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Unformatted text preview: Math 113 (Spring 2009) YumTong Siu Solutions of Problems of Math 113 Final Examination May 20, 2009, 9:15 a.m.  12:15 p.m. Science Center A Total Number of Points = 200 1 Problem 1. Evaluate the following two deﬁnite integrals by using the theory of residues. 2π dθ (a) (15 points) 2. θ=0 (5 + 4 cos θ ) (b) (15 points)
∞ x=−∞ x sin(2x) dx. x2 + 4 Solution. (a) Rewrite the integral as 2π 2π dθ 1 dθ 5 2 2= 16 θ=0 θ=0 (5 + 4 cos θ ) + cos θ 4 This is the same as Exercise #7 from Stein & Shakarchi, p.104, asking for the computation of 2π dθ 2π a (†) 3 2= (a + cos θ) 0 (a2 − 1) 2 with a > 1 when a = 5 . To evaluate the integral (†) with a general a > 1, 4 we use the parametrization a = eiθ for 0 ≤ θ ≤ 2π so that 1 1 dz cos θ = z+ , dθ = . 2 z iz Hence dθ (a + cos θ)2 θ=0 dz = −i 2 1 1 z =1 z a + z+z 2
2π Math 113 (Spring 2009) YumTong Siu = −i = −i 4zdz (2az + z 2 + 1)2 4zdz 2 z =1 = −4i z =1 For the meromorphic function 2 (z + a)2 − (a2 − 1) zdz √ √ 2 2 . z + a + a2 − 1 z + a − a2 − 1
z =1 z f (z ) = √ √ 2 2 z + a + a2 − 1 z + a − a2 − 1 √ √ the only poles are at −a − a2 − 1 √ −a + a2 − 1, both double poles. and Since a > 1, clearly the pole −a − a2 − 1 is outside the unit circle. The √ point −a + a2 − 1 has absolute value < 1, because we can look at the right√ angled triangle with two sides of length 1 and a2 − 1 and with hypothenuse a. The diﬀerence of the length of the hypothenuse a and the length of one side abutting the right angle is less than the length of the other side which √ is 1. The residue of f at the double pole −a + a2 − 1 is computed by 2 √ d 2−1 √ res−a+ a2 −1 f = z+a− a f (z ) √ dz z =−a+ a2 −1 d z = √ dz z + a + a2 − 1 2 √ z =−a+ a2 −1 √ −z + a + a2 − 1 a = = √ 3 3. √ z + a + a2 − 1 4 (a2 − 1) 2
z =−a+ a2 −1 Thus 2π 0 dθ 2π a √ 3. 2 = (−4i)2π i res−a+ a2 −1 f = (a + cos θ) (a2 − 1) 2
2π With a =
2π 5 4 we get 5
4 θ=0 dθ 1 2= 16 (5 + 4 cos θ) dθ + cos θ θ=0 2 = 2π 5 1 10 π 4 3 = 27 . 16 2 52 −1 4 Math 113 (Spring 2009) YumTong Siu (b) The change of variables u = 2x transforms the integral ∞ x sin(2x) dx 2 x=−∞ x + 4 to
u sin u 2 2 u u=−∞ +4 2 ∞ 3 d This is the same as Exercise #4 from Stein & Shakarchi, p.103, asking for the computation of ∞ x sin x (∗∗) dx 2 2 x=−∞ x + a for a > 0 when a = 4. To compute (∗∗), we let f (z ) = zeiz . z 2 + a2 u 2 = ∞ u=−∞ u sin u du. u2 + 16 The poles of f (z ) are the zeroes of z 2 + a2 which are ai and −ai. Only ai is in the upper halfplane. We now integrate f (z )dz over the boundary of the upper halfdisk of radius R centered at the origin and let R → ∞. The integral f (z )dz
CR over the upper halfcircle CR of radius R centered at 0 in the counterclockwise sense goes to 0 as R → ∞, because integration by parts by integrating eiz yields z=−R zeiz zeiz d z iz dz = − e dz z 2 + a2 i (z 2 + a2 ) z=R dz i (z 2 + a2 ) CR z=−R zeiz a2 − z 2 = − eiz dz i (z 2 + a2 ) z=R i (z 2 + a2 )2 CR CR and because eiz  ≤ 1 for z ∈ CR and z=−R zeiz 2R for R > a ≤ 2 i (z 2 + a2 ) z=R R − a2 Math 113 (Spring 2009) YumTong Siu and (a2 + R2 ) π R a2 − z 2 e dz ≤ for R > a. i (z 2 + a2 )2 (R2 − a2 )2 CR
iz 4 The residue resai f of f at the simple pole ai of f is given by zeiz zeiz e−a resai f = lim (z − ai) 2 = = . z →ai z + a2 z + ai z=ai 2 By applying the theory of residues to the integration of f (z )dz over the boundary of the upper halfdisk of radius R centered at the origin and letting R → ∞, we get ∞ ∞ x sin xdx xeix dx = Im 2 2 2 2 x=−∞ x + a x=−∞ x + a = Im f (z )dz = Im (2π i resai f ) R e−a = Im 2π i = π e−a . 2 With a = 4 we get
∞ x=−∞ x sin x π dx = 4 . x2 + 16 e Problem 2. Evaluate the following two deﬁnite integrals by using the theory of residues and branchcuts. ∞ log x (a) (16 points) dx. 2 x=0 x + 4 (b) (16 points)
