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1                                             CAAM 335 Matrix Analysis:                                                   ...
2                    Clearly, their sum is indeed 0.                (c) f (z) = 5 sin(2z) = (− 5 cos(2z)) . Hence,        ...
3                                 sin(πz2 )+cos(πz2 )     The function f (z) =              (z−2)           is differentia...
4         Lecture Notes Section 8.5 (P. 93)Exercise [1] (20 points) Let us confirm the representation (8.7) in the matrix c...
5                                                                                                    1                   ...
6Let C be a sufficiently large circle that encircles the 3 poles at λ3 = −1, λ2 = −1 + i and λ3 = −1 − i.                  ...
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Sol7

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Sol7

  1. 1. 1 CAAM 335 Matrix Analysis: Solutions to HW 7 RProblem 1 (5+5+5=15 points) (a)Compute the integral C f (z)dz for f (z) = z C = t 2 + it : t ∈ [0, 2] . (b)-(c)Verify Cauchy’s theorem for the functions f (z) = 3z2 + iz − 4, f (z) = 5 sin(2z), if C is the square with vertices 1 ± i, −1 ± i. (a) Z Z 2 Z 2 f (z)dz = (t 2 − it)(2t + i)dt = 2t 3 + t + i(t 2 − 2t 2 )dt C 0 0 Z 2 1 1 2 1 2 8 = 2t 3 + t − it 2 dt = t 4 + t 2 |t=0 − i t 3 |t=0 = 10 − i 0 2 2 3 3 (b)-(c) A parametrization of the square with vertices 1 ± i, −1 ± i is given by C = {z(t) = x(t) + iy(t) : t ∈ [0, 4)} , where    −1 + 2t  t ∈ [0, 1),  −1  t ∈ [0, 1), 1 t ∈ [1, 2), −1 + 2(t − 1) t ∈ [1, 2),   x(t) = y(t) =  1 − 2(t − 2)  t ∈ [2, 3),  1  t ∈ [2, 3), −1 t ∈ [3, 4), 1 − 2(t − 3) t ∈ [3, 4),   Let C j = {z(t) = x(t) + iy(t) : t ∈ [ j − 1, j)} for j = 1, 2, 3, 4. Then Z 4 Z 4 C f (z)dz = ∑ f (z)dz = ∑ (g(z( j)) − g(z( j − 1)). j=1 C j j=1 where g(z) is an anti-derivative of f (z) so that f (z) = g (z). (b) f (z) = 3z2 + iz − 4 = (z3 + iz2 /2 − 4z) . Hence Z f (z)dz = (1 − i)3 + i(1 − i)2 /2 − 4(1 − i) − [(−1 − i)3 + i(−1 − i)2 /2 − 4(−1 − i)] = −2 − 4i C1 Z f (z)dz = (1 + i)3 + i(1 + i)2 /2 − 4(1 + i) − [(1 − i)3 + i(1 − i)2 /2 − 4(1 − i)] = 14 C2 Z f (z)dz = (−1 + i)3 + i(−1 + i)2 /2 − 4(−1 + i) − [(1 + i)3 + i(1 + i)2 /2 − 4(1 + i)] = −2 + 4i C3 Z f (z)dz = (−1 − i)3 + i(−1 − i)2 /2 − 4(−1 − i) − [(−1 + i)3 + i(−1 + i)2 /2 − 4(−1 + i)] = 10 C4
