SOLUTION SET IV FOR 8.75 FALL 4.. Residues... Functions of a Complex Variable In the following, I use the notation Res zz f(z) Res(z ) Res[f(z), z ], where Res is the residue of f(z) at (the isolated singularity) z. 8. Evaluate the integral dz C z when C is the curve sketched in Figure.. Solution. has two simple poles. One is at z, the other is at z. It s easy to z check that Res[ z, ], and Res[ z, ]. The pole at z is encircled in the counterclockwise (positive) sense, while the pole at z is encircled in the clockwise sense. Hence, C dz z ires[ z, ] ires[ z, ] i ( i) i. 88 Determine the residue of each of the following functions at each singularity: (a) e z, (b) e z, (c) cos z, (d) ( + z )e z. Solution. (a) We have So e z e z n z n n! has an essential singularity at z, and (b) We have e z n + z + Res[e z, ]. z n n! + z + n n z n n!. z n n!. Date: October 9,.
SOLUTION SET IV FOR 8.75 FALL 4 So e z has an essential singularity at z, and (c) We have (Note that the coefficient of z (d) We have Res[e z, ]. cos z ( ) n n (z ) n (n)!. n is.) So cos Res[cos ( + z )e z ( + z ) z, ]. z n z n n! ( + z )( + z + z + z 3 6 + has an essential singularity at z, and n4 3 + 7 6 z + [higher powers of z ]. z n n! ) So ( + z )e z has an essential singularity at z, and Res[( + z )e z, ] 7 6..3..3 Evaluation of Real Definite Integrals.. 9. Use residue calculus to evaluate the following integrals: (a) A+B sin θ A (A > B ), B (b) a +sin θ a +cos θ a (a > ), a + (c) sin4 θ cos4 θ 3 6, (d) sin θ 5+4 cos θ 4. Solution. (a) First make the substitution: z e iθ, dz ie iθ. Now the complex z describes the unit circle C in the positive sense as θ varies from to. So, as was discussed in class, the integral becomes A + B sin θ C C dz iz A + B z iz dz Bz + iaz B
SOLUTION SET IV FOR 8.75 FALL 4 3 The poles of the integrand are simple and occur when Bz + ia z B, which in turn gives Furthermore, z ± ia ± B A B z + z A B + A B B. Therefore, z + is a (simple) pole inside the unit circle. Now using the known formula g(z) Res zz h(z) g(z ) h (z ), where z : simple zero of h(z), we get: Res(z + ) Thus, by the residue theorem, A + B sin θ Bz + + ia ia + i A B + ia i A B (b) We manipulate the integrand as follows: So, with the new variable ϕ θ, a + sin θ a cos θ + a + sin θ 4 i i A B A B a + cos θ. a + cos θ dϕ a + cos ϕ dϕ a + cos ϕ dϕ a + sin ϕ, where we shifted the integration variable by / in the integral of the third line and used the periodicity of the integrand. By using the result of part (a) above with A a + and B, we get dϕ a + sin ϕ (a + ) a a +.
4 SOLUTION SET IV FOR 8.75 FALL 4 Thus, and, hence, a + cos θ a + sin θ 3 a a +, a + sin (θ ) a + sin θ a + sin θ a a +. The integral of the third line ensues from the periodicity of the integrand. (c) By subtracting the given integrals, we get So, Then, sin 4 θ sin 4 θ 4 cos 4 θ. sin 4 θ (sin 4 θ cos 4 θ) (sin θ cos θ)(sin θ + cos θ) (sin θ cos θ) sin θ cos θ sin 4 θ + sin 4 θ + cos 4 θ. 3 sin 4 θ + sin 4 θ 3 [sin 4 θ + sin 4 (θ + ) + sin4 (θ + ) + sin 4 (θ + 3 (sin 4 θ + cos 4 θ) sin 4 θ. )]
SOLUTION SET IV FOR 8.75 FALL 4 5 This shows sin 4 θ cos 4 θ 4 sin 4 θ. But sin 4 θ ( z dz )4 C iz iz (z 6i C ) 4 z 5 dz i 6i Res[(z ) 4 z 5, ] i 6i 6 3 4, where C is the unit circle with center at origin. The residue was found easily by noticing that (z ) 4 by which the coefficient of z is 6. So, z 5 z8 4z 6 + 6z 4 4z + z 5, sin 4 θ cos 4 θ 3 6. (d) Again, by the usual replacement z e iθ, sin θ 5 + 4 cos θ C (z ) (iz) 5 + 4 z + z dz iz C (z ) iz (z + 4z + 4) dz The simple poles of this integrand occur when 4z + z + 4, i.e., when z or z, while a double pole occurs at z. Since z is not within the unit circle, we disregard it.
