Residues and Contour Integration Problems

Transcription

1 Residues and ontour Integration Problems lassify the singularity of f(z) at the indicated point.. f(z) = cot(z) at z =. Ans. Simple pole. Solution. The test for a simple pole at z = is that lim z zcot(z) exists and is not. We can use L Hôpital s rule: zcos(z) limzcot(z) = lim z z sin(z) Thus the singularity is a simple pole. = lim z cos(z) zsin(z) cos(z) =. 2. f(z) = +cos(z) (z π) 2 at z = π. Ans. Removable. Solution. Power series is the simplest way to do this. We can expand cos(z) in a Taylor series about z = π. To do so, use the trig identity cos(z) = cos(z π). Next, expand cos(z π) in a power series in z π: +cos(z) = cos(z π) = 2! (z π)2 4! (z π)4 + From this, we get +cos(z) (z π) 2 = (z π)2 ( 2 4! (z π)2 + ) (z π) 2 = 2 4! (z π)2 +, which is the Laurent series for +cos(z) (z π) 2. Since there are no negative powers in the series, the singularity is removable. 3. f(z) = sin(/z). Ans. Essential singularity. 4. f(z) = z2 z +2z+ at z =. Ans. Pole of order 2.. f(z) = z 3 sin(z) at z =. Ans. Pole of order f(z) = csc(z)cot(z) at z =. Ans. Pole of order 2.

2 Find the residue of g(z) at the indicated singulatity. 7. g(z) = + at z = i. Ans. Res i(g) = 2 i. Solution. Since g(z) = (z i)(z+i), we have that (z+i)g(z) = z i, which is analytic and nonzero at z = i. Hence, g(z) has a simple pole at z = i. The residue is thus Res i (g) = lim z i (z +i)g(z) = 2i = 2 i 8. g(z) = ez z 3 at z =. Ans. Res (g) = 2. Solution. Using the power series for e z, we see that the Laurent series for g(z) about z = is e z z = +z + 2! z2 + 3! z3 + 4! z4 + 3 z 3 = z 3 +z 2 + 2! z + 3! + 4! z+ Thetheresidueisa, thecoefficientofz. Hence, Res (g) = a = g(z) = tan(z) at z = π/2. Ans. Res π/2 (g) =.. g(z) = z+2 ( 2z+) 2 at z =. Ans. Res (g) =.. g(z) = f(z)/h(z) at z = z, given that f(z ), h(z ) =, and h (z ). Show that z = z is a simple pole and find Res z (g). Ans. Res z (g) = f(z )/h (z ). The singularities for the functions below are all simple poles. Find all of them and use exercise above to find the residues at them. 2. g(z) = z2. Ans. The singularities are at i and 4i and the residues iz 4 are Res i (g) = 2i and Res 3 4i(g) = 7i. 3 Solution. The singularities are the roots of iz 4 =, which are i and 4i. In our case, the functions f and h in exercise are f(z) = andh(z) = iz 4,andf(z)/h (z) = ( )/(2z i). It immediately follows that The other residue follows similarly. Res i (g) = i2 2i i = 2 3i = 2 3 i. 2

3 3. g(z) = tan(z). Ans. The singularities are at z n = (n + )π, where 2 n =,±,±2,..., and the residues at z n are Res zn (g) =. 4. g(z) = z2 z 3 8. Ans. The singularities are at the roots of z3 8 =. There are three of these: 2, 2e 2iπ/3 and 2e 4iπ/3. The residues at these three points are all /3.. g(z) = ez. Ans. The singularities are at the roots of sin(z) =, sin(z) which are nπ, n =,±,±2,..., and the residues there are Res nπ (g) = ( ) n e nπ. 6. g(z) = sin(z) 3z+2. Ans. Thesingularitiesareattherootsofz2 3z+2 =, which are and 2. The residues are Res (g) = sin() and Res 2 (g) = sin(2). 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 3/2. Ans. 2πi/3. Solution. From exercise 4, g(z) has three singularities, located at 2, 2e 2iπ/3 and 2e 4iπ/3. A simple sketch of shows that only 2 is inside of. Thus, by the residue theorem and exercise 4, we have z 3 8 dz = 2πiRes 2(g) = 2πi/3 = 2πi/3. 8. dz, where is the counterclockwise oriented circle with radius z and center. Ans. 2πi. 9. dz, where is any simple closed curve that is positively 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). 2. e z dz, where is the positively traversed rectangle with corners sin(z) π/2 i, π/2 i, π/2+2i and π/2+2i. Ans. 2πi( e π +e 2π ). 3

4 2. z+2 dz, where is the positively oriented semicircle that is ( 2z+) 2 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 +2 ( 2z +) 2dz = 2πiRes (g) = 2πi =. 22. dz, where is the negatively oriented (i.e., clockwise) ( +)( +4) semicircle that is located in the upper half plane and has center, radius R > 2, and diameter located on the real axis. Ans. π/6. Find the values of the definite integrals below by contour-integral methods π. Ans. π/2. 3sin(θ) 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, sin(θ) = (z z )/(2i). We thus have 2π 3sin(θ) = dz iz( 3z 3/z ) = 2i ( 2)dz 3 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 f(z)/(z i/3), where f is analytic at i/3. Using exercise, we see that ( 2) 2 Res i/3 = 3 iz 3 6z i = 2 6i/3 i = i/4. The residue theorem then implies that ( 2)dz 3 iz 3 = 2πiRes ( 2 i/3 = π/2. 3 iz 3 4

5 24. 2π. Ans. 2π/. 3 2cos(θ) 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, cos(θ) = (z +z )/2. We thus have 2π 3 2cos(θ) = dz iz(3 z /z) = idz 3z +. The integrand has singularities at z ± = (3 ± )/2. Only z = (3 )/2 is inside. It is a simple pole because the integrand has the form f(z)/(z (3 )/2)), where f is analytic at (3 )/2. Using exercise, we see that i Res 3 )/2 3z + = i 2z 3 = i. The residue theorem then implies that idz 3z + = 2πiRes i 3 )/2 3z π 26. 2π. Ans. 2π/3. 4sin(θ) = 2π. cos(θ). Ans. 4π/. (This has two simple poles within the 3+2cos(θ) contour.) 27. dx. Ans. π/6. (Hint: reverse the contour in exercise 22 (x 2 +)(x 2 +4) and let R.)