APPM 436/536 Homework Assignment #6 Solutions Spring 8 Problem # ( points: onsider f (zlog z ; z r e iθ. Discuss/explain the analytic continuation of the function from R R R3 where r > and θ is in the regions: R: π/ θ π/; R: π/3 θ π/3; R3: π/ θ π. In particular, does f (r,θ π/4 f (r,θ 7π/4? Explain. Solution: f (zlog z log z. We have to find such branches of log z, f (z in R, f (z in R and f 3 (z in R3 that an analytic function f (z can be defined in R R R3 by f (z, f (z f (z, f 3 (z, z R z R z R3 This is possible if f (z f (z in R R and f (z f 3 (z in R R3. First, we have to choose f (z as a branch analytic in R. Define f (zlog z for <r <, π θ π (this is the principal branch of log z with the branch cut on (,]. Then the range [ π,π] of θ covers the interval [π/3,π/3]. So we can take f (z f (z (the analytic continuation from R to R is trivial. Now f 3 (z must be analytic in R3 and f 3 (z f (z for π/ θ π/3. Such function is f 3 (zlog z for <r <, θ π (this is the principal branch of log z with the branch cut on [,+. We check that f ( π/4 f ( π/4(logr iπ/4logr iπ/ and f (7π/4 f 3 (7π/4(logr + 7iπ/4logr + 7iπ/. Thus, f (r e iπ/4 f (r e 7iπ/4. Problem # (3 points: Evaluate the integral I f (zd z, πi where is the unit circle centered at the origin, for the following f (z: z (a f (z z 3, < a< + a3 log(z b (b f (z, a>, b>, principal branch z+ /a (c tan(z Solution: (a There are singular points at z a, ae iπ/3, and ae iπ/3 ; all of these are inside the unit circle, so I Res(f ; a+res(f ; ae iπ/3 +Res(f ; ae iπ/3 a ( a ae iπ/3 ( a ae iπ/3 + ae iπ/3 + (ae iπ/3 + a(ae iπ/3 ae iπ/3 + ae iπ/3 + (ae iπ/3 + a(ae iπ/3 ae iπ/3 a(+cos(π/3 + + eiπ/3 (e iπ/3 + e iπ/3 (e iπ/3 + i a sin(π/3(+cos(π/3 i sin(π/3 + 3a 6i a sin(π/3.
(b This f has a branch point at z b, make the cut on [b,+ and, for the principal branch, when z x > b, log(z b log x b. Then z /a is a simple pole inside. Since we are integrating over the unit circle, I Res(f ; /alog( /a bln (ab+ a + iπ. (c The singular points are those where cos(z, i.e. z z k π/4+πk/, k Z. Two such points, z ±π/4 are inside. Using that tan(z sin((z k+ u cos((z k + u sin(z kcos(u cos(u sin(z k sin(u sin(u we get (u /+... u (u 3 /6+... ( (u /+...(+(u /6+... u u + u 3 +..., I Res(f ;π/4+res(f ; π/4. Or, shorter, using residue formula, I Res(f ;π/4+res(f ; π/4 sin(z cos (z + sin(z zπ/4 cos (z z π/4. Problem #3 ( points: Let be the unit circle centered at the origin. Evaluate the integral I f (zd z, πi for the following f (z in two ways: (i enclosing the singular points inside and (ii enclosing the singular points outside (by including the point at infinity. Do you get the same result in both cases? f (z z + z 3 Solution: (i Since z is inside, I Res(f ;. (ii Here, I Res(f (z; z Res(t f (t ; t Res((+ t /t; t. Problem #4 (5 points: Determine the type of singular point each of the following functions have at infinity. z N (a, a>, N > M positive integers. z M +a (b log(z+ a, a> (c sin z Solution: (a z N, a>, N > M positive integers: let z /t, then z M +a z N z M + a t M t N (+ at M, which has a pole of order N M at t of strength, so pole of order N M and strength.
(b log(z+ a, a> : log(z+ alog z+ log+log(+ a/(z, so z is a logarithmic branch point. (c sin z: sin z n ( n z n+ /(n+! so z is essential singularity. Problem #5 ( points: Assume that f and g are analytic outside a circle of radius R centered at the origin and lim f (z and lim zg (z, z z where and are constants. Show that πi Solution: Since f and g are analytic outside, πi g (ze f (z d z e. g (ze f (z d z Res(g (ze f (z ;. To find the residue, we deduce the following: Since f (z as z, e f (z e. Since zg (z as z, g (z as z and g (ze f (z as z since e is finite. If h(z as z, then Res(h(z; lim z zh(z. Thus, as we wanted to show. Res(g (ze f (z ; lim z zg (ze f (z e, Problem #6 (5 points: Evaluate the following real integral: (a Solution: (a x + α (x + β d x, α>, β> x + α (x + β d x, α>, β> Since integrand is an even function, x + α (x + β d x x + α (x + β d x. Since integrand is a rational function with degree of denominatordegree of numerator+, we close the contour by large semicircle in the upper half-plane and use residues: ( πi x + α (x + β d x πi Res(f (z;iβ d z + α πi d z (z+ iβ ziβ z (z+ iβ (z + α (z+ iβ 3 ( πi iβ + α β 4iβ 3 3 ziβ π(α + β 4β 3.
