Ma 416: Complex Variables Solutions to Homework Assignment 6

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1 Ma 46: omplex Variables Solutions to Homework Assignment 6 Prof. Wickerhauser Due Thursday, October th, 2 Read R. P. Boas, nvitation to omplex Analysis, hapter 2, sections 9A.. Evaluate the definite integral 2π (2 + sin θ) 2 dθ. Solution: get Substitute z = e iθ, dz = ie iθ dθ dθ = (/iz) dz, and sin θ = (z /z)/2i to 2π dθ (2 + sin θ) 2 = dz 4iz dz ( ) iz 2 + z /z 2 = (z 2 + 4iz ) 2 = 2i 4iz dz (z z + ) 2 (z z ) 2, where is the positively oriented unit circle {z = e iθ : θ 2π}, and z ± = ( 2 ± )i are the roots of the denominator quadratic polynomial. Since z + = ( 2 + )i lies in the region { z < } enclosed by, it is a pole of order 2 for the integrand f(z) def = 4iz(z z + ) 2 (z z ) 2. The enclosed region contains no other poles of f, since z = 2 + >. By the residue theorem, the integral equals 2πiRes(z + ), where Res(z + ) is the residue of f at z +. Evaluate this residue with auchy s formula on page 7 of our text, with n = 2: ( ) d (n ) Res(z + ) = lim [f(z)(z z + ) n ] = lim z z + (n )! dz z z + = lim z z + 4i(z + z ) (z z ) = 4i(z + + z ) (z + z ) = [ d dz 2 i. 4iz (z z ) 2 Hence 2πiRes(z + ) = 4π is the value of the original integral. 2. Evaluate the improper integral (x4 + ). Solution: Write z 4 + = (z 2 + i)(z 2 i) = (z z )(z z 2 )(z z )(z z 4 ), where z = ( + i)/ 2, z 2 = ( + i)/ 2, z = ( i)/ 2, and z 4 = ( i)/ 2 are the four 4 th roots of. Next, fix big R > 2, let = h a v be the simple closed piecewise smooth quartercircle contour in the first quadrant consisting of the horizontal line segment h = {z = x : x [, R]}, the quarter-arc a = {Re iθ : θ π/2}, and the vertical line segment v = {z = (R y)i : y [, R]}. As parametrized, has positive orientation. ]

2 The region enclosed by will contain the simple pole z of f(z) def = (z 4 + ) and no other singular points of f. Hence by the residue theorem, 2πi Res(z ), where Res(z ) is the residue of f at z. But since z is a simple pole, Res(z ) = lim z z (z z )f(z) = [(z z 2 )(z z )(z z 4 )] = Thus π ( + i) 2 for any R > 2. Also, h f(z) dz + a f(z) dz + v f(z) dz, where h v a R R π/2 x 4 + ; di(r y) R (R y) 4 + = i ire iθ dθ R 4 e 4iθ + f(z) dz π a 2 dt t 4 +, R R 4 as R. The integral over the arc a must therefore vanish as R, leaving lim ( i) R x i( + i) 2. after y (R t); The limit π/[( + i) 2] on the left-hand side is attained as soon as R > 2. Hence x 4 + = π ( i)( + i) 2 = π Evaluate the principal value integral PV (x ). Solution: Note that f(z) def = (z ) has simple poles at the three cube roots of, namely z = e 2πi/ = ( + i )/2, z 2 = e 4πi/ = ( i )/2, and z = e 6πi/ =. Fix R > 2 and < ɛ < and consider the piecewise differentiable curve = ɛ 2 R defined by = {z = x : R x ɛ}; ɛ = {z = ɛe iθ : θ π}; 2 = {z = x : + ɛ x R}; R = {z = Re iθ : θ π}. Then is a simple closed curve that encloses one pole of f (namely z ), skirts within ɛ of another (z = ), and completely excludes the third (z 2 ). Note that is positively oriented. By the residue theorem, f(z) = 2πiRes(z ) = 2πi lim z z (z z )f(z) = 2 2πi (z z 2 )(z z ) = 4π (i ).

