Grátis

52 pág.

# Stein & Shakarchi - Complex Analysis - Solutions

- Denunciar

## Pré-visualização | Página 1 de 12

Chapter 2.6, Page 64 Exercise 1: Prove that∫ ∞ 0 sin(x2)dx = ∫ ∞ 0 cos(x2)dx = √ 2pi 4 . These are the Fresnel integrals. Here, ∫∞ 0 is interpreted as limR→∞ ∫ R 0 . Solution. Let f(z) = eiz 2 . We integrate f(z) around a circular sector of radiusR running from θ = 0 to pi4 . Along the x axis the integral is ∫ R 0 eix 2 dx. Along the curved part we have z = Reiθ and the integral is∫ pi/4 0 eiR 2e2iθ iReiθdθ = iR ∫ pi/4 0 e−R 2 sin(2θ)ei(θ+iR 2 cos(2θ))dθ. Finally, along the segment at angle pi4 we have z = re ipi/4 and the integral is ∫ 0 R e−r 2 eipi/4dr. The total integral is zero since f is analytic everywhere. As R→∞, the integral over the third piece approaches −eipi/4 ∫ ∞ 0 e−x 2 dx = −eipi/4 √ pi 2 = − √ 2pi 4 − √ 2pi 4 i. To estimate the integral over the curved piece, we used the fact that sin(2φ) ≥ 4φpi for 0 ≤ φ ≤ pi4 ; this follows from the concavity of sin(2φ). Using this,∣∣∣∣∣iR ∫ pi/4 0 e−R 2 sin(2θ)ei(θ+iR 2 cos(2θ))dθ ∣∣∣∣∣ ≤ R ∫ pi/4 0 ∣∣∣e−R2 sin(2θ)ei(θ+iR2 cos(2θ))∣∣∣ dθ = R ∫ pi/4 0 e−R 2 sin(2θ)dθ ≤ R ∫ pi/4 0 e−4R 2θ/pidθ = − pi 4R e−4R 2θ/pi ∣∣∣pi/4 0 = pi(1− e−R2) 4R . As R→∞, this approaches zero and we are left with∫ ∞ 0 eix 2 dx− √ 2pi 4 − √ 2pi 4 i = 0. Taking real and imaginary parts, we have∫ ∞ 0 cos(x2)dx = ∫ ∞ 0 sin(x2)dx = √ 2pi 4 . � Exercise 2: Show that ∫∞ 0 sin x x dx = pi 2 . 1 2 Solution. We integrate f(z) = e iz z around an indented semicircular contour bounded by circles of radius � and R in the upper half plane. The integrals along the two portions of the real axis add up to∫ � −R cos(x) + i sin(x) x dx+ ∫ R � cos(x) + i sin(x) x dx = 2i ∫ R � sin(x) x dx because cosine is even and sine is odd. The integral around the arc of radius R tends to zero as R → ∞, by the Jordan lemma; since this lemma isn’t mentioned in the book, here’s a proof for this specific case: On this arc, z = Reiθ so the integral is∫ pi 0 eiRe iθ Reiθ iReiθdθ = i ∫ pi 0 e−R sin(θ)eiR cos(θ)dθ. The absolute value of this integral is at most∫ pi 0 e−R sin(θ)dθ = 2 ∫ pi/2 0 e−R sin(θ)dθ by symmetry. Now sin(θ) ≥ 2θpi for 0 ≤ θ ≤ pi2 by the concavity of the sine function, so this is at most 2 ∫ pi/2 0 e−2Rθ/pidθ = −pie −2Rθ/pi R ∣∣∣pi/2 0 = pi(1− e−R) R which tends to 0 as R→∞. Finally, the integral over the inner semicircle tends to −pii; this is an immediate consequence of the fractional residue theorem, but since that doesn’t seem to be mentioned in this book either (gosh!), we can also see it from the fact that e iz z = 1 z + O(1) as z → 0, and since the length of the semicircle is tending to zero, the integral over it approaches the integral of 1 z over it, which is∫ 0 pi 1 �eiθ i�eiθdθ = − ∫ pi 0 1dθ = −pii. Putting the pieces together and letting R→∞ and �→ 0, we have 2i ∫ ∞ 0 sin(x) x dx− pii = 0⇒ ∫ ∞ 0 sin(x) x dx = pi 2 . � Exercise 3: Evaluate the integrals∫ ∞ 0 e−ax cos bx dx and ∫ ∞ 0 e−ax sin bx dx, a > 0 by integrating e−Ax, A = √ A2 +B2, over an appropriate sector with angle ω, with cosω = aA . Solution. As indicated, we integrate f(z) = e−Az around a circular sector of radius R with 0 ≤ θ ≤ ω, where ω = cos−1( aA ) is strictly between 0 and 3 pi 2 . (Here we assume b 6= 0 since otherwise the integrals are trivially equal to 1a and 0 respectively). The integral along the x axis is ∫ R 0 e−Axdx→ ∫ ∞ 0 e−Axdx = 1 A as R → ∞. To estimate the integral over the curved part we use the fact that cos(θ) ≥ 1− 2θpi for 0 ≤ θ ≤ pi2 , which follows from the concavity of the cosine in the