Problem #7 (45 points: Evaluate the following real integrals: (a (b (c Solution: (a I (b I x sin xd x x + a, a> x coskx x + 4x+ 5 d x, k > coskx cosmx x + a d x, a>, k, m real x sin xd x x + a, a>. onsider ze i z d z J f (zd z z + a, Then, by Jordan lemma, I Im(lim R J. Since so I πe a. x coskx x d x, k >. One can verify that + 4x+ 5 [ R,R] {Re iθ, θ π} J πi Res(f (z;i aπi i ae a i a iπe a, x + 4x+ 5(x+ +i(x+ i. onsider ze i kz d z J f (zd z z + 4z+ 5, [ R,R] {Re iθ, θ π} Then, by Jordan lemma, I Re(lim R J. Since J πi Res(f (z; +i so k( +i ( +iei πi π( +ie i k k, i I πe k ( cos(k+sin(k. (c coskx cosmx x + a d x, a>, k, m real. Recall the identity coskx cosmx cos((k+mx+cos((k mx. Using this, we have coskx cosmx x + a d x Re ( e i (k+mx x + a d x+ e i (k mx x + a d x [Re(I (k+ m+re(i (k m] 4
where For s>, consider e i sx I (s x + a d x e i sz d z J + f (zd z z + a, [ R,R] {Re iθ, θ π} Then, by Jordan lemma, I (slim R J +. Now J + πi ( Res(f (z;i a ( e sa πi i a πe sa a I (s. For s<, consider e i sz d z J f (zd z z + a, [ R,R] {Re iθ, θ π} Then, by Jordan lemma, I (slim R J. Now Thus, in general, I (sπe s a /a and J πi ( Res(f (z; i a ( e sa sa πe πi i a a I (s. coskx cosmx x + a d x π a (e k+m a + e k m a. Problem #8 (5 points: Show that coshπx d x sec ( a, a <π. Use a rectangular contour with corners at±r and±r+ i. Solution: Let + + 3 + 4 be the rectangular contour and n, n,...,4 are its sides in counterclockwise order, [ R,R]. Then I coshπx d x lim R coshπx d x lim f (xd x. R Inside, the integrand f (z is analytic except points where coshπz, i.e. z i /. Thus f (zd z πi Res(f (z;i / On the other hand, πi cosh az πsinhπz cos(a/. zi / lim f (zd z lim f (zd z, R R 4 5
because a < π. Besides, Thus, so I sec(a/ and lim f (zd z lim R 3 R lim R R R R R cosh a(x+ i coshπ(x+ i d x coshi a+ sinh ax sinhi a d x coshπx coshiπ cos a lim R R R d x I cos a. coshπx f (zd z cos(a/ I + I cos a I cos (a/, coshπx d x I sec(a/. Problem #9 (5 points: onsider a rectangular contour with corners at b± i R and b+ ±ir. Use this contour to show that b+i R e az lim R πi sinπz d z π(+e a, where <b<, Ima <π. b i R Solution: Let + + 3 + 4 be the clockwise rectangular contour and n, n,...,4 are its sides in clockwise order, [b i R,b+ i R]. Then b+i R e az I lim R πi sinπz d z lim f (zd z, R πi b i R where f (z e az sinπz. Inside, the integrand f (z is analytic except points where sinπz, i.e. only z. Thus f (zd z πi Res(f (z; e az πi πcosπz i e a. z On the other hand, because Ima < π. Besides, lim f (zd z lim f (zd z, R R 4 R a(b++i y e lim f (zd z lim R 3 R R sinπ(b+ +i y i d y Thus, and I π(+e a. R e a a(b+i y e lim R R sinπ(b+ i ycosπ i d y e a I. (+e a I i e a πi e a π Problem # (5 points: Use a sector contour with radius R as in figure 4..6, centered at the origin with angle θ π 5, to find, for a>, d x x 5 + a 5 π 5a 4 sin π. 5 6
Solution: ontour x + + L ; on x, z x, x R; on, z Re iθ, θ π 5 ; on L, z r e πi /5, r R. Then d x I x 5 + a 5 lim f (zd z, R x lim R f (zd z lim L R f (zd z π/5 R e πi /5 dr r 5 + a 5 eπi /5 I, Rdθ R 5 a 5 R, and, since the only s.p. inside is z ae iπ/5, f (zd z πi Res(f (z; ae iπ/5 πi 5a 4 e 4iπ/5. Thus, so I ( e πi /5 I πi 5a 4 e 4iπ/5, πi 5a 4 e 4iπ/5 ( e πi /5 πi 5a 4 (i sin(π/5 π 5a 4 sin(π/5. Extra-redit Problem # ( points: onsider a rectangular contour with corners at (±R, and (±R, a. Show that R R e z d z e x d x e (x+i a d x+j R, R R where J R a e (R+i y i d y a e ( R+i y i d y Show lim R J R, whereupon we have e (x+i a d x e x d x π, and consequently, deduce that e x cosax d x πe a. Solution: Let + + 3 + 4 be the counterclockwise rectangular contour and n, n,...,4 are its sides in counterclockwise order, [ R,+R]. Then R e z d z e x d x, e z d z 3 e z d z 4 e z d z R R a a R e (R+i y i d y, e (x+i a d x e ( R+i y i d y R R a e (x+i a d x, e ( R+i y i d y, and, by auchy theorem, e z d z, which shows the first point. We have J R a e (R+i y i d y + ( a e R e y d y+ a 7 a e y d y e ( R+i y i d y R,
which proves that e (x+i a d x e x d x π. Finally, e (x+i a d x e a e x e i ax d x e a e x cosax d x, the last equality because the integral of imaginary (odd part is zero. 8