  2. 2. 2 Clearly, their sum is indeed 0. (c) f (z) = 5 sin(2z) = (− 5 cos(2z)) . Hence, 2 5 5 Z f (z)dz = − cos(2(1 − i)) + cos(2(−1 − i)) = 5 sin(2) sinh(2)i C1 2 2 5 5 Z f (z)dz = − cos(2(1 + i)) + cos(2(1 − i)) = −5 sin(2) sinh(2)i C2 2 2 5 5 Z f (z)dz = − cos(2(−1 + i)) + cos(2(1 + i)) = 5 sin(2) sinh(2)i C3 2 2 5 5 Z f (z)dz = − cos(2(−1 − i)) + cos(2(−1 + i)) = −5 sin(2) sinh(2)i C4 2 2 Again, they sums up to 0.Problem 2 (5+5+10 =20 points) Find the residues of the following functions at 0: (z2 + 1)/z, ez /z2 , (2z + 1)/(z(z3 − 5)). Let us call the three functions f j , j = 1, 2, 3. Then by the residue formula 1 z2 + 1 res( f1 , 0) = lim z = 1, z→0 (1 − 1)! z 1 d ez res( f2 , 0) = lim z2 2 = lim ez = 1, z→0 (2 − 1)! dz z z→0 1 z2 + 1 res( f3 , 0) = lim z = −1/5. z→0 (1 − 1)! z(z3 − 5)Problem 3 (5+10=20 points) Let C = 3eit : t ∈ [0, 2π) . Compute the two integrals sin(πz2 ) + cos(πz2 ) e2z Z Z dz, 4 dz. C (z − 1)(z − 2) C (z + 1) You may use results from Chapter 8 to arrive quickly at the solution. Be sure to explain your rationale. Solution 1: • Let C1 be the circle around a = 1 with radius r = 1/2 and C2 be the circle around a = 2 with radius r = 1/2. Then, sin(πz2 ) + cos(πz2 ) sin(πz2 ) + cos(πz2 ) sin(πz2 ) + cos(πz2 ) Z Z Z dz = dz + dz C (z − 1)(z − 2) C1 (z − 1)(z − 2) C2 (z − 1)(z − 2)
  3. 3. 3 sin(πz2 )+cos(πz2 ) The function f (z) = (z−2) is differentiable on and inside C1 . By the Cauchy Integral Formula, sin(πz2 ) + cos(πz2 ) sin(π12 ) + cos(π12 ) Z dz = 2πi f (1) = 2πi = −2πi(sin(π) + cos(π)) = 2πi. C1 (z − 1)(z − 2) (1 − 2) sin(πz2 )+cos(πz2 ) The function f (z) = (z−1) is differentiable on and inside C2 . By the Cauchy Integral Formula, sin(πz2 ) + cos(πz2 ) sin(π22 ) + cos(π22 ) Z dz = 2πi f (1) = 2πi = 2πi(sin(4π) + cos(4π)) = 2πi. C2 (z − 1)(z − 2) (2 − 1) Consequently, sin(πz2 ) + cos(πz2 ) Z dz = 2πi + 2πi = 4πi. C (z − 1)(z − 2) • We use the equation dn f n! f (z) Z (a) = dz dan 2πi C (z − a)n+1 with a = −1, n = 3, f (z) = e2z . We can apply this formula for n = 4 and a = −1, since a = −1 is inside C and f (z) = e2z is differentiable on C and inside C. f (z) = e2z , f (z) = 2e2z , f (z) = 4e2z , f (z) = 8e2z . e2z 2πi 2πi 2(−1) 16πi −2 8πi −2 Z 4 dz = f (−1) = 8e = e = e . C (z + 1) 3! 3! 6 3Solution 2: Alternatively, we can also use the Residue Theorem to compute both integrals. • For the first function, sin(πz2 ) + cos(πz2 ) sin(π) + cos(π) res(1) = lim (z − 1) = =1 z→1 (z − 1)(z − 2) −1 sin(πz2 ) + cos(πz2 ) sin(4π) + cos(4π) res(2) = lim (z − 2) = = 1. z→2 (z − 1)(z − 2) 1 Hence sin(πz2 ) + cos(πz2 ) Z dz = 2πi(1 + 1) = 4πi. C (z − 1)(z − 2) • For the second function, 1 d3 e2z 1 3 2z 4 −2 res(−1) = lim (z + 1)4 = lim 2 e = e . z→−1 (4 − 1)! dz3 (z + 1)4 z→−1 6 3 Hence e2z 8πi −2 Z 4 dz = 2πi res(−1) = e . C (z + 1) 3
