Residues and Contour Integration Problems

1 Residues and Contour Integration ProblemsClassify the singularity off(z)at the indicated (z) = cot(z) atz= 0. Ans. Simple The test for a simple pole atz= 0 is that limz 0zcot(z)exists and is not 0. We can use L H opital s rule:limz 0zcot(z) = limz 0zcos(z)sin(z)= limz 0cos(z) zsin(z)cos(z)= the singularity is a simple (z) =1+cos(z)(z )2atz= . Ans. Power series is the simplest way to do this. We can expandcos(z) in a Taylor series aboutz= . To do so, use the trig identitycos(z) = cos(z ). Next, expand 1 cos(z ) in a power seriesinz :1 + cos(z) = 1 cos(z ) =12!(z )2 14!(z )4+ From this, we get1 + cos(z)(z )2=(z )2(12 14!(z )2+ )(z )2=12 14!(z )2+ ,which is the Laurent series for1+cos(z)(z )2.

2 Since there are no negativepowers in the series, the singularity is (z) = sin(1/z). Ans. Essential (z) =z2 zz2+2z+1atz= 1. Ans. Pole of order (z) =z 3sin(z) atz= 0. Ans. Pole of order (z) = csc(z) cot(z) atz= 0. Ans. Pole of order the residue ofg(z)at the indicated (z) =1z2+1atz= i. Ans. Res i(g) = Sinceg(z) =1(z i)(z+i), we have that (z+i)g(z) =1z i, whichis analytic and nonzero atz= i. Hence,g(z) has a simple pole atz= i. The residue is thus Res i(g) = limz i(z+i)g(z) =1 2i= (z) =ezz3atz= 0. Ans. Res0(g) = Using the power series forez, we see that the Laurent seriesforg(z) aboutz= 0 isezz3=1 +z+12!z2+13!z3+14!z4+ z3=z 3+z 2+12!z 1+13!+14!z+ The the residue isa 1, the coefficient ofz 1.

3 Hence, Res0(g) =a 1= (z) = tan(z) atz= /2. Ans. Res /2(g) = (z) =z+2(z2 2z+1)2atz= 1. Ans. Res1(g) = (z) =f(z)/h(z) atz=z0, given thatf(z0)6= 0,h(z0) = 0, andh (z0)6= 0. Show thatz=z0is a simple pole and find Resz0(g). (g) =f(z0)/h (z0).The singularities for the functions below are all simple poles. Find all ofthem and use exercise 11 above to find the Residues at (z) =z2 1z2 5iz 4. Ans. The singularities are atiand 4iand the residuesare Resi(g) = 23iand Res4i(g) = The singularities are the roots ofz2 5iz 4 = 0, whichareiand 4i. In our case, the functionsfandhin exercise 11 aref(z) =z2 1 andh(z) =z2 5iz 4, andf(z)/h (z) = (z2 1)/(2z 5i).It immediately follows thatResi(g) =i2 12i 5i= 2 3i= other residue follows (z) = tan(z).

4 Ans. The singularities are atzn= (n+12) , wheren= 0, 1, 2,.., and the Residues atznare Reszn(g) = (z) =z2z3 8. Ans. The singularities are at the roots ofz3 8 = are three of these: 2, 2e2i /3and 2e4i /3. The Residues at thesethree points are all 1 (z) =ezsin(z). Ans. The singularities are at the roots of sin(z) = 0,which aren ,n= 0, 1, 2,.., and the Residues there are Resn (g) =( 1)nen . (z) =sin(z)z2 3z+2. Ans. The singularities are at the roots ofz2 3z+2 = 0,which are 1 and 2. The Residues are Res1(g) = sin(1) and Res2(g) =sin(2).Use the residue theorem to evaluate the Contour intergals below. Where pos-sible, you may use the results from any of the previous Cz2z3 8dz, whereCis the counterclockwise oriented circle with radius1 and center 3/2.

5 Ans. 2 From exercise 14,g(z) has three singularities, located at 2,2e2i /3and 2e4i /3. A simple sketch ofCshows that only 2 is inside ofC. Thus, by the residue theorem and exercise 14, we have Cz2z3 8dz= 2 iRes2(g) = 2 i/3 = 2 Cz2z3 8dz, whereCis the counterclockwise oriented circle with radius3 and center 0. Ans. 2 Cz2 1z2 5iz 4dz, whereCis any simple closed curve that is positivelyoriented ( , counterclockwise) and encloses the following points: (a)onlyi; (b) only 4i; (c) bothiand 4i; (d) neitherinor 4i. Ans. (a)4 /3. (b) 34 /3. (c) 10 . (d) Cezsin(z)dz, whereCis the positively traversed rectangle with corners /2 i, 5 /2 i, /2 + 2iand 5 /2 + 2i. Ans. 2 i(1 e +e2 ).321. Cz+2(z2 2z+1)2dz, whereCis the positively oriented semicircle that islocated in theright half planeand has center 0, radiusR >1, anddiameter located on the imaginary axis.

6 Ans. From exercise 10, the only singularity of the integrand is at1. By the residue theorem and exercise 10, we have Cz+ 2(z2 2z+ 1)2dz= 2 iRes1(g) = 2 i 1 = 2 C1(z2+1)(z2+4)dz, whereCis thenegativelyoriented ( , clockwise)semicircle that is located in theupper half planeand has center 0,radiusR >2, and diameter located on the real axis. Ans. the values of the definite integrals below by Contour -integral 2 0d 5 3 sin( ). Ans. Begin by converting this integral into a Contour integral overC, which is a circle of radius 1 and center 0, oriented positively. To dothis, letz=ei . Note thatdz=iei d =izd , sod =dz/(iz). Also,sin( ) = (z z 1)/(2i). We thus have 2 0d 5 3 sin( )= Cdziz(5 3z 3/z2i)= C( 2)dz3z2 10iz integrand has singularities atz = (10i 8i)/6 ={3 =i/3 is insideC.}

7 It is a simple pole because the integrand has theformf(z)/(z i/3), wherefis analytic ati/3. Using exercise 11, wesee thatResi/3(( 2)3z2 10iz 3)= 26z 10i= 26i/3 10i= residue theorem then implies that C( 2)dz3z2 10iz 3= 2 iResi/3(( 23z2 10iz 3)= 2 0d 3 2 cos( ). Ans. 2 / Begin by converting this integral into a Contour integral overC, which is a circle of radius 1 and center 0, oriented positively. To dothis, letz=ei . Note thatdz=iei d =izd , sod =dz/(iz). Also,cos( ) = (z+z 1)/2. We thus have 2 0d 3 2 cos( )= Cdziz(3 z 1/z)= Cidzz2 3z+ integrand has singularities atz = (3 5)/2. Onlyz = (3 5)/2 is insideC. It is a simple pole because the integrand has theformf(z)/(z (3 5)/2)), wherefis analytic at (3 5)/2.

8 Usingexercise 11, we see thatRes3 5)/2(iz2 3z+ 1)=i2z 3= i residue theorem then implies that Cidzz2 3z+ 1= 2 iRes3 5)/2(iz2 3z+ 1)=2 2 0d 5 4 sin( ). Ans. 2 2 0cos( )d 13+12 cos( ). Ans. 4 /15. (This has two simple poles within thecontour.)27. 1(x2+1)(x2+4)dx. Ans. /6. (Hint: reverse the Contour in exercise 22and letR .)5