first quadrant. Then we have ∣∣∣∣∫ ω 0 e−ARe iθ Reiθdθ ∣∣∣∣ ≤ ∫ ω 0 ∣∣∣e−AReiθReiθ∣∣∣ dθ = R ∫ ω 0 e−AR cos(θ)dθ ≤ R ∫ ω 0 e−ARe2ARθ/pidθ = Re−AR pi 2AR e2ARθ/pi ∣∣∣ω 0 = pi 2A ( e−AR(1− 2ω pi ) − e−AR ) . Since 1− 2ωpi is a positive constant, this tends to 0 as R →∞. Finally, on the segment with θ = ω, z = reiω = r a+biA , so the integral is ∫ 0 R e−Ar(a+ib)/A a+ ib A dr = a+ ib A ∫ 0 R e−are−ibrdr. Putting the pieces together and letting R→∞, we have a+ ib A ∫ 0 ∞ e−axe−ibxdx+ 1 A = 0⇒ ∫ ∞ 0 e−axeibxdx = 1 a+ ib = a− ib a2 + b2 . Comparing the real and imaginary parts, we have ∫ ∞ 0 e−ax cos(bx)dx = a a2 + b2 and ∫ ∞ 0 e−ax sin(bx)dx = b a2 + b2 . � Exercise 4: Prove that for all ξ ∈ C we have e−piξ2 = ∫∞−∞ e−pix2e2piixξdx. 4 Solution. Let ξ = a+ bi with a, b ∈ R. Then∫ ∞ −∞ e−pix 2 e−2piixξdx ∫ ∞ −∞ e−pix 2 e−2piix(a+bi)dx = ∫ ∞ −∞ e−pi(x 2−2bx)e−2piiaxdx = epib 2 ∫ ∞ −∞ e−pi(x−b) 2 e−2piiaxdx = epib 2 e−2piiab ∫ ∞ −∞ e−pi(x−b) 2 e−2piia(x−b)dx = epib 2 e−2piiab ∫ ∞ −∞ e−piu 2 e−2piiaudu = epib 2 e−2piiabe−pia 2 = e−pi(a+bi) 2 = e−piξ 2 . � Exercise 5: Suppose f is continuously complex differentiable on Ω, and T ⊂ Ω is a triangle whose interior is also contained in Ω. Apply Green’s theorem to show that ∫ T f(z)dz = 0. This provides a proof of Goursat’s theorem under the additional assumption that f ′ is continuous. Solution. Write f(z) as f(x, y) = u(x, y)+iv(x, y) where u, v are real-valued and z = x+ iy. Then dz = x+ idy so∮ T f(z)dz = ∮ T (u(x, y) + iv(x, y))(dx+ idy) = ∮ T u dx− v dy + i ∮ T v dx+ u dy = ∫∫ ( ∂v ∂x + ∂u ∂y ) dxdy + i ∫∫ ( ∂u ∂x − ∂v ∂y ) dxdy = 0 by the Cauchy-Riemann equations. (The double integrals are, of course, taken over the interior of T .) � Exercise 6: Let Ω be an open subset of C and let T ⊂ Ω be a triangle whose interior is also contained in Ω. Suppose that f is a function holomorphic in Ω except possibly at a point w inside T . Prove that if f is bounded near w, then ∫ T f(z)dz = 0. 5 Solution. Let γ� be a circle of radius � centered at w, where � is sufficiently small that γ� lies within the interior of T . Since f is holomorphic in the region R between T and γ�,∫ ∂R f(z)dz = ∫ T f(z)dz − ∫ γ� f(z)dz = 0. Thus, ∫ T f(z)dz = ∫ γ� f(z)dz. But f is bounded near w and the length of γ� goes to 0 as � → 0, so ∫ γ� f(z)dz → 0 and therefore ∫ T f(z)dz = 0. (Note: If we’re not allowed to use Cauchy’s theorem for a region bounded by two curves, one can use a “keyhole contour” instead; the result is the same.) � Exercise 7: Suppose f : D → C is holomorphic. Show that the diameter d = supx,w∈D |f(z)− f(w)| of the image of f satisfies 2|f ′(0)| ≤ d. Moreover, it can be shown that equality holds precisely when f is linear, f(z) = a0 + a1z. Solution. By the Cauchy derivative formula, f ′(0) = 1 2pii ∮ Cr f(ζ) ζ2 dζ where Cr is the circle of radius r centered at 0, 0 < r < 1. Substituting −ζ for ζ and adding the two equations yields 2f ′(0) = 1 2pii ∮ Cr f(ζ)− f(−ζ) ζ2 dζ. Then |2f ′(0)| ≤ 1 2pi ∮ |f(ζ)− f(−ζ)| r2 dζ ≤ Mr r ≤ d r , where Mr = sup |ζ|=r |f(ζ)− f(−ζ)|. Letting r → 1 yields the desired result. � Exercise 8: If f is a holomorphic function on the strip −1 < y < 1, x ∈ R with |f(z)| ≤ A(1 + |z|)η, η a fixed real number for all z in that strip, show that for each integer n ≥ 0 there exists An ≥ 0 so that |f (n)(x)| ≤ An(1 + |x|)η, for all x ∈ R. Solution. For any x, consider a circle C centered at x of radius 12 . Applying the Cauchy estimates to this circle, |f (n)(x)| ≤ n!‖f‖C (1/2)n where ‖f‖C = supz∈C |f(z)|. Now for z ∈ C, 1 + |z| ≤ 1 + |x| + |z − x| = 3 2 + |x| < 2(1 + |x|), so |f(z)|