6 SOLUTION SET IV FOR 8.75 FALL 4 Res( ) lim z (z + ) (z ) iz (z + )(z + ) ( 3 4 ) i( 3 ) 3i 6 [ ] d Res() ( )! dz (z f(z)) z [ ( d (z ) )] dz i(4z + z + 4) z [ 4i(z + 5z + )(4z 3 4z) (z 4 z ] + )( 6iz i) ( 4i) (z + 5z + ) 5i 6 Alternatively, you may expand the integrand in z (considering z small ) and find the coefficient of z. (Try it for practice!) Thus, sin ( θ 3i i 5 + 4 cos θ 6 5i ) 6 4 z 9. Use residue calculus to evaluate the following integrals: (a) (x+b) +a a (a > ), (b) (a >, b > ), (c) (d) (x +a )(x +b ) ab(a+b) (a > ), x 4 +4a 4 8a 3 (a > ). (x +a ) 4a 3 Solution. (a) The degree of the denominator is greater than the degree of the numerator and the function is finite for all real values of x. Thus, we can employ the strategy given in class by closing the original path with a large semicircle in the upper half plane (or lower half plane). By shifting the integration variable by b, we get (x + b) + a x + a F (z)dz i Res(z k ) C k where the points z k are the poles of F (z) z +a in the upper half-plane.
SOLUTION SET IV FOR 8.75 FALL 4 7 The (simple) poles occur when z + a, that is when z ±i (a ) ±ia (since a > ). So there is one (simple) pole in the upper half-plane, namely, at z ia. Thus, Res(z ) z ia. (x + b) + a i ia a. (b) has two singularities on the upper half plane. One of these is at z ai, (z +a )(z +b ) the other is at z bi; both of them are simple poles. Note that the denominator of is of degree 4. Accordingly, (x +a )(x +b ) (x + a )(x + b ) Since (x +a )(x +b ) (c) z 4 +4a 4 i (Res[ (z + a )(z + b ), ai] + Res[ (z + a )(z + b ), bi]) i ( ai(b a ) + bi(a b ) ) ab(a + b). is even, we get (x + a )(x + b ) (x + a )(x + b ) ab(a + b). has two singularities on the upper half plane. One of these is at z e i 4 a, the other is at z e 3i 4 a. Both of them are simple poles. Note that the degree of the denominator of is 4. Accordingly, Since x 4 +4a 4 x 4 +4a 4 x 4 + 4a 4 i (Res[ z 4 + 4a 4, e i 4 a] + Res[ z 4 + 4a 4, e 3i 4 a]) e i 4 3i 4 i ( 8 a 3 i e 8 a 3 i ) 4 a [( 3 i) ( 4a 3. is even, we have x 4 + 4a 4 i)] x 4 + 4a 4 8a 3. (d) The degree of the denominator is greater than twice the degree of the numerator and the function is finite for all real values of x. We can once again employ the strategy given in class. Also, note that the integrand is even so that (x + a ) (x + a )
8 SOLUTION SET IV FOR 8.75 FALL 4 The poles occur when (z + a ), that is when z ±ia. Thus, there is one pole in the upper half-plane, i.e., at z ia, and it is a double pole. Therefore, Res(ia) [ d (z ia) ] dz (z + a ) [ ] d dz (z + ia) [ ] (z + ia) 3 zia 4ia 3 zia zia (x + a ) i 4ia 3 4a 3 9. Use residue calculus to evaluate the following integrals: (a) x sin mx a +x e am (a >, m > ), (b) (x +a )(x +b ) (b a ) ( e am a e bm b ) (a >, b >, m, b a), (c) (x+b) +a a e am cos bm (a >, m ), sin mx (x+b) +a a e am sin bm (a >, m ), (d) e am ( + am) (a >, m ), (x +a ) 4a 3 (e) e am (cos am + sin am) (a >, m ), x 4 +4a 4 8a 3 (f) x 3 sin mx x 4 +4a 4 e am cos am (a >, m > ). Solution. (a) z a +z has a simple pole in the upper half plane, which is at z ai. x a + x + i x sin mx a + x xe mxi a + x i Res[ zemzi a + z, ai] i aie am ai i e am, where we close the path in the upper half plane for the last integral involving e imz, since m >. The first integral in the first line is of course because the integrand is odd. So, x sin mx a + x e am.