∞ x=0 xα dx for − 1 < α < 1. x2 + 9 Solution. (a) This is the same as Exercise #10 from Stein & Shakarchi, p.104, asking for the computation of ∞ log x (‡) dx x2 + a 2 0 Math 113 (Spring 2009) YumTong Siu 5 for a > 0 when a = 2. To evaluate (‡), we choose a branch of log z = log z  + i arg z by specifying the range − π < arg z < 3 and choose 2 2 f (z ) = log z . z 2 + a2 The contour is the boundary, in the counterclockwise sense, of the set which is equal to the upper halfdisk of radius R > 0 minus the upper disk of radius ε with 0 < ε < R. The integral over the upper halfcircle CR of radius R centered at the origin approaches 0 as R → ∞, because log R + π f (z )dz ≤ 2 π R → 0 as R → ∞. R − a2 CR The integral over the upper halfcircle Cε of radius ε centered at the origin approaches 0 as r → 0, because − log ε + π f (z )dz ≤ πε → 0 as ε → 0. a2 − ε2 Cε On the upper halfplane there is only one pole for f (z ) which is a simple pole at ai. The residue of f (z ) at ai is computed by log z resai f = lim (z − ai) 2 z →ai z + a2 log a + π i log z 2 = = . z + ai z=ai 2ai The theory of residues yields ∞ 0 π log a + π i log x dx (log(−x) + π i) dx 2 + = 2π i resai f = . 2 2 x2 + a 2 a x=0 x + a x=−∞ Taking the real part of both sides, we get ∞ 0 log x dx log(−x) dx π log a + = . 2 + a2 2 + a2 x a x=0 x x=−∞ Changing the sign of the variable x from x to −x in the second integral on the lefthand side, we get ∞ log x dx π log a 2 = 2 2 a x=0 x + a Math 113 (Spring 2009) YumTong Siu and 6 ∞ x=0 ∞ log x dx π log a = . x2 + a2 2a When a = 2, we get x=0 log x π log 2 dx = . 2+4 x 4 (b) Use the change of variables x = 3u to transform the integral to ∞ ∞ ∞ xα (3u)α uα α−1 dx = d (3u) = 3 du 2 2 2 x=0 x + 9 u=0 (3u) + 9 u=0 u + 1 so that we need only consider the simpler integral ∞ xα dx. 2 x=0 x + 1 To compute this simpler integral, we deﬁne z α by choosing the numerical value of the polarcoordinate angle of z to be in the open interval (0, 2π ). απ i For this branch the value of z α at z = i is e 2 and the value of z α at z = −i 3απ i is e 2 . We denote by Cs the circle of radius s > 0 centered at the origin. We integrate the holomorphic function f (z ) := zα z2 + 1 over the contour which starts from the real point r, goes along the real axis from r to R and then follows the curve CR in the counterclockwise sense to get from the real point R back to the same real point R and then goes along the real axis from R to r and then follows the curve Cr in the clockwise sense to get from the real point r back to the same real point r. We now calculate the two residue Resz=i f (z ) and Resz=−i f (z ). For the ﬁrst residue we have απ i 2 zα =e . Resz=i f (z ) = z + i z=i 2i For the second residue we have 3απ i zα e2 Resz=−i f (z ) = = . z − i z=−i −2i Math 113 (Spring 2009) YumTong Siu By the residue theorem over the contour described above, R R iα2π α xα e x f (z )dz − f (z )dz − dx 2 dx + 2 1 + x2 r (1 + x ) CR Cr r = 2π i (Resz=i f (z ) + Resz=−i f (z )) απi 3απ i 2 2 απ i e e = 2π i + = π e 2 1 − eαπi . 2i −2i 7 The reason for the last integral on the lefthand side of the equation is that the value of z α is xα eiα2π at the real point x for that integrand, because for that integrand at the real point x the angle θ for its polar representation is 2π . As R → ∞, Rα f (z )dz ≤ sup f (z ) · (length of CR ) ≤ · 2π R z∈C 1 + R2 CR R approaches 0, because α < 1. As r → 0, rα f (z )dz ≤ sup f (z ) · (length of Cr ) ≤ · 2π r z∈C 1 + r2 r Cr approaches 0, because α > −1. Hence ∞ xα απ i iα2π 1−e dx = π e 2 1 − eαπi 1 + x2 0 and
∞ 0 π πe 2 π 2 = = −απi . απ i = 1 − eiαπ cos απ e 2 −e 2 2 xα π e 2 (1 − eαπi ) dx = 1 + x2 1 − eiα2π
