Homework # Solutions Math 8, Fall 03 Instructor: Dr. Doreen De Leon HW #a Find the singularities and determine if they are isolated for. f cos cos has one singularity, 0 0. and g is entire. cos is analytic for all 0, and 0 0 is an isolated singularity.. f 3 + 4 f is a quotient of two functions, and the denominator is 0 only if 3 + 4 0, so the singularities of f are 3 0 0 + 4 0 4 Since f is analytic for all 0, 4, 0 0 and 0 4 are isolated singularities. 3. f cos π The denominator is 0 only if π cos 0, which is true only if π cos. From previous work, we know that this is true only if π nπ, n Z, n 0, and 0. 4n So, the singluar points are 0, ± 4, ±,.... The singularities, n ±, ±,... are 8 4n isolated singular points, since there is a deleted neighborhood of each point throughout which f is analytic.
0 0 is not an isolated singluarity, because every deleted neighborhood of 0 contains at least one point of the form and so, contains a singularity. Why? Given ɛ > 0, we 4n can find a value N such that 4N < ɛ just let N > and thus, is inside the deleted 4ɛ 4N ɛ-neighborhood of 0. HW #b. p. 39:,. Find the residue at 0 of the function a + + + + 3 + Res 0 +. b cos Res 0 cos c sin sin Res 0 cos. sin 0. + +.! + 4! 4 + 4! 3. sin 3! 3 + 5! 5 3! 3 5! 5 + 3! 5! 4 +.
d cot 4 cot cos sin! + 4! 4 3! 3 + 5! 5. e So, 3! 3 + 5! 5 + cot Res 0 4 45. sinh 4 3 45 3 + + 0! + 4! 4 + 0 5 + 6! 7 + 0 3! + 0 + 5! 4 + 0 5 + 7! 7 + 3 + 0 3 + 30 4 + 0 5 6! + 7! 6 + 3 + 0 3 + 8 4 + 0 5 360 6 + 45 4 + 0 5 [ 6! + 7! 360] 6 + cot 4 4 3 45 3 + 5 3 3 45 +. sinh + 3! 3 + 5! 5 + sinh 4 4 + 3! 3 + 5! 5 + + + 4 + 6 + 4 + 7 6 3 + 47 40 5 + 3 + 7 6 + 47 40 +. sinh Res 0 4 7 6.. Use auchy s residue theorem to evaluate the integral of each of these functions around the circle 3 in the positive sense. a e The only singularity of the function is 0 0. So, we need to determine the residue at 0 0. e +! 3! 3 + +! 3! +.. 3
b e Res 0, and 3 e d πires 0 e πi πi. e The only singularity of the function is 0. So, we need to determine the residue at 0. e e + e e e +! 3! 3 + e e + e! e Res e, and 3 e +. 3! e e d πires πi πi e e. c e d The only singularity of e is 0 0. So, we need to determine the residue at 0 0. e + +! + 3! 3 + 4! 4 + + +! + 3! + 4! +. Res 0 e 3! 6, and + 3 e d πires 0 e πi πi 6 3. f + +, so f has two singularities, 0 0 and 0, both of which are interior to the contour. This means that we need to find the residue at both points. 4
0 0: So, 0 : So, + + + + 4 + 3 8 + 4 8 3 6. 4 8 3 6 4 8 6. + 3 4 3 8 3 6. + Res 0. + + 3 + 3. + 3 + +. 3 3 3 + 3 3 + +. 3 3 + 3 4 3 8 + Res 3 4 + 3 8 3 6 +. + 3. 5
3 + + d πi Res 0 πi + 3 πi. + Res +. Problem Evaluate the following integrals a d 3 The singularities of 3 are 0 0 and 0. Only 0 0 is inside the circle. in order to evaluate the integral, we need to find the residue at 0 0. 3 3 3 + + + 3 + + + + + + + + + + + 3 + + + + + 3 + 6 +. So, 3 + 3 + 6 + Res 0 3. b d πi πi. 3 e 3 d From part a, we see that 0 0 is the only singularity inside. So, we need to 6
find the residue of the function at 0 0. e 3 e d πi πi. 3 3 e + 3 + 6 + + 4 + 9 + Res 0 e 3. + +! + 3 HW #c 3. p. 43:,. In each case, write the principal part of the function at the isolated singular point and identify the type of singularity. a e The only singular point is 0 0. And, e + +! + 3! 3 + + +! + 3! 3 +. the principal part of e is! + 3! 3 + n +! n, n and 0 0 is an essential singularity. b + The only singularity is 0. And, + the principal part of + is +, and 0 is a simple pole. + + + + + + + + +. 7
c sin The only singularity is 0 0. And, sin the pricnipal part of sin d cos 3! 3 + 5! 5 3! + 5! 4. The only singularity is 0 0. And, cos is 0, and 0 0 is a removable singularity.! + 4! 4! + 4! 3. the pricnipal part of cos is, and 0 0 is a simple pole. e 3 0 is the only singularity. The principal part is and so, 0 is a pole of order 3. 3,. Show that the singular point is a pole. Determine the order m of that pole and the corresponding residue. a cosh 3 0 0 is the only singularity. Since 3 cosh 3 +! + 4! 4 + 3! 4! 4! 4!, and the principal part is!, 0 0 is a simple pole so, m, and the resiude is B. b e 4 0 0 is the only singularity. Since, e 4 4 + +! + 3! 3 + 4! 4 + 4 4 3 3 3 4 3 4 3 3, 8