Note that (b) get z +a x sin mx a +x is even, we have SOLUTION SET IV FOR 8.75 FALL 4 9 x sin mx a + x x sin mx a + x e am. has a simple pole on the upper half plane, which it at z ai. Since m, we x + a + i sin mx x + a where we close the path in the upper half plane. vanishes because the integrand is odd. So, Since x +a Similarly, So, is even, we have x + a (x + a )(x + b ) e mxi x + a e mxi i Res[ z + a, ai] i e am ai e am e am x. + a a, a The second integral in the first line e am x + a a. e bm x. + b b b ( a x + a x + b ) x + a b a ( (e am b a a (e am (b a ) a e bm ) b e bm ). b x + b ) (c) The given integrals are evaluated by the standard prescription as follows. (x + b) Re + a sin mx (x + b) Im + a e imx (x + b) Re + a e imx (x + b) Im + a e im(x b) x + a e im(x b) x + a,
SOLUTION SET IV FOR 8.75 FALL 4 where Hence, e imx e imz x + a i Res zia z + a a e ma. ( (x + b) + a Re e imb a e ma) a e ma cos mb, sin mx ( (x + b) + a Im e imb a e ma) a e ma sin mb. (d) We can calculate the real part of e imx (x +a ) by noticing that our function is even. As in question 9(d), we have one pole in the upper half-plane. This is a double pole at z ia. Res(ia) [ ( d (z ia) e imz )] dz (z + a ) zia [ ( d e imz )] dz (z + ia) zia [ (z + ia) ime imz e imz ] (z + ia) (z + ia) 4 (ia) ime ma e ma 4ia (ia) ( 4 ) ma + ie ma 4a 3 zia Therefore, (x + a ) [ Re e imx ] (x + a ) [ ( ma + Re i ie ma 4a ( ) 3 ma + e ma 4a 3 )] (e) Clearly, x 4 + 4a 4 Re e imx x 4 + 4a 4. We calculate the last integral involving e imz by closing the path in the upper half plane since m >. The poles of the integrand occur at z 4 + 4a 4 and are all simple. The
SOLUTION SET IV FOR 8.75 FALL 4 poles that lie in the upper half plane are z e i/4 a and z e 3i/4 a. Accordingly, It follows that e imx e x 4 i (Res imz + 4a4 zz z 4 + 4a 4 + Res zz ( e imz i 4z 3 + eimz 4z 3 ) e imz z 4 + 4a 4 i [ (cosma + i sin ma)( + i) + (cos ma i sin ma)( i)] 6a3 (cos ma + sin ma) 4a3 x 4 + 4a 4 (cos ma + sin ma). 8a3 (f) Once again, we can calculate the imaginary part of ) x 3 e imx x 4 +4a 4 noting that our function is even and m >. The poles occur when z 4 + 4a 4 z ± ±ia, as in part (e) above. Thus the roots are: a ia a ( + i ) a + ia since a >. a a a ia. a 3 ia ia a. a 4 a 3 ia + a. Thus there are two poles in the upper half-plane: a and a 3 (both simple poles). Res(a ) lim z a (z a )f(z) (z a ia)z 3 e imz lim z a z 4 + 4a z 3 e imz lim z a (z + a + ia)(z + ia ) (a + ia) e im(a+ia) 8ia z 3 e imz Res(a 3 ) lim z a3 (z + ia a)(z ia ) (ia a) e ima ma 8ia
SOLUTION SET IV FOR 8.75 FALL 4 Therefore, x 3 sin mx x 4 + 4a 4 [ Im x 3 e imx ] x 4 + 4a 4 ( (a + ia) [ Im e ima ma (ia a) e ima ma )] 8a [ ] Im i(eima e ma + e ima e ma ) ( e ima + e ima ) e ma e ma cos ma