MATH 185, HOMEWORK 9: SOLUTIONS TO SELECTED PROBLEMS Exercise 1. (For emphasis.) Let g and h be analytic in an open set containing z0 and assume that h has a simple zero at z0 . Prove that ¸ · g(z0 ) g(z) = 0 Resz=z0 . h(z) h (z0 ) Solution. Since h has a simple zero at z0 , we can write h(z) = (z − z0 )H(z) where H is analytic at z0 and H(z0 ) 6= 0. Then we have that g(z) H(z) is analytic at z0 , so it has a Taylor series at z0 : ∞ X g(z) ak (z − z0 )k . = H(z) k=0 Then ∞ g(z) X ak (z − z0 )k−1 = h(z) k=0 is the Laurent expansion of g/h centered at z0 . The residue of g/h at z0 is the coefficient of 1/(z − z0 ) in this expansion, which is a0 = So we have g(z0 ) . H(z0 ) · ¸ g(z) g(z0 ) Resz=z0 = . h(z) H(z0 ) Now we would like to write it in terms of h. By the product rule, h0 (z) = H(z) + (z − z0 )H 0 (z), so h0 (z0 ) = H(z0 ). Thus · ¸ g(z) g(z0 ) Resz=z0 . = 0 h(z) h (z0 ) Note: This is true whether or not g is zero at z0 . 1 2 MATH 185, HOMEWORK 9: SOLUTIONS TO SELECTED PROBLEMS Also note: An analytic function h has a simple zero at z0 if and only if (h(z0 ) = 0 and h0 (z0 ) 6= 0). Proof. If h(z0 ) = 0, then h(z) = (z − z0 )H(z) where H is analytic at z0 . By the product rule, h0 (z0 ) = H(z0 ), so H(z0 ) = 0 if and only if h0 (z0 ) = 0. ¤ Exercise 2. Prove that the integral Z ∞ 0 ¯ ¯ ¯ sin x ¯ ¯ ¯ ¯ x ¯ dx diverges. Solution. This is actually a real analysis question; it could very well have appeared in your Math 104 class. (If you manage to find a complex-analytic proof, please let me know!) By definition, ¯ ¯ Z ∞¯ Z R¯ ¯ sin x ¯ ¯ sin x ¯ ¯ ¯ ¯ ¯ lim ¯ x ¯ dx = R→∞ ¯ x ¯ dx 0 0 ¯ N Z (k+1)π ¯ X ¯ sin x ¯ ¯ ¯ = lim ¯ x ¯ dx. N →∞ k=0 kπ But elementary estimates show that ¯ Z (k+1)π ¯ Z kπ+ 3π ¯ sin x ¯ 4 1 ¯ ¯ dx ≥ | sin x| dx ¯ x ¯ (k + 1)π kπ+ π4 kπ µ ¶ 1 ³π ´ 1 √ ≥ . (k + 1)π 2 2 So we have Z (N +1)π 0 ¯ ¯ N X ¯ sin x ¯ 1 ¯ ¯ dx ≥ √ ¯ x ¯ 2(k + 1) 2 k=0 N +1 1 X1 . = √ 2 2 k=1 k But this series (the harmonic series) diverges as N → ∞, so ¯ Z ∞¯ ¯ sin x ¯ ¯ ¯ diverges. ¯ x ¯ dx 0 MATH 185, HOMEWORK 9: SOLUTIONS TO SELECTED PROBLEMS Exercise 3. Derive the formula Z ∞ 3 cos x π dx = cosh x cosh(π/2) −∞ by integrating the function eiz / cosh z around the rectangle with vertices −R, R, R + πi, −R + πi, and letting R → ∞. Solution (Sketch). Call the given rectangle ΓR , and let f (z) = One can show that Z Z −π lim R→∞ eiz . cosh z f (z) dz = (1 + e ∞ ) ΓR −∞ eix dx . cosh x As usual in this class, we can use the Residue Theorem to find another way to compute the integral over ΓR . For that, we need to find the singularities of f in the rectangle. That is, we must find the zeros of h(z) = ez + e−z in the rectangle. But h(z) = 0 exactly when ¶ µ 2k + 1 πi, k ∈ Z. z= 2 (Moreover, these are simple zeros since h0 (z) = ez − e−z is not zero when h(z) = ez + e−z is zero.) The only one in the rectangle ΓR is πi/2, so Z f (z) dz = 2πi Resz=πi/2 [f (z)]. ΓR By Exercise 1 (or other methods), Resz=πi/2 [f (z)] = −ie−π/2 . Putting it all together, we get the desired answer. (Remember, this is just a sketch. Please fill in the details yourself if you haven’t already.) Exercise 4. Evaluate Z ∞ 0 by integrating the function around the same contour Γ²,R Solution. I got sin2 x dx x2 1 − e2iz z2 as used in integrating sin x/x. Z ∞ 0 sin2 x π dx = . 2 x 2 4 MATH 185, HOMEWORK 9: SOLUTIONS TO SELECTED PROBLEMS Is that what you got? Book Problems: page 267 (5, 6, 7, 8) page 275 (1, 2, 7, 9)