advertisement

Last name: name: 1 Final Exam M601 (Notes, books, and calculators are not authorized) Show all your work in the blank space on the exam sheet. Answers with no justification will not be graded. Question 1: Let z = x + iy, were x, y ∈ R. Is the function f (z) = cos(y) + i sin(x) holomorphic in C? (Justify rigorously your answer.) Let P (x, y) = cos(y) and Q(x, y) = sin(x) be the real and imaginary parts of f . The functions P and Q are twice continuously differentiable functions with respect to x and y over R2 . Let us verify whether the Cauchy Riemann relations hold: ∂x P (x, y) = 0, ∂y P (x, y) = − sin(y), ∂y Q(x, y) = 0. ∂y Q(x, y) = cos(x). The function f is not holomorphic since ∂y P (x, y) 6= −∂x Q(x, y). Question 2: Using the notation z = reiθ , were r ≥ 0, θ ∈ [−π, π), we define the function log(z) as follows log(z) := log(r) + iθ. Prove that log(z) is not continuous across the axis {y = 0, x < 0}. Let −r0 be an arbitrary number on the negative real axis {y = 0, x < 0}. Let us set z+ (π − ) = r0 ei(π−) , z− () = r0 ei(−π+) with > 0. It is clear that lim z+ () = −r0 = lim z− (), →0 →0 i.e., both z+ () and z− () converge to −r0 as goes to zero. If the complex log function defined above was continuous, the two quantities log(z+ ) and log(z− ) should converge to the same value as goes to zero. But using the above definition of log(z) we have lim log(z+ ) = lim log(r0 ) + i(π − ) = log(r0 ) + iπ →0 →0 and lim log(z− ) = lim log(r0 ) + i(−π + ) = log(r0 ) − iπ. →0 →0 In conclusion lim→0 log(z+ ) 6= lim→0 log(z− ), thereby proving that log(z) as defined above is not continuous across the axis {y = 0, x < 0}. 2 Final Exam, December 12, 2012 Question 3: Let z = x + iy, were x, y ∈ R. Let ez be defined by ez := ex (cos(y) + i sin(y)). (a) Prove that eiz = e−iz by using the definition of ez . By definition eiz = eix−y = e−y (cos(x) + i sin(x)) = e−y (cos(x) − i sin(x)) = e−y−ix = e−i(x−iy) . As a result eiz = e−iz (b) Define sin(z) := eiz −e−iz . 2i Compute sin(iy) for y ∈ R. Using the definition of sin(z), we infer that sin(iy) = Question 4: Evaluate R C 1 −y (e − ey ) = −i sinh(y). 2i 3z 2 dz along the curve C = γ([0, 1]) where γ(t) = t2 . Answer 1: Since dz = γ 0 (t)dt = 2tdt, we have Z 3z 2 dz = C Z 1 3t4 2tdt = 0 1 6t5 dt = t6 |10 = 1. 0 0 2 Z 3 Answer 2: Observe that 3z = f (z) where f (z) = z ; as a result Z Z 3z 2 dz = f 0 (z)dz = f (1) − f (0) = 1 C C R Question 5: Evaluate C cos(z)dz along the curve C = γ([0, 3]) where γ(t) = it for 0 ≤ t ≤ 1, γ(t) = t − 1 + i(1 + sin(50π(t − 1))) for 1 ≤ t ≤ 2, and γ(t) = (1 + i)(3 − t) + π(t − 2) for 2 ≤ t ≤ 3. Observe that cos(z) = f 0 (z) where f (z) = sin(z) and f is clearly holomorphic over C, (known from class). As a result Z Z cos(z)dz = f 0 (z)dz = f (γ(3)) − f (γ(0)) = f (π) − f (0) = sin(π) − sin(0) = 0. C C Last name: name: 3 Question 6: Let f (z) = (2z 2 + 3z + 1)2 z −3 . Compute 2 1 3 π centered at 2 traversed counterclockwise. R C f (z)dz where C is the circle of radius f (z) = (2z 4 + 9z 2 + 1 + 12z 3 + 4z 2 + 6z)z −3 = (2z 4 + 12z 3 + 13z 2 + 6z + 1)z −3 is holomorphic except at 0 and zero is a pole of third order. The residue of f at zero is Res(f, 0) = 13. The open domain limited by the circle of radius 32 π centered at 12 contains the pole of f ; as a result Z f (z)dz = 2iπRes(f, 0) = 26iπ. C R 2π iθ Question 7: Compute the integral I = 0 5−4dθ cos(θ) . (Hint: Use the parameterization z = e and observe that cos(θ) = 21 (z + z1 ), dz = izdθ. Conclude with the residue Theorem.) We use the hint. We set γ(θ) = eiθ and C = γ(0, 2π), and observe that dz := γ 0 (θ)dθ = ieiθ dθ = 1 dz. Note that C is the unit circle centered at 0. We have izdθ, i.e., dθ = iz Z 2π I= 0 Let us define f (z) = dθ = 5 − 4 cos(θ) −1 i(2z 2 −5z+2) = Z C dz = iz(5 − 2(z + z −1 )) −1 . 2i(z− 12 )(z−2) Z C dz i(5z − 2z 2 − 2) This function is holomorphic in C expect at and 2. We apply the residue theorem after observing that only the pole at 1 I = 2iπRes(f, ). 