απ i απ i The ﬁnal answer is ∞ 0 The ﬁnal answer is ∞ xα (1 − a)π π. dx = 2 1+x 4 cos a2 3α−1 π xα 2 dx = πα . 2+9 x cos 2 x=0 Math 113 (Spring 2009) YumTong Siu 8 Problem 3 (20 points). Let D be a connected open subset of C. A holomorphic function g (z ) on D is said to be univalent if g (z1 ) = g (z2 ) for any z1 = z2 in D. Suppose fn (z ) is a sequence of univalent holomorphic functions on D such that fn (z ) approaches some nonconstant function f (z ) uniformly on D as n → ∞. Prove that f (z ) is a univalent holomorphic function by using Rouch´’s theorem to show that if f (z1 ) = f (z2 ) for some z1 = z2 in e D, then there exist some positive integer N and some 0 < ε < z1 − z2  such that for n ≥ N there exists ζ2 ∈ D with ζ2 − z2  < ε and fn (z1 ) = fn (ζ2 ). Solution. Since f (z ) is the uniform limit D of the sequence of holomorphic functions fn (z ) on D, it follows for any closed solid triangle T in D f (z )dz = lim fn (z )dz = 0
∂T n→∞ ∂T (where ∂ T is the boundary of T ) and by Morera’s Theorem f (z ) is holomorphic on D. Assume that f (z ) is not univalent on D so that f (z )1 = f (z2 ) for some z1 = z2 in D and we are going to derive a contradiction. Since f (z ) is nonconstant on D and D is connected, f (z ) − f (z1 ) is not identically zero and there exists some 0 < ε < z1 − z2  which is less than the distance from z2 to the boundary of D such that the holomorphic function f (z ) − f (z1 ) is nowhere zero on z − z2  = ε. Let η > 0 such that f (z ) − f (z1 ) ≥ η on z − z2  = ε. There exists some positive integer N such that f (z ) − fn (z ) < η on z − z2  = ε for n ≥ N and f (z1 ) − fn (z1 ) < η for n ≥ N . Then 2 2 (f (z ) − f (z1 )) − (fn (z ) − fn (z1 )) ≤ f (z ) − fn (z ) + f (z1 ) − fn (z1 ) ηη < + = η ≤ f (z ) − f (z1 ) 22 on z − z2  = ε for n ≥ N . By Rouch´’s theorem, fn (z ) − fn (z1 ) and f (z ) − e f (z1 ) have the same number of solutions in z − z2  < ε for n ≥ N . Since z2 is a zero of f (z ) − f (z1 ) on z − z2  < ε, we conclude that for n ≥ N there exists some ζ2 ∈ D with ζ2 − z2  < ε such that ζ2 is a root of fn (z ) − fn (z1 ). This contradicts the univalent property of fn (z ) and we conclude that f (z ) must be univalent on D. Problem 4. Let τ be a complex number with Im(τ ) > 0. Math 113 (Spring 2009) YumTong Siu 9 (a) (14 points). Use the theory of residues to verify that the Fourier transform ˆ f (ξ ) = f (x)e−2πiξx dx
x∈R of the function f (x) = 1 (τ +x)2 is zero for ξ < 0 and is of the form ˆ f (ξ ) = A ξ e2πiξτ for ξ > 0, where A is a real number. Determine the real number A. (b) (10 points). Apply the Poisson summation formula to the function f (x) = 1 and use (a) to compute (τ +x)2 1 (τ + n)2 n=−∞ in terms of sin(πτ ). Hint: Expand eiπτ + e−iπτ e2iπτ + 1 2 cot πτ = i iπτ = i 2iπτ =i 1− e − e−iπτ e −1 1 − e2iπτ as a power series in e2iπτ and diﬀerentiate both sides with respect to τ . Solution. (a) For ξ < 0 and R, b > 0 let Γ+ be the closed rectangle whose R,b four vertices are −R, R, R + bi, −R + bi. Since e−2πiξz (τ + z )2 is holomorphic at every point of Γ+ as a result of Im tau > 0 and Im (tau + z ) > R,b 0 for z ∈ Γ+ , it follows that R,b e−2πiξz 2 dz = 0, ∂ Γ+ (τ + z ) R,b where ∂ Γ+ is the boundary of Γ+ in the counterclockwise sense. We have R,b R,b −2πiξz e e2πξ Im z (τ + z )2 = τ + z 2 ,
∞ Math 113 (Spring 2009) YumTong Siu which implies that, for ξ < 0, lim
R→∞ R→∞ 10 lim z =R+η i, 0≤η ≤b e−2πiξz dz = 0, (τ + z )2 e−2πiξz dz = 0 (τ + z )2 e−2πiξz dz. (τ + z )2 z =−R+η i, 0≤η ≤b and ˆ f (ξ ) = x∈R f (x)e −2π iξ x dx = z =x+ib, −∞<x<∞ On the other hand, ∞ e−2πiξz dx 2πξ b 2 dz ≤ e 2 2 z=x+ib, (τ + z ) x=−∞ (x + Re τ ) + (Im τ ) −∞<x<∞ ˆ approaches 0 as b → ∞, because ξ < 0. Hence f (ξ ) = 0 for ξ < 0. ˆ For the computation of f (ξ ) for ξ > 0 we introduce the closed rectangle − ΓR,b for R, b > Im τ whose four vertices are −R, R, R − bi, −R − bi. Since e−2πiξz (τ + z )2 is holomorphic at every point of Γ− except at the point z = −τ which is a R,b double pole, it follows that e−2πiξz e−2πiξz dz = 2π i Resz=−τ , 2 (τ + z )2 ∂ Γ− (τ + z ) R,b where ∂ Γ− is the boundary of Γ− in the counterclockwise sense. We have R,b R,b −2πiξz e e2πξ Im z (τ + z )2 = τ + z 2 , e−2πiξz dz = 0, (τ + z )2 which implies that, for ξ > 0, lim