  4. 4. 4 Lecture Notes Section 8.5 (P. 93)Exercise [1] (20 points) Let us confirm the representation (8.7) in the matrix case. More precisely, if Φ(z) ≡ (zI − B)−1 is the resolvent associated with B then (8.7) states that h mj Φ j,k Φ(z) = ∑ ∑ (z − λ j )k j=1 k=1 where 1 Z Φ j,k = Φ(z)(z − λ j )k−1 dz. 2πi Cj Compute the Φ j,k per (8.15) for the B in (7.13). Confirm that they agree with those appearing in (7.16). B = [1 0 0; 1 3 0; 0 1 1]; syms z; inv(z*eye(3)-B) ans = [ 1/(z-1), 0, 0] [ 1/(z-1)/(z-3), 1/(z-3), 0] [ 1/(z-1)ˆ2/(z-3), 1/(z-1)/(z-3), 1/(z-1)] Thus   (z − 1)(z − 3) 0 0 1 (zI − B)−1 =  (z − 1) (z − 1)2 0 . (z − 1)2 (z − 3) 1 (z − 1) (z − 1)(z − 3) and the eigenvalues of B are λ1 = 1 with multiplicity m1 = 2 and λ2 = 3 with multiplicity m2 = 1. We use the Cauchy’s Theorem (for differentiable terms) and the Residue Theorem (for terms with singularities) to compute the following integrals:  1  (z−1) 0 0 1 1 Z Z 1 1 Φ1,1 = Φ(z)(z − 1)0 dz = 0  dz   (z−1)(z−3) (z−3) 2πi C(1,1) 2πi C(1,1)  1 1 1 (z−1)2 (z−3) (z−1)(z−3) (z−1) 1 0 0     1 1 0 0 =  (z−3) 0 0  =  −1 0 0   2 1 1 1 −1 4 1 −2 1 (z−3) (z−3) z=1   1 0 0 1 1 Z Z 1 (z−1) Φ1,2 = Φ(z)(z − 1)1 dz = 0  dz  (z−3) (z−3) 2πi 2πi   C(1,1) C(1,1) 1 (z−1) (z−1)(z−3) (z−3) 1     0 0 0 0 0 0 = 0 0 0  = 0 0 0  1 − (z−3) 0 0 z=1 −1 2 0 0
  5. 5. 5  1  (z−1) 0 0 1 1 Z Z 1 1 Φ2,1 = Φ(z)(z − 3)0 dz = 0  dz   (z−1)(z−3) (z−3) 2πi 2πi  C(3,1) C(3,1) 1 1 1 (z−1)2 (z−3) (z−1)(z−3) (z−1)   0 0 0   0 0 0 1 =  (z−1) 1 0  = 1 1 0 .  2 1 1 1 1 (z−1)2 (z−1) 0 4 2 0 z=3Exercise [2] (10 points) Use (8.14) to compute the inverse Laplace transform of 1/(s2 + 2s + 2). Recall h 1 Z (L −1 q)(t) ≡ q(z)ezt dz = ∑ res(λ j ) 2πi C j=1 where C is a simple closed curve that encloses the poles of q(z), in this case at z = −1 + i and z = −1 − i. 1 1 ezt Z L −1 2 + 2s + 2 (t) = dz = res(−1 + i) + res(−1 − i) s 2πi C (z − (−1 + i))(z − (−1 − i)) ezt ezt e(−1+i)t e(−1−i)t = + = + z − (−1 − i) z − (−1 + i) 2i −2i z=−1+i z=−1−i eit − e−it = e−t = e−t sin(t) 2iExercise [3] (20 points) Use the result of the previous exercise to solve, via the Laplace transform, the differential equation x (t) + x(t) = e−t sint, x(0) = 0. Hint: Take the Laplace transform of each side. Taking the Laplace transform of each side, and using the result from the previous exercise, L (x (t) + x(t)) = L (e−t sint) 1 sL x − x(0) + L x = s2 + 2s + 2 1 (s + 1)L x = 2 (s + 2s + 2) 1 Lx = (s + 1)(s2 + 2s + 2) Now, we use the inverse Laplace transform to determine x(t).
  6. 6. 6Let C be a sufficiently large circle that encircles the 3 poles at λ3 = −1, λ2 = −1 + i and λ3 = −1 − i. 3 1 1 ezt Z L −1 (t) = dz = ∑ res(λ j ) (s + 1)(s2 + 2s + 2) 2πi C (z + 1)(z − (−1 + i))(z − (−1 − i)) j=1 ezt ezt ezt = + + z2 + 2z + 2 (z + 1)(z − (−1 − i)) (z + 1)(z − (−1 + i)) z=−1 z=−1+i z=−1−i e−t e(−1+i)t e(−1−i)t −eit − e−it = + + = e−t 1 + = e−t (1 − cos(t)) 1 (2i)(i) (−2i)(−i) 2Therefore, the solution to the differential equation is x(t) = e−t (1 − cos(t)).

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