2 with 1 −1 1 1 = Res(f, ) = lim1 (z − )f (z) = 1 2 2 3i 2i( 2 − 2) z→ 2 Finally, I = 2π 1 2π = 3i 3 1 2 is in the unit circle: 1 2 4 Final Exam, December 12, 2012 Question 8: Is the matrix A = 2+i 7−i 0 diagonalizable. Justify your answer. 2+i There is a unique eigenvalue λ = 2 + i. Answer 1: If A was diagonalizable it could be written as follows A = P DP −1 where D = (2 + i)I. That would mean that A = (2 + i)P IP −1 = (2 + i)P P −1 = (2 + i)I, which is wrong. In conclusion A is not diagonalizable. 1 2 0 Question 9: Consider the matrix 0 3 0. The eigenvalues of A are 1, 2, 3. Is the matrix 2 −4 2 A diagonalizable? Be very accurate in your statements. The 3×3 matrix A has three distinct eigenvalues, hence A is diagonalizable. (Theorem from class) Question 10: Let du dt = F (u) be five-dimensional nonlinear dynamical system. Assume that u0 ∈ R5 solves F (u0 ) = 0 and assume that the dynamics in the neighborhood of u0 is described √1 , − 1 + 3i, by dz dt = Az where A is 5 × 5 real-valued matrix with eigenvalues −2 + 4i, −2 − 4i, 3 10 − 13 − 3i. Is u0 a stable equilibrium solution? Justify your answer. The real part of the eigenvalue √1 10 is positive. As a result u0 is a unstable equilibrium point. Last name: name: 5 Question 11: Let A be a n×n complex-valued Hermitian matrix (i.e., AT = A), n ≥ 2. Let M be a real-valued symmetric positive definite matrix. We define a new complex inner product hv, wi := v T M w and the associated norm kvk2 := v T M v. Assume that AM = M A. (a) Show that hv2 , Av1 i = hAv2 , v1 i, for all vectors v1 , v2 in Cn . Give all the details. hv2 , Av1 i = v2T M Av1 = v2T M Av1 = v2T M T AT v1 = v2T AT M T v1 = (Av2 )T M v1 = hAv2 , v1 i. (b) Prove that the eigenvalues of A are real. Be very accurate in your statements. Let (v, λ) be an eigenpair. Then hAv, vi = (Av)T M v = λv T M v = λkvk2 and hv, Avi = v T M Av = v T M λv = λkvk2 . From (a) we know that hAv, vi = hv, Avi; as a result λkvk2 = λkvk2 , which implies λ = λ since v 6= 0. (c) Let λ1 and λ2 be two different eigenvalues of A and let v1 , v2 be two associated eigenvectors. Prove that v1 and v2 are orthogonal with respect to the complex inner product hv, wi := v T M w. We have hv2 , Av1 i = hv2 , λ1 v1 i = v2T M (λ1 v1 ) = λ1 v2T M v1 , = λ1 hv2 , v1 i, hAv2 , v1 i = hλ2 v2 , v1 i = (λ2 v2 )T M v1 = λ2 v2T M v1 = λ2 hv2 , v1 i. From (a) and (b) we deduce that 0 = (λ1 − λ2 )hv2 , v1 i = (λ1 − λ2 )hv2 , v1 i, which proves that v1 and v2 are orthogonal since λ1 6= λ2 . 6 Final Exam, December 12, 2012 Question 12: The following λ1 := 1, v1 := (1, 1, 0), λ2 := −2, v2 := (1, 0, −1), and λ3 := 3, v3 := (1, 2, 3) are eigenpairs of a matrix A. (a) What are the fundamental solutions of the ODE system du(t) dt = Au(t)? The fundamental solutions have been defined in class and are e t v1 , e−2t v2 , and e3t v3 . This is all I wanted you to write. Recall that upon defining the matrix P := [v1T v2T v3T ], we have A = P DP −1 where D is the diagonal matrix composed of the eigenvalues. Then the ODE can be re-written as follows: d(P −1 u) = DP −1 u. dt = Dw by defining w(t) := P −1 u(t). Then w(t) is given by α1 et 1 0 0 w(t) = α2 e−2t = α1 et 0 + α2 e−2t 1 + α3 e3t 1 , 0 0 0 α3 e3t This can also be re-written dw dt where α1 , α2 , α3 are any real numbers. As a result, 0 1 u(t) = P w(t) = α1 et P 0 + α2 e−2t P 1 + α3 e3t P 0 0 0 1 = α1 et v1 + α1 e−2t v2 + α1 e3t v3 . 0 In conclusion, u(t) is a linear combination of the fundamental solutions et v1 , e−2t v2 , e3t v3 . (b) What is the solution of du(t) dt = Au(t), with u(0) = (6, 5, 1)? We know that u(t) is a linear combination of the fundamental solutions: u(t) = α1 et v1 + α2 e−2t v2 + α3 e3t v3 . The condition u(0) = (6, 5, 1) implies (6, 5, 1) = α1 v1 + α2 v2 + α3 v3 . The corresponding augmented 1 1 1 6 1 1 1 0 2 5 ∼ 0 −1 0 −1 0 −1 3 1 matrix is 1 6 1 1 −1 ∼ 0 3 1 0 1 1 0 1 −1 2 6 1 1 ∼ 0 2 0 The solution is (α1 , α2 , α3 ) = (3, 2, 1). This means u(t) = 3et v1 + 2e−2t v2 + e3t v3 . 1 1 0 0 0 1 5 1 2 ∼ 0 1 0 0 1 0 0 0 1 3 2 . 1