R→∞ z =R−η i, 0≤η ≤b Math 113 (Spring 2009) YumTong Siu e−2πiξz dz = 0 (τ + z )2 11 R→∞ lim z =−R−η i, 0≤η ≤b and ˆ f (ξ ) =
x∈R f (x)e −2π iξ x dx = z =x−ib, −∞<x<∞ e−2πiξz e−2πiξz dz + 2π i Resz=−τ . (τ + z )2 (τ + z )2 On the other hand, ∞ e−2πiξz dx −2πξ b 2 dz ≤ e 2 2 z=x−ib, (τ + z ) x=−∞ (x + Re τ ) + (Im τ ) −∞<x<∞ approaches 0 as b → ∞, because ξ > 0. Hence
−2π iξ z ˆ(ξ ) = 2π i Resz=−τ e f (τ + z )2 for ξ > 0. We now compute the residue e−2πiξz d −2πiξz Resz=−τ e = 2π i ξ e2πiξτ 2= dz (τ + z ) z =−τ ˆ f (ξ ) = −4π 2 ξ e2πiξτ and conclude that for ξ > 0. This means that the constant A = −4. (b) We expand eiπτ + e−iπτ e2iπτ + 1 2 cot πτ = i iπτ = i 2iπτ =i 1− e − e−iπτ e −1 1 − e2iπτ as a power series in e2iπτ to get cot πτ = i 1 − 2
∞ n=0 e2niπτ . We diﬀerentiate both sides with respect to τ to get ()
∞ −π = 4π n e2niπτ . sin2 (πτ ) n=0 Math 113 (Spring 2009) YumTong Siu By Poisson’s summation formula we have
∞ 12 ˆ f (n) = n=−∞ n=−∞ n f (n), which by the result from (a) means −4π From () it follows that 1 π2 = . (τ + n)2 sin2 (πτ ) n=−∞ Problem 5 (20 points). Use M¨bius transformations and the Schwarz lemma o to show that if (z ) is a holomorphic map from the open unit disk D = f z ∈ C z  < 1 to itself, then (∗) f (z ) 1 2≤ 1 − f (z ) 1 − z 2 for z ∈ D, where f (z ) is the derivative of f (z ). Moreover, if the lefthand side and the righthand side of (∗) holds at some point z of D, show that f is a biholomorphic map from D to itself. Solution. We will use a M¨bius transformation to reduce the inequality for o a general point z ∈ D and a general holomorphic map f : D → D to the special case f (0) = 0 and for the special point z = 0. For that purpose, we are going to see how the quotient dz  1 − z 2 transforms in a biholomorphic selfmap of D. Consider a general biholomorphic selfmap of D which must be of the form ζ = eiα z−a az − 1 ¯
∞ 2 ∞ n=1 ne 2π inτ = 1 (τ + n)2 n=−∞ ∞ Math 113 (Spring 2009) YumTong Siu with α ∈ C and a ∈ C with a < 1. Since z − a 2 1 − ζ 2 = 1 − az − 1 ¯ az − 12 − z − a2 ¯ = az − 12 a2 z 2 − az − az + 1 − (z 2 − az − az + a2 ) ¯ ¯ ¯¯ = 2 az − 1 (1 − a2 )(1 − z 2 ) = az − 12 and dζ −(1 − a2 ) = eiα , dz (¯z − 1)2 a dζ  dz  = . 1 − ζ 2 1 − z 2 dz  1 − z 2 13 it follows that This means that the expression is invariant under any biholomorphic selfmap of D. Take any point a ∈ D. Besides the two complex variables z and w used in w = f (z ), we are going to introduce two more complex variables ζ and ω . Let z = S (ζ ) = and w = T (ω ) = ζ +a 1 + aζ ¯ . ω + f (a) 1 + f (a)ω Let ω = h(ζ ) be deﬁned by h = T −1 ◦ f ◦ S −1 so that f = T ◦ h ◦ S . Then h : D → D is holomorphic and h(0) = 0. Since the quotient dz  1 − z 2 Math 113 (Spring 2009) YumTong Siu is invariant under any biholomorphic selfmap of D, it follows that dζ  dz  = 2 1 − ζ  1 − z 2 and dζ  dz  = 2 1 − ζ  1 − z 2 14 At z = a we have ζ = 0 and w = f (a) and ω = 0 and dz  dζ  = 1 − z 2 z=a ζ =0 and dω 
ω =0 The inequality at z = a is equivalent to f (z ) 1 2≤ 1 − f (z ) 1 − z 2 dw = . 1 − w2 w=f (a) which in turn is equivalent to dw dz  ≤ dζ  = 1 − w2 w=f (a) 1 − z 2 z=a ζ =0 dζ 
