Residues and ontour Integration Problems lassify the singularity of fz at the indicated point.. fz = cotz at z =. Ans. Simple pole. Solution. The test for a simple pole at z = is that lim z z cotz exists and is not. We can use L Hôpital s rule: z cosz lim z cotz = lim z z sinz Thus the singularity is a simple pole. = lim z cosz z sinz cosz =.. fz = +cosz at z = π. Ans. Removable. z π Solution. Power series is the simplest way to do this. We can expand cosz in a Taylor series about z = π. To do so, use the trig identity cosz = cosz π. Next, expand cosz π in a power series in z π: + cosz = cosz π =! z π 4! z π4 + From this, we get + cosz z π = z π 4! z π + z π = 4! z π +, which is the Laurent series for +cosz z π. Since there are no negative powers in the series, the singularity is removable. 3. fz = sin/z. Ans. Essential singularity. 4. fz = z z z +z+ at z =. Ans. Pole of order.. fz = z 3 sinz at z =. Ans. Pole of order. 6. fz = cscz cotz at z =. Ans. Pole of order.
Find the residue of gz at the indicated singulatity. 7. gz = z + at z = i. Ans. Res ig = i. Solution. Since gz = z iz+i, we have that z + igz = z i, which is analytic and nonzero at z = i. Hence, gz has a simple pole at z = i. The residue is thus Res i g = lim z i z + igz = i = i 8. gz = ez z 3 at z =. Ans. Res g =. Solution. Using the power series for e z, we see that the Laurent series for gz about z = is e z z = + z +! z + 3! z3 + 4! z4 + = z 3 +z + 3 z 3! z + 3! + 4! z+ The the residue is a, the coefficient of z. Hence, Res g = a =. 9. gz = tanz at z = π/. Ans. Res π/ g =.. gz = z+ z z+ at z =. Ans. Res g =.. gz = fz/hz at z = z, given that fz, hz =, and h z. Show that z = z is a simple pole and find Res z g. Ans. Res z g = fz /h z. The singularities for the functions below are all simple poles. them and use exercise above to find the residues at them. Find all of. gz = z. Ans. The singularities are at i and 4i and the residues z iz 4 are Res i g = i and Res 3 4ig = 7i. 3 Solution. The singularities are the roots of z iz 4 =, which are i and 4i. In our case, the functions f and h in exercise are fz = z and hz = z iz 4, and fz/h z = z /z i. It immediately follows that The other residue follows similarly. Res i g = i i i = 3i = 3 i.
3. gz = tanz. Ans. The singularities are at z n = n + π, where n =, ±, ±,..., and the residues at z n are Res zn g =. 4. gz = z z 3 8. Ans. The singularities are at the roots of z3 8 =. There are three of these:, e iπ/3 and e 4iπ/3. The residues at these three points are all /3.. gz = ez. Ans. The singularities are at the roots of sinz =, sinz which are nπ, n =, ±, ±,..., and the residues there are Res nπ g = n e nπ. 6. gz = sinz z 3z+. Ans. The singularities are at the roots of z 3z+ =, which are and. The residues are Res g = sin and Res g = sin. Use the residue theorem to evaluate the contour intergals below. Where possible, you may use the results from any of the previous exercises. 7. dz, where is the counterclockwise oriented circle with radius z 3 8 and center 3/. Ans. πi/3. z Solution. From exercise 4, gz has three singularities, located at, e iπ/3 and e 4iπ/3. A simple sketch of shows that only is inside of. Thus, by the residue theorem and exercise 4, we have z z 3 8 dz = πi Res g = πi/3 = πi/3. 8. dz, where is the counterclockwise oriented circle with radius z 3 8 3 and center. Ans. πi. z 9. z dz, where is any simple closed curve that is positively z iz 4 oriented i.e., counterclockwise and encloses the following points: a only i; b only 4i; c both i and 4i; d neither i nor 4i. Ans. a 4π/3. b 34π/3. c π. d.. e z dz, where is the positively traversed rectangle with corners sinz π/ i, π/ i, π/ + i and π/ + i. Ans. πi e π + e π. 3
. z+ dz, where is the positively oriented semicircle that is z z+ located in the right half plane and has center, radius R >, and diameter located on the imaginary axis. Ans.. Solution. From exercise, the only singularity of the integrand is at. By the residue theorem and exercise, we have z + z z + dz = πi Res g = πi = πi.. dz, where is the negatively oriented i.e., clockwise z +z +4 semicircle that is located in the upper half plane and has center, radius R >, and diameter located on the real axis. Ans. π/6. Find the values of the definite integrals below by contour-integral methods. 3. π. Ans. π/. 3 sinθ Solution. Begin by converting this integral into a contour integral over, which is a circle of radius and center, oriented positively. To do this, let z = e iθ. Note that dz = ie iθ = iz, so = dz/iz. Also, sinθ = z z /i. We thus have π 3 sinθ = dz iz 3z 3/z = i dz 3z iz 3. { 3i The integrand has singularities at z ± = i ± 8i/6 = Only i/3. z = i/3 is inside. It is a simple pole because the integrand has the form fz/z i/3, where f is analytic at i/3. Using exercise, we see that Res i/3 = 3z iz 3 6z i = 6i/3 i = i/4. The residue theorem then implies that dz 3z iz 3 = πi Res i/3 = π/. 3z iz 3 4
4. π. Ans. π/. 3 cosθ Solution. Begin by converting this integral into a contour integral over, which is a circle of radius and center, oriented positively. To do this, let z = e iθ. Note that dz = ie iθ = iz, so = dz/iz. Also, cosθ = z + z /. We thus have π 3 cosθ = dz iz3 z /z = idz z 3z +. The integrand has singularities at z ± = 3 ± /. Only z = 3 / is inside. It is a simple pole because the integrand has the form fz/z 3 /, where f is analytic at 3 /. Using exercise, we see that i Res 3 / z 3z + = i z 3 = i. The residue theorem then implies that idz z 3z + = πi Res i 3 / z 3z +. π 6. π. Ans. π/3. 4 sinθ = π. cosθ. Ans. 4π/. This has two simple poles within the 3+ cosθ contour. 7. dx. Ans. π/6. Hint: reverse the contour in exercise x +x +4 and let R.