complex_ch10

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1. Chapter 10 Exercise 1. (c.f. Chapter 9, Ex9. Ex11. Ex15) (a) I. 0 is a double pole; Res 1 z4 + z2 ; 0 = lim z→0 1 1! d dz z2 1 z4 + z2 = lim z→0 d dz 1 z2 + 1 = lim z→0 −2z (z2 + 1)2 = 0. II. i is a simple pole; Res 1 z4 + z2 ; i = lim z→i (z − i) 1 z4 + z2 = lim z→i 1 z2(z + i) = − 1 2i . III. −i is a simple pole; Res 1 z4 + z2 ; −i = lim z→−i (z + i) 1 z4 + z2 = lim z→−i 1 z2(z − i) = 1 2i . (b) kπ is a simple pole for k ∈ Z; since cot z = cos z sin z and d dz sin z = cos z, then Res cot z; kπ = cos z cos z z=kπ = 1, for k ∈ Z. (c) kπ is a simple pole for k ∈ Z; since csc z = 1 sin z and d dz sin z = cos z, then Res csc z; kπ = 1 cos z z=kπ = (−1)k, for k ∈ Z. (d) I. 1 is a simple pole; Res e1/z2 z − 1 ; 1 = lim z→1 (z − 1) e1/z2 z − 1 = lim z→1 e1/z2 = e. II. 0 is an essential singularity; Res e1/z2 z − 1 ; 0 = C−1 = −e + 1. (from Ex11.b of Chapter 9) (e) 1 z2 + 3z + 2 = 1 (z + 1)(z + 2) I. Since lim z→−1 (z + 1) 1 z2 + 3z + 2 = lim z→−1 1 z + 2 = 1 = 0 and lim z→−1 (z + 1)2 1 z2 + 3z + 2 = 0, 1
2. then by Theorem 9.5, −1 is a simple pole, and Res 1 z2 + 3z + 2 ; −1 = lim z→−1 (z + 1) 1 z2 + 3z + 2 = 1. II. Since lim z→−2 (z + 2) 1 z2 + 3z + 2 = lim z→−2 1 z + 1 = −1 = 0 and lim z→−2 (z + 2)2 1 z2 + 3z + 2 = 0, then by Theorem 9.5, −2 is a simple pole, and Res 1 z2 + 3z + 2 ; −2 = lim z→−2 (z + 2) 1 z2 + 3z + 2 = −1. (f) Let the Laurent expansion of sin 1 z is ∞ k=−∞ ck zk. Since sin 1 z = ∞ n=−∞ (z−1)2n+1 (2n + 1)! , then sin 1 z has a essential singularity at z = 0, and Res sin 1 z ; 0 = C−1 = 1. (g) Let the Laurent expansion of ze3/z is ∞ k=−∞ ck zk. Since ze3/z = z ∞ n=0 1 n! ( 3 n )n = ∞ n=0 3n n! z1−n = 1 k=−∞ 31−k (1 − k)! zk, then ze3/z has a essential singularity at z = 0, and Res ze3/z; 0 = C−1 = 31−(−1) [1 − (−1)]! = 9 2 . (h) 略 Exercise 2. (c.f. Ex1.) (a) From Ex1(b), kπ is a simple pole and Res(cot z; kπ) = 1 for k ∈ Z. Since n(|z| = 1, kπ) =    1, if k = 0 0, otherwise , then |z|=1 cot zdz = 2πi ∞ k=−∞ n(|z| = 1, kπ) Res(cot z; kπ) = 2πi. 2