ζ =0 which in turn is equivalent to ≤ dω  ω =0 h (0) ≤ 1 which now follows from Schwarz’s lemma. If the inequality f (z ) 1 2≤ 1 − f (z ) 1 − z 2 becomes an equality at z = a, then we have h (0) = 1 and by the Schwarz lemma h(z ) = eiβ z for some real number β and is a biholomorphic selfmap of D, which forces f to be the composite of three biholomorphic selfmaps S , h, T of D and be a biholomorphic selfmap of D. Math 113 (Spring 2009) YumTong Siu 15 Problem 6 (32 points). Use argument functions (in a way analogous to their use in the derivation of SchwarzChristoﬀel transformations) and follow the steps given below to show that for any integer n ≥ 3 the map z → F (z ) deﬁned by z dζ F (z ) = 2 1 (1 − ζ n ) n maps the open unit disk D = z ∈ C z  < 1 onto the interior of a regular polygon of n sides whose total perimeter is given by π dθ A 2 0 (sin θ ) n for some positive rational number A. Find the rational number A explicitly as a function of n. Step One (6 points of the 32 points of the Problem). Let C be the unit circle which is the boundary of D. Let n ≥ 3. Let m be the largest integer < n . 2 2k π i For any nonnegative integer k let ωk = e n . For any nonnegative integers k and the function z arg (z − ωk )2 is constant for z in the open arc of the unit circle C from ω to ω+1 in the counterclockwise sense. Step Two (8 points of the 32 points of the Problem). Let k be any nonnegative integer. For z = eiθ the function 1 dz arg 2 (1 − z n ) n dθ is constant for z in the open arc of the unit circle C from ωk to ωk+1 in the counterclockwise sense. Step Three (6 points of the 32 points of the Problem). Let k be any nonnegative integer and let arg (z − ωk ) be deﬁned with the branchcut r ωk r ≥ 1 . Then as z goes counterclockwise on the unit circle C past ωk , there is a jump of −π in the value of arg (z − ωk ) at the point ωk . Math 113 (Spring 2009) YumTong Siu 16 Step Four (6 points of the 32 points of the Problem). For any nonnegative integer k , let ωk ωk+1 be the open arc on the unit circle C from ωk to ωk+1 in the counterclockwise sense. Then dζ 2 ω (1 − ζ n ) n ω
k k+1 is equal to π dθ 0 (sin θ) n 2 times a positive constant which is independent of the nonnegative integer k . Final Step (6 points of the 32 points of the Problem). Finish the proof after the preceding four steps. Solution. This is the same as Exercise #23 from Stein & Shakarchi, p.254.
2π Proof of Step One. Since z must be of the form e(θ+ n )i for some 0 < θ < we can rewrite 2π z e(θ+ n )i arg = arg 2π 2kπ i 2 (z − ωk )2 e(θ+ n )i − e n 2π , n = arg e(θ+ 2(+k)π n 2 )i e 1 (θ+ 2(−k)π )i − e −1 (θ+ 2(−k)π )i 2 n 2 n e−
2kπ n 2π e(θ+ n )i i π which is independent of θ, because 0 < θ < 2n and 1 2( − k )π 2 sin θ+ > 0. 2 n = arg e− = arg
2k π n i 2 (θ+ 2(−k)π )i − e −1 (θ+ 2(−k)π )i n 2 n e −2kπi −1 = arg −e n 4 sin2 1 θ + 2(−k)π 2 n
1 2 Math 113 (Spring 2009) YumTong Siu 17 Proof of Step Two. The statement in Step Two is the same as the function n−1 n dz 1 dz dθ arg = arg (1 − z n )2 dθ (z − ωj )2 j =0 being constant for z in the open arc ωk ωk+1 of the unit circle C from ωk to ωk+1 in the counterclockwise sense. Using z = eiθ , we have dz = i eiθ = i z dθ and the statement to be veriﬁed is the same as the function n−1 n−1 n 1 dz iz iz arg = arg = arg (1 − z n )2 dθ (z − ωj )2 (z − ωj )2 j =0 j =0 being constant for z in the open arc ωk ωk+1 of the unit circle C from from ωk to ωk+1 in the counterclockwise sense. Since we have already veriﬁed earlier in Step One that each summand iz arg (z − ωj )2 is constant for z in the open arc ωk ωk+1 of the unit circle C from from ωk to ωk+1 in the counterclockwise sense, it follows that for z = eiθ the function 1 dz arg 2 (1 − z n ) n dθ is constant for z in the open arc ωk ωk+1 of the unit circle C from ωk to ωk+1 in the counterclockwise sense. Proof of Step Three. Let k be a nonnegative integer. Let P be a point on the unit circle C which is very close to ωk such that to go from P to ωk requires a very small counterclockwise movement. Let Q be a point on the unit circle C which is very close to ωk such that to go from ωk to Q requires a very small counterclockwise movement. Then the direction of the vector ωk P (joining ωk to P ) rotates to the direction of the vector O ωk (joining the origin O to ωk ) by an angle π + εP for some very small positive number 2