3 The four path integrals comprising f(z) dz may be written as follows: ɛ R 2 R x + +ɛ x ; π iɛe iθ dθ π ɛ [ ɛe iθ ] = i dθ + ɛe iθ ɛ 2 e 2iθ ; π ire iθ dθ R R e iθ. Evidently ɛ f(z) dz iπ/ as ɛ, and R f(z) dz as R. As ɛ and R, the path integral 2 f(z) dz converges to the principal value we are seeking. ombining these calculations gives PV (x ) = 4π (i ) iπ = π Note that the negative values of the integrand in < x < outweigh the positive values in < x <. 4. Evaluate the integral e x/n e x, for n =, 2,,.... (Hint: use rectangular contours of fixed height 2π with one side lying on the x-axis.) Solution: Following the hint, fix R > 2 and < ɛ < and let = B ɛb U T ɛt D be the piecewise differentiable rectangular contour with two nibbles near z = and z = 2πi defined as follows: B = {z = x : R x ɛ and ɛ x R}; T = {z = 2πi x : R x ɛ and ɛ x R}; U = {z = R + iy : y 2π}; D = {z = R + i(2π y) : y 2π}; ɛb = {z = ɛe iθ : θ π}; ɛt = {z = 2πi + ɛe iθ : θ π}. t may be seen that is a simple closed curve. For each integer n >, the function f(z) = e z/n /( e z ) only has simple poles at z k = 2kπi for any integer k, so it has no singular points in the region enclosed by. The nibbles ɛb and ɛt avoid the poles z = and z = 2πi, respectively. By the residue theorem,. Now B ɛ R e x/n e x + R ɛ e x/n def e x = e x/n e x, where = [ R, ɛ] [ɛ, R]. This tends to our desired integral as R and ɛ. But also, T e [2πi x]/n d[2πi x] e [2πi x] = e 2πi/n e x/n d( x) e x = e 2πi/n e x/n e x,

4 after the substitution x x. Thus ( e 2πi/n ) B T e x/n e x. The values for ɛb f(z) dz and ɛt f(z) dz are computed as in Exercise 9. on pages 84 8 of our text. They are partial residues, exactly half the residues at z and z, respectively, and negative because the poles are outside the contour : ɛb 2 [2πi]Res(z ) = πi; ɛt 2 [2πi]Res(z ) = e 2π/n πi; Here we have computed the residues in the usual way: ze z/n Res(z ) = Res() = lim z e z = ; Res(z ) = Res(2πi) = lim z 2πi The remaining contour integrals may be estimated: f(z) dz U f(z) dz D 2π 2π e R/n e iy/n dy e R e iy e R/n e iy/n dy e R e iy (z 2πi)e z/n e z = e 2πi/n 2πeR/n e R ombining these computations gives the value of the integral:, as R ; 2πe R/n, as R. e 2 e x/n e x = ( + e2πi/n )πi e 2πi/n = (e πi/n + e πi/n )πi (e πi/n e πi/n ) = π cot(π/n). Note that this integral must be interpreted in the principal value sense.. For each relation below, find all the complex numbers z satisfying it: (a) z n = for a fixed n {2,, 4,...} (b) e z = e (c) e z = i (d) tan z = + i. Solution: (a) z {e 2kπi/n : k =, 2,..., n}. There are n distinct values in this set, the n n th roots of unity, equidistributed around the unit circle. (b) Since e = e iπ e = e +iπ and the exponential function is 2πi-periodic, the solution set is { + πi + 2kπi = + (2k + )πi : k Z}. (c) Since e iπ/2 = i and the exponential function is 2πi-periodic, the solution set is {z 2 : z = iπ/2 + 2kπi = (2k + 2 )πi : k Z} = { (2k + 2 )2 π 2 : k Z}. 4

5 (d) Use the idea in the solution to Exercise. on page 9 of our text: + i = tan z = sin z cos z = e iz e iz i e iz + e iz (2 i)e iz = ie iz e 2iz = i 2 i = + 2 i = ( + 2 i The final expression on the right-hand side is in the form re iθ with r = / and θ R satisfying cos θ = /, sin θ = 2/. From this we conclude that R(2iz) = 2(z) = ln(r).847 (z).694; (2iz) = 2R(z) = θ {arctan( 2) + π + 2kπ : k Z} R(z) {.72 + kπ : k Z}, using the approximation arctan( 2).7, the principal value of arctangent which lies in [ π 2, π 2 ]. We must add π to this principal value to get an angle θ 2.44 in the second quadrant satisfying sin θ >, cos θ <. Hence the solution set may be written approximately as { i + kπ : k Z}. ). 6. Evaluate the integral ( + x 2 ) x. Solution: This is a special case of Exercise.(a) on page 97 of the textbook: x λ a 2 + x 2, a =, λ = 2. We follow the model solution on page 294 of the textbook, substituting x 2 a 2 t = t to obtain the equivalent integral t +(+λ)/2 dt 2 aλ = t /4 dt + t 2 + t, = t 4 dt 2 + t, which evaluates to 2 aλ ( 4 ) = π 2 sin 4 π = π, 2 using the notation and technique in Section A, pages 9 97 of the textbook.