Problems 3 (due 27 January)

Transcription

1 Problems 3 due 7 January). You previously found singularities of the functions below. Here, identify the poles, order of these poles and residues for these functions: a) tanhz) = ez e z e z + e z Solution: The singularities occur along the line x = 0, at the points where y = m + /)π for m = 0, ±, ±.... Along the line x = 0, e z + e z = cosy). This has zeros at y = m + /)π, and these are all of order, since cosy) has a finite derivative ±) at these points. Thus, the singularities are all simple poles. In order to find the residues, note that tanhz) = sinh z cosh z = sinh z 0 + cosh z 0 z z 0 ) + cosh z 0 + sinh z 0 z z 0 ) +, where the first-order terms in the Taylor series for sinh z and cosh z about the point z 0 can be found by using d sinh z/dz = cosh z and d cosh z/dz = sinh z. Near the singularities identified above, where cosh z vanishes, tanhz) = sinh z 0 + orderz z 0 ) sinh z 0 z z 0 ) + orderz z 0 ) 3 = z z 0 +orderz z 0 ). b) Thus, the residues at all the singularities are equal to. ze z ) ) Solution: Consider the function gz) = e z = e z/ e z/ e z/). Apart from a multiplicative factors of e z/ and i, neither of which has any zeros or singularities, gz) e z/ e z/ sinhz/). For similar reasons as in part a), this can only vanish along the line x = 0, where gz) siny/) for z = x + iy. The zeros of siny/) occur at y = mπ, where m = 0, ±, ±.... Again, the derivative at these points is finite, so the zeros are of order. Thus, for all m 0,

2 the singularities of Eq. ) are simple poles at z = imπ. There is also an additional order pole at z = 0, due to the additional factor of z in the denominator of Eq. ). In order to find the residues, consider first the singularities at z 0. Here, we can write gz) above as gz) = e z/ sinh z/ and perform a Taylor series for sinh z/ about one of its zeros at z 0 : [ gz) = e z/ sinh z 0 / + ] coshz 0/) z z 0 ) + orderz z 0 ) = e z0/ coshz 0 /) z z 0 ) + orderz z 0 ) Thus, the residue at z m = imπ for m = ±, ±... is R m = z m e zm/ coshz m /). For the singularity at z = 0, we can expand the denominator of Eq. ) as z e z ) = z z + ) z + = z + ) z +. From this, we find z e z ) = z z + ), c) since / + z) = z +. Thus, the residue at z = 0 is /. z + a ) Solution: z + a ) = z ia) z + ia). So, there are obviously poles at z = ±ia, and both of these poles are order. In order to find the residue, we can follow an approach similar to that of Eq. A9.8) of Appendix A in the textbook. Let z = ia + ζ

3 near the pole at ia. z + a ) = ia + ζ) 4 ζ) [ ia i ζ = a iζ/a) 4 4ζ a iζ/a) 4 ] = a ) 4ζ i ζ a ) + i ζ a + i ζ )...) a = a 4iζ/a + orderζ) ) 4ζ + iζ/a + ) = a 4ζ 3i ζa ) + orderζ) = a 4ζ + 3ia 4ζ +. So, the residue is 3ia 4. Similarly, the residue at z = ia is. Evaluate the following integrals: 3ia 4. a) dx x 4 + = π Solution: There are poles at the 4 distinct /4th roots of -: For example, z m = e iπ/4+im )π/ m =,, 3, 4. z = + i)/ and z = + i)/. Each of these poles is a simple, order pole. The residues R, at z, are R = z z )z z 3 )z ) = i + i) and R = z z )z z 3 )z ) = i + i). 3

4 We can evaluate the integral using a contour in the upper half plane y 0). This contour only contains the poles at z,. Thus, dx x 4 + = πi R + R ) = π b) cos ax x + dx = πe a a > 0) Solution: First of all, note that e iax = cosax)+i sinax) and sinax) is an odd function. Combined with the even function /x + ), the addition of sinax) does not change the integral. Thus, we can evaluate this integral as e iax x + dx. As discussed in class see also the Problem A.5 on page 509), this integral can be extended to a contour in the upper half plane since the exponential vanishes in the upper half plane for large y = Imz) > 0. The value of the resulting integral for the closed contour is πi e a i = πe a, since the residue at the pole at z = i is e a /ia. 3. Integrals of the following sort frequently arise in doing inverse) Fourier transformations. Evaluate e iωt dω a > 0) a iω for both the case that t > 0 and t < 0. Hint: consider contours in both the upper and lower half plane.) Solution: First of all, note that the only singularity of the integrand as a function of complex) ω is at ω = ia. This is the simple order ) pole, and the residue is R = ie at. For t < 0, the integral satisfies the conditions for Jordan s Lemma as discussed in class, or see Problem A.5 on page 509) for a semi-circular contour in the upper half plane y = Imz) > 0). In this upper half plane, however, there are no singularities. Thus, the integral is 0. 4

5 For t > 0, consider a semicircular contour in the lower half plane, in a clock-wise direction. In this case, the integral is πir = πe at. The minus sign is due to the orientation of the contour in this case.) In other words, for a > 0, { e iωt a iω dω = πe at for t > 0). 0 for t < 0) The integral is discontinuous and ill-defined at t = 0. 5