−→ −→ Math 113 (Spring 2009) YumTong Siu
−→ 18 εP . The direction of the vector O ωk (joining the origin O to ωk ) rotates to the direction of the vector ωk Q (joining ωk to Q) by an angle
−→ −→ π 2 + εQ for
−→ some very small positive number εQ . Thus the direction of the vector ωk P (joining ωk to P ) rotates to the direction of the vector ωk Q (joining ωk to Q) by the angle π + εP + εQ . However, since we are using the branchcut r ωk r ≥ 1 for the function arg (z − ωk ), we are not allowed to rotate the direction of
−→ −→ the the vector ωk P (joining ωk to P ) in the counterclockwise direction to the direction of the vector ωk Q (joining ωk to Q). We can only do this in the clockwise direction in order for the variable z representing P and moving to Q along C not to cross the branchcut other than at the point ωk . The angle needed for this clockwise rotation is π − εP − εQ . Thus as z goes counterclockwise on the unit circle C past ωk , there is a jump of −π in the value of arg (z − ωk ) at the point ωk , because of the need to use the clockwise direction. Proof of Step Four. Since d 1 F (z ) = 2 dz (1 − z n ) n and along the unit circle C with z = eiθ it follows that d 1 dz F (z ) = , 2 dθ (1 − z n ) n dθ d arg F (z ) = arg dθ dz 2 (1 − z n ) n dθ 1 is constant when z is on the open arc ωk ωk+1 from ωk to ωk+1 along the unit circle in the counterclockwise sense. This means that the open arc ωk ωk+1 from ωk to ωk+1 along the unit circle in the counterclockwise sense is mapped by F (z ) to an open linesegment Lk . The length of the image linesegment Lk is given by dζ 2 ζ=eiθ , 2kπ <θ< 2(k+1)π (1 − ζ n ) n
n n Math 113 (Spring 2009) YumTong Siu which after a change of variables ζ → ζ e− ı becomes dζ 2 , n n ζ =eiθ ,0<θ < 2π (1 − ζ ) n
2kπ 19 n 2k π because ζ e− ı = ζ n . Thus the length of the image linesegment Lk is given by 2π 2π 2π n n n dθ dθ dθ = = nθ 2 2 2 2 nθ n θ=0 (1 − einθ ) n θ=0 nθ θ=0 2 n sin n 2 e 2 i − e− 2 i which becomes
2 n 2 θ=0 sin n θ n after we replace θ by 2 θ. Thus the total perimeter (which is equal to n times the length of the image linesegment Lk ) is equal to π 2 dθ 1− n 2 2 θ=0 sin n θ and the number A is equal to 21− n . Final Step of the Solution of Problem 5. In the deﬁnition z dζ F (z ) = 2 1 (1 − ζ n ) n for the map F (z ), for some real number α we can rewrite the integrand as z dζ iα () F (z ) = e 2. n−1 1 (ζ − ωk ) n k=0
2 2 n π dθ
2 Step Two tells us that as z goes along the open arc ωk ωk+1 of the unit circle C from ωk to ωk+1 in the counterclockwise sense, the value F (z ) stays in a straight line segment Lk . From () Step Three tells us that as z along C passes ωk in the counterclockwise sense, its image F (z ) turns an angle equal 2 π to − n (−π ) which is equal to 2n . By Step Four the length of the linesegments Math 113 (Spring 2009) YumTong Siu 20 Lk is independent of k and the sum of such lengths over 0 ≤ k ≤ n − 1 is equal to π 2 dθ 1− n 2 . 2 θ=0 sin n θ Hence the image of the closed unit disk under z → F (z ) is a regular polygon of n sides whose total parameter is equal to π 2 dθ 1− n 2 . 2 θ=0 sin n θ Problem 7. (a) (10 points). Let z = sin w with z = x + iy and w = u + iv and x, y, u, v ∈ R. Verify that x2 y2 − =1 sin2 u cos2 u by using the addition formula for the sine function of a complex variable, expressed in terms of the sine function, the cosine function, the hyperbolic sine function, and the hyperbolic cosine function. The hyperbolic sine and cosine functions are deﬁned respectively by sinh ξ = Hence show that 1 ξ e − e−ξ , 2 cosh ξ = 1 ξ e + e−ξ . 