c and the principal part is 4 3 3, 0 0 is a pole of order 3 so, m 3, and the residue is B 4 3. e 0 is the only singularity. Since e e + e e, e e + +! + 3! 3 + e + e + e + 4 3 e +, the principal part is e + e. 0 is a pole of order so, m and the residue is B e. 3. p. 48:, 3, 4. In each case, show that the singularity is a pole. Determine the order m and the residue B of each pole. a + + 0 is the only singularity, and + + + φ. b Since φ + 3 0, 0 is a pole of order, and B φ 3. 3 + 0 is the only singularity, and 3 + 3 + 3 3 3 + 3 8 +. 3 φ 3 8. 3 9
Then, 0 is a pole of order m 3. We have that c 0 πi: e + π Since φ 3 8 φ 3 8, φ 3 4. B φ! e + π there are two singularities, 0 πi and 0 πi. e + π So, 0 πi is a pole of order m, and 3 4 3 6. e + πi πi, e πi + πi. φ e πi. 0 πi: B φ πi πi π i. e πi πi πi e + π So, 0 πi is a pole of order m, and e +πi πi. φ e + πi. B φπi πi π i. eπi πi + πi 0
3. Find the value of taken counterclockwise aorund the circle 3 3 + + 9 d a The singularities of the integrand are 0, ±3i. Only 0 lies inside. So, writing 3 3 + 33+ f + 9 +9, we see that φ 33 + + 9, and 0 is a simple pole. and Res f φ 33 + + 9, f d πi πi. b 4 In this case, all three singularities are interior to. 0 : Already done above. 0 3i: So, 0 3i is a simple pole, and f φ 3 3 + 3i + 3i 3 3 + 3i + 3i 3 3 + 3i. Res f φ 3i 3i 3 3i 3 + 3i 3i 3i + 8i 6i + 3i + 8i 6i + 3i 3i 3i 45 + 75i 6i0 45 + 75i 60i 5 4 49 i..
0 3i: So, 0 3i is a simple pole, and f 3 3 + + 3i 3i 3 3 + +3i 3i. φ 3 3 + + 3i. So, 4. Find the value of the integral Res f φ3i 3i 33i 3 + 3i 3i + 3i 8i 6i + 3i 8i 3i 6i + 3i 3i 45 + 75i 6i0 45 + 75i 60i 5 4 + 49 i. f d πi d 3 + 4 + 5 4 49 5 i + 4 + 49 i 6πi. taken counterclockwise around the circle. a The singularities of the integrand are 0 0 and 0 4. Only 0 0 is interior to. f 3 + 4 +4 3, and 0 0 is a pole of order 3. So, and φ + 4, Res f φ 0. 0! φ + 4 φ + 4 3 φ 0 4 3.
and so Res f 0 64, f πi π 64 3 i. b + 3 In this case, both singularities are inside. We already determined Res f 0 64, so we need only find Res f. Since 4 f 3 4 3 4, 0 4 is a simple pole, and so Then, φ 3, Res f φ 4 4 64. f d πi 64 + 0. 64 4 HW #d p. 55:,, 4. Show that the point 0 is a simple pole of f csc sin by appealing to and that the residue there is a Theorem in Sec. 76. Since f sin p q, we see that p and q are entire; p0 0; q0 0; and q 0 cos 0 0. p0 Res f 0 q 0. b the Laurent series for csc found in Ex. in Sec. 67.. Show that csc + 3! + [ 3! Res f.. 0 ] 3 + 5! 3
sinh a Res πi sinh i π e t sinh sinh p q p, q are entire. pπi πi sinhπi πi i sinπ πi 0. qπi πi sinhπi 0. q sinh + cosh q πi πi sinhπi + πi coshπi π cosπ π 0. sinh Res πi sinh pπi q πi πi π i π. e t b Res πi sinh + Res cosπt πi sinh First, we write 0 πi: so we see that p and q are entire. Then e t sinh p q, pπi e πit cosπt + i sinπt 0. qπi sinhπi 0. q cosh q πi 0. e t Res πisinh pπi qπi cosπt i sinπt. 0 πi: p πi e πit cosπt i sinπt 0. q πi sinh πi 0. q cosh q πi 0. Res πi Res πi e t sinh p πi cosπt + i sinπt. q πi e t sinh + Res πi e t sinh 4. Let denote the positively oriented circle and evaluate 4 cosπt.
a tan d p and q are entire. The singularities are where cos 0, or are 0 π and 0 π. f tan sin cos p q n + π, n Z. The only singularities inside 0 π : π π p sin 0. π q 0. q sin q π Res tan. π. 0 π : p π q π sin π 0. 0. q sin q π. Res tan. π b So, d sinh The singularities of f tan d πi + 4πi. Z. The only singularities inside are 0 0, 0 ± π i. Since sinh occur where sinh 0 nπ i, n we see that both p and q are entire. f sinh p q, 5
0 0: p0 0. q0 0. q cosh q 0. 0 π i: p0 Res f 0 q 0. π p i 0. π q i 0. q cosh q π i coshπi 0. Res tan π i. 0 π i: p π i 0. q π i 0. q cosh q π i cosh πi 0. So, Res tan π i. d sinh πi + + πi. 6