + φ′(x) = −3xy2 + x3 + 6 φ′(x) = x3 + 6 φ(x) = ∫ (x3 + 6) dx φ(x) = 1 4 x4 + 6x + C Therefore, the family of harmonic conjugates of u(x, y) are v(x, y) = 1 4 (x4 + y4) − 3 2 x2y2 + 2y + 6x + C 7. Consider the integral I = ∫ C ( z̄2 − z̄ ) dz where C is the circle |z| = 2 oriented counterclockwise. (a) Use the ML-Bound formula to find an upper bound on |I|. (b) Find the exact value of |I|. Solution: (a) First, the length of the contour is L = 2πr = 2π(2) = 4π. Next, we find an upper bound on |f(z)| for all z on C using the Triangle Inequality. ∣ ∣z̄2 − z̄ ∣ ∣ ≤ |z̄2| + |z̄| = |z|2 + |z| = 22 + 2 = 6 Therefore, we let M = 6 and we get the following upper bound on |I|: |I| ≤ ML = 6(4π) = 24π (b) The function f(z) = z̄2 − z̄ is analytic nowhere. So we have to parametrize the