2 (x + 1)2 + y 2 − (x − 1)2 + y 2 = 2 sin u for 0 ≤ u ≤ π by using the characterization of a hyperbola as the locus of a 2 point whose distances from two ﬁxed points have a constant diﬀerence. (b) (8 points). Let w = sin z . Use the mapping behavior of the map 1 1 w= ζ+ 2 ζ from the polar coordinates for ζ ∈ C and the Cartesian coordinates for w ∈ C and use ζ = eiZ and z = π − Z to show that w = sin z maps the semiinﬁnite 2 strip π π − ≤ Re z ≤ , Im z ≥ 0 2 2 bo the upper halfplane { Im w ≥ 0 }. Math 113 (Spring 2009) YumTong Siu 21 (c) (6 points). Use (a) and (b) to ﬁnd a bounded harmonic function U (x, y ) on the open upper halfplane which is of the form −1 2 + y2 − 2 + y2 U (x, y ) = a + b sin c (x + 1) (x − 1) for some real numbers a, b, c such that (i) u has the boundary value 0 on {x < −1, y = 0}, (ii) u has boundary value 1 on {x > 1, y = 0}, and (iii) u has normal derivative zero at the boundary points on {−1 < x < 1, y = 0}. Solution. (a) The addition formula sin (w1 + w2 ) = sin w1 cos w2 + cos w1 sin w2 which holds when both w1 , w2 are real holds also for w1 , w2 ∈ C, because, for ﬁxed w2 real, the diﬀerence of both sides is an entire function in w1 which vanishes identically for w1 ∈ R must also vanish for w1 ∈ C and because we can also hold a complex w1 ﬁxed and argue that the diﬀerence of both sides as an entire function in w2 which vanishes for w2 real must also vanish for w2 complex. For z = x + iy and w = u + iv with z = sin w we have x+iy = sin (u + iv ) = sin u cos(iv )+cos u sin(iv ) = sin u cosh v +i cos u sinh v, because 1 −v 1 i(iv) e + e−i(iv) = e + ev = cosh v, 2 2 1 i(iv) 1 −v 1 sin(iv ) = e − e−i(iv) = e − ev = i ev − e−v = i sinh v. 2i 2i 2 cos(iv ) = x = sin u cosh v, and y = cos u sinh v Hence x2 y2 − = cosh2 v − sinh2 v = 1. 2 2u sin u cos x2 y 2 − 2 =1 a2 b A hyperbolic Math 113 (Spring 2009) YumTong Siu 22 can be alternatively deﬁned by as the locus of a point whose distances to the √ two foci (c, 0) and (−c, 0) with c = a2 + b2 have a diﬀerence equal to 2a. In other words, the equation of the hyperbola can be rewritten as 2 2 2− 2 = 2a, (x + c) + y (x − c) + y as one can see by using the squaring process twice to get rid of the square roots in the last equation. For our case, a = sin u and b = cos u so that √ c = a2 + b2 = 1 and the equation x2 y2 − = cosh2 v − sinh2 v = 1. sin2 u cos2 u is equivalent to 2 2 (x + 1) + y 2 − (x − 1) + y 2 = 2 sin u. When 0 ≤ u ≤ π , the value sin u ≥ 0 and as a result we can take away the 2 absolute value in the above equation and get 2 2− (x + 1) + y (x − 1)2 + y 2 = 2 sin u. (b) Use the Cartesian coordinates (u, v ) for w = u + iv with u, v ∈ R and use the polar coordinates (ρ, ϕ) for ζ = ρeiϕ with ρ ≥ 0 and ϕ ∈ R. From 1 1 1 1 −iϕ iϕ w= ζ+ = ρe + e 2 ζ 2 ρ it follows that 1 1 u= ρ+ cos ϕ, 2 ρ 1 1 v= ρ− sin ϕ. 2 ρ The upper halfcircle with radius > 1 deﬁned by ζ = ρeiϕ Math 113 (Spring 2009) YumTong Siu for ρ > 1 and 0 ≤ ϕ ≤ π is mapped to the upper halfellipse
1 2 23 u2 ρ+
1 ρ 1 in the upper halfplane with semimajor axis 1 ρ + ρ and semiminor axis 2 1 1 1 ρ − ρ . As ρ → 1+ , the semiminor axis 1 ρ − ρ → 0+ . Hence the map 2 2 ζ → W maps oneone the exterior of the unit disk in the upper halfplane of ζ to the upper halfplane of W , (i) with the upper half unit circle in the variable ζ mapped to the interval [−1, 1] in the variable w, (ii) the halfline [1, ∞) in the variable ζ mapped to the the halfline [1, ∞) in the variable w, and (iii) the halfline (−∞, −1] in the variable ζ mapped to the the halfline (−∞, −1] in the variable w. With ζ = eiZ the exterior of the the unit disk in the upper halfplane of ζ corresponds to