Class Notes (838,457)
MATH 316 (5)
John Toth (5)
Lecture

# solution_homework_feb12a.pdf

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School
Department
Mathematics & Statistics (Sci)
Course
MATH 316
Professor
John Toth
Semester
Fall

Description
Math 113 (Spring 2009) Yum-Tong Siu 1 Solution of Homework Assigned on February 5, 2009 due February 17, 2009 Problem 1 (from Stein & Shakarchi, pp.30-31, #25). (a) Evaluate the integral Z z dz ° for all integers n (positive, negative, or zero). Here ° is any circle centered at the origin with the positive (counterclockwise) orientation. (b) Same question as before, but with ° any circle not containing the origin. Hint: Use the parametrization z = a+re for 0 • µ • 2… for the circle of center a 2 C and radius r > 0. Find a complex-valued function F(µ) of the real variable µ for 0 • µ • 2… such that d ¡ ¢ n d ¡ ¢ F(µ) = a + re iµ a + reiµ dµ dµ for 0 • µ • 2…. Consider F(2…) ¡ F(0). Distinguish between the case where jaj < r and the case where jaj > r. (c) Show that if jaj < r < jbj, then Z 1 2…i dz = ; °(z ¡ a)(z ¡ b) a ¡ b where ° denotes the circle centered at the origin, of radius r, with the positive orientation. Hint: Use the decomposition of 1 into partial fractionsA + B (z¡a)(z¡b) z¡a z¡b (where A and B are complex numbers) and use Part (b). Solution of Problem 1. (a) Use the parametrization z = e iµ for0 • µ • 2… for ° to compute the integral Z Z 2… Z 2… n inµ ¡ iµ i(n+1)µ z dz = e d e = ie dµ Z° µ=0 Z µ=0 2… 2… = ¡ sin((n + 1)µ)dµ + i cos((n + 1)µ)dµ µ=0 µ=0 which yields 2…i for n = ¡1 and 0 for n 6= ¡1. Math 113 (Spring 2009) Yum-Tong Siu 2 (b) We use the parametrization z = a + re iµ for 0 • µ • 2… for the circle °. Then Z Z 2… n ¡ iµn iµ z dz = a + re ire dµ: ° µ=0 For n 6= ¡1 we can use the explicit primitive 1 ¡ iµ¢n+1 F(µ) = a + re n + 1 ¡ ¢n for a + re iµ ireiµas a function of µ. Since F(2…) = F(0), it follows that the integral Z Z 2…¡ ¢n z dz = a + reiµ ire dµ = F(2…) ¡ F(0) = 0 ° µ=0 for n 6= ¡1. For n = ¡1, we can use the primitive ¡ ¢ ¡ ¢ def ﬂ ﬂ Im a + re iµ F(µ) = log a + re iµ = log a + re iﬂ+ itan ¡1 ; Re(a + re )µ where the numerical value of ¡ iµ¢ ¡1 Im a + re tan iµ Re(a + re ) is chosen as a continuous function of µ for 0 • µ • 2…. The integral Z dz = F(2…) ¡ F(0) z ° is equal to 2…i for jaj < r and to 0 if jaj > r. Instead of using this primitive for n = ¡1 we can also compute by brute force the integral Z Z dz 2… ire dµ = iµ ° z µ=0 a + re as follows by changing the interval of the parameter to ¡… • µ • … and using the rational parametrization of the unit circle 1 ¡ t2 cosµ = ; 1 + t2 Math 113 (Spring 2009) Yum-Tong Siu 3 sinµ = 2t ; 1 ¡ t 2dt dµ = 1 + t2 µ with t = tan 2 Before we perform this brute-force computation, we assume that a 6= 0 and make a coordinate change in z by replacing z byjajso that with respect to this new coordinate system a = jaj > 0. By multiplying both the denominator and the numerator of the integrand by the complex conjugate of the denominator to make the denominator real, we get Z Z … iµ Z 2… dz ire dµ acosµ + r + iasinµ = iµ= ir 2 2dµ ° z µ=¡… aZ+ re µ=0 a + 2ar cosµ + r … acosµ + r = ir dµ µ=¡… a + 2ar cosµ + r2 ‡ · because sin(¡µ) = ¡sinµ Z 1 a 1¡t + r = ir 1+t 2dt a + 2ar 1¡t + r21 + t2 Zt=¡1 1+t 1 (r + a) + (r ¡ a)t dt = 4ir t=0 (r + a) + (r ¡ a) t 1 + t Z 1 ˆ r¡a ! 1 r+a 1 = 2i ¡r¡a ¢2 + 2 dt t=0 r + a 1 + r+a t 1 + t • µ ¶ ‚ 1 ¡1 r ¡ a ¡1 = 2i tan r + a t + tan t •µ ¶ t=0 ‚ r ¡ a ‡… · ‡… · = 2i the sign of ¡ 0 + ¡ 0 r + a 2 2 ‰ = 2…i for a < r 0 for a > r; where r¡a ‰ r ¡ a ﬂr+aﬂ +1 for a < r the sign ofr + a = ﬂr¡aﬂ = ¡ 1 for a > r: r+a (c) By using Z Z dz dz = 2…i = 0 ° z ¡ a °z ¡ b Math 113 (Spring 2009) Yum-Tong Siu 4 and the partial fraction decomposition µ ¶ 1 1 1 1 = ¡ ; (z ¡ a)(z ¡ b) b ¡ a z ¡ b z ¡ a we get µ ¶ Z dz 1 Z dz Z dz 2…i = ¡ = : ° (z ¡ a)(z ¡ b) b ¡ a °z ¡ b °z ¡ a a ¡ b Problem 2 (from Stein & Shakarchi, p.64, #1). Prove that Z Z p 1 ¡ 2¢ 1 ¡ 2 2… sin x dx = cos x dx = : x=0 x=0 4 R R These are the Fresnel integrals. Here, is interpreted as lim R. 0 R!1 0 ¡z2 Hint: Integrate the holomorphic function f(z) = eover the path which is the boundary of ﬂ n iµﬂ … o z = re ﬂ0 < r < R; 0 < µ < 4 R1 2 p and use ¡1 e¡xdx = … which can be derived by squaring and using polar 2 coordinates in R . Solution of Problem 2. Denote by C tRe arc n iµ … o z = Re ﬂ0 • µ • 4 2 in the counter-clockwise sense. Let f(z) = . We are going to check that Z lim f(z)dz = 0: R!1 CR First we observe that 1 sin’ ‚ ’ 2 … 1 for 0 • ’ • 3 because if we set g(’) = sin’ 2 ’, then from g(0) = 0 and 0
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