Re(iZ ) > 0 and 0 < Im(iZ ) < π in the variable iZ , which in turn corresponds to Im z < 0 and 0 < Re z < π . We have 1 1 1 iZ w= ζ+ = e + e−iZ = cos Z 2 ζ 2 and it maps { Im Z < 0, 0 < Re Z < π } to the upper halfplane in the variable w, (i) with the interval [0, π ] in the variable Z corresponds to the interval [−1, 1] in the variable w, (ii) the halfray { Re Z = 0, Im Z ≤ 0 } in the variable z corresponds to the interval [1, ∞) in the variable w, and (iii) the halfray { Re Z = π , Im Z ≤ 0 } in the variable z corresponds to the interval (−∞, −1] in the variable w. 2 +
1 2 v2 ρ−
1 ρ 2 = 1 Math 113 (Spring 2009) YumTong Siu 24 Finally the transformation z = π − Z gives w = sin z which maps the 2 π π Im z > 0, − < Re z < 2 2 to the upper halfplane in the variable w. Moreover, (i) the interval − π , π in the variable z corresponds to the interval [1, ∞) 22 in the variable w, (ii) the halfray x = π , y ≥ 0 in the variable z corresponds to the inter2 val (−∞, −1] in the variable w, and (iii) the halfray x = − π , y ≥ 0 in the variable z corresponds to the 2 interval [−1, 1] in the variable w. (c) From the mapping behavior of w = sin z with w = u + iv and z = x + iy we know that the function u as a function of (x, y ) on the upper halfplane assumes the constant boundary value − π on {x > 1, y = 0} and the constant 2 boundary value π on {x < −1, y = 0} and has derivative zero at the boundary 2 1 points on {−1 < x < 1, y = 0}. Thus U (x, y ) = 1 + π u(x, y ) assumes the 2 constant boundary value 1 on {x > 1, y = 0} and the constant boundary value 0 on {x < −1, y = 0} and has derivative zero at the boundary points on {−1 < x < 1, y = 0}. Because of the result in (a), in terms of (x, y ) the function U (x, y ) can be written as 11 1 U (x, y ) = + sin−1 (x + 1)2 + y 2 − (x − 1)2 + y 2 . 2π 2 1 1 Hence a = 2 and b = π and c = 1 . 2 Problem 8. Let ω1 , ω2 ∈ C be Rlinearly independent. Let η1 , η2 be nonzero complex numbers and ξ1 , ξ2 ∈ C. Suppose f (w) is an entire function on C such that Let c ∈ C such that f is nowhere zero on the four sides of the parallelogram Ω with vertices a, a + ω1 , a + ω2 , a + ω1 + ω2 . (a) (9 points) Use the argument principle to show that the case η1 = 0 and η2 = i ω1 can never occur. (b) (9 points) Use the argument principle to compute the number of zeroes of f (with multiplicities counted) in the interior of the parallelogram Ω if ω1 = π and ω2 = πτ and η1 = 0 and η2 = −8i. f (w + ων ) = eην w+ξν f (w) for ν = 1, 2 and w ∈ C. Math 113 (Spring 2009) YumTong Siu Solution. From f (w + ων ) = eην w+ξν f (w) for ν = 1, 2 and w ∈ C it follow that d d log f (w + ων ) = ην + log f (w) for ν = 1, 2 and w ∈ C dw dw and the integral of equal to w=a+ω1
d dw 25 log f (w) along the boundary of the parallelogram Ω is w=a+ω2 d d log f (w) dw + log f (w + ω1 ) dw dw dw w =a w =a w=a+ω1 w=a+ω2 d d − log f (w + ω2 ) dw − log f (w)dw dw dw w =a w=a w=a+ω1 d d =− log f (w + ω2 ) − log f (w) dw dw dw w=a w=a+ω2 d d + log f (w + ω1 ) − log f (w) dw dw dw w=a 1 (−ω1 η2 + ω2 η1 ) 2π i which is equal to −ω1 η2 + ω2 η1 and by the argument principle is equal to the number of zeroes of f (w) in the open parallelogram Ω and must be a nonnegative integer. (a) When η1 = 0 and η2 = i ω1 , we have 1 1 ω1 2 (−ω1 η2 + ω2 η1 ) = (−ω1 i ω1 ) = − <0 2π i 2π i 2π which is a negative number, giving a contradiction. (b) When ω1 = π and ω2 = πτ and η1 = 0 and η2 = −8i, we have 1 1 (−ω1 η2 + ω2 η1 ) = (−π (−8i)) = 4, 2π i 2π i from which we conclude that there are four zeroes of f (with multiplicities counted) in the interior of the parallelogram Ω if ω1 = π and ω2 = πτ and η1 = 0 and η2 = −8i. ...
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 Spring '09
 Integrals

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