3. (b) (Another version) Let f(z) = (z3 − 1)(z − 4) = (z − 1) z − −1 + √ 3i 2 z − −1 − √ 3i 2 (z − 4), and let z1 = 1, z2 = −1 + √ 3i 2 , z3 = −1 − √ 3i 2 , z4 = 4. For z1 : Since lim z→z1 (z − z1 ) 1 (z − 4)(z3 − 1) = lim z→z1 1 (z − 4)(z − z2 )(z − z3 ) = 1 (1 − 4)(1 − −1+ √ 3i 2 )(1 − −1− √ 3i 2 ) = − 1 9 and lim z→z1 (z − z1 )2 1 (z − 4)(z3 − 1) = lim z→z1 z − z1 (z − 4)(z − z2 )(z − z3 ) = 0, then z1 is a simple pole. For z2 : Since lim z→z2 (z − z2 ) 1 (z − 4)(z3 − 1) = lim z→z2 1 (z − 4)(z − z1 )(z − z3 ) = 1 (−1+ √ 3i 2 − 4)(−1+ √ 3i 2 − 1)(−1+ √ 3i 2 − −1− √ 3i 2 ) = 1 9 − 6 √ 3i and lim z→z2 (z − z2 )2 1 (z − 4)(z3 − 1) = lim z→z2 z − z2 (z − 4)(z − z1 )(z − z3 ) = 0, then z2 is a simple pole. For z3 : Since lim z→z3 (z − z3 ) 1 (z − 4)(z3 − 1) = lim z→z3 1 (z − 4)(z − z1 )(z − z2 ) = 1 (−1− √ 3i 2 − 4)(−1− √ 3i 2 − 1)(−1− √ 3i 2 − −1+ √ 3i 2 ) = 1 9 + 6 √ 3i and lim z→z3 (z − z3 )2 1 (z − 4)(z3 − 1) = lim z→z3 z − z3 (z − 4)(z − z1 )(z − z2 ) = 0, 3
4. then z3 is a simple pole. For z4 : Since lim z→z4 (z − z4 ) 1 (z − 4)(z3 − 1) = lim z→z4 1 z3 − 1 = 1 43 − 1 = 1 63 and lim z→z4 (z − z4 )2 1 (z − 4)(z3 − 1) = lim z→z4 z − z4 z3 − 1 = 0, then z4 is a simple pole. Next, we calculate the winding numbers and residues of f(z) for zk , k = 1, 2, 3, 4. Winding numbers: n(|z| = 2, zk ) =    1, if k = 1, 2, 3 0, if k = 4 . Residues: Res(f; z1 ) = lim z→z1 (z − z1 ) 1 (z − 4)(z3 − 1) = − 1 9 ; Res(f; z2 ) = lim z→z2 (z − z2 ) 1 (z − 4)(z3 − 1) = 1 9 − 6 √ 3i ; Res(f; z3 ) = lim z→z3 (z − z3 ) 1 (z − 4)(z3 − 1) = 1 9 + 6 √ 3i ; Res(f; z4 ) = lim z→z4 (z − z4 ) 1 (z − 4)(z3 − 1) = 1 63 . Hence, |z|=2 dz (z − 4)(z3 − 1) = 2πi 4 k=1 n(|z| = 2, zk ) Res(f; zk ) = 2πi − 1 9 + 1 9 − 6 √ 3i + 1 9 + 6 √ 3i = − 2πi 63 . Specially, |z|=r dz (z−4)(z3−1) = 0, for r > 4. (c) From Ex1(f), 0 is an essential singularity and Res(sin 1 z ; 0) = 1. Since n(|z| = 1, 0) = 1, then |z|=1 sin 1 z dz = 2πi × n(|z| = 1, 0) × Res(sin 1 z ; 0) = 2πi. 4
5. (d) From Ex1(g), 0 is an essential singularity and Res(ze3/z; 0) = 9 2 . Since n(|z| = 2, 0) = 1, then |z|=2 ze3 z dz = 2πi × n(|z| = 2, 0) × Res(ze3 z ; 0) = 9πi. Exercise 3. When z = 2πki, k ∈ Z, the function (1−e−z)n is zero, then 1 (1−e−z)n has a pole at z = 2πki, k ∈ Z. Let C be any regular closed curve surrounding z = 0 and not surrounding any of the other singularities: z = 2πki, k = ±1, ±2, · · · . Then it leads to n(C, 2πki) =    1, if k = 0 0, otherwise , and C dz (1 − e−z)n = 2πi k∈Z n(C, 2πki)Res 1 (1 − e−z)n ; 2πki = 2πiRes 1 (1 − e−z)n ; 0 . Next, letting ω = 1−e−z, we have e−z = 1−ω ⇒ −e−zdz = −dω ⇒ dz = dω e−z = dω 1−ω , it implies C dz (1 − e−z)n = C∗ dω ωn(1 − ω) , where C∗ is the image under ω = 1 − e−z of C. Note that: 1. 1 ωn(1−ω) has a pole at ω = 0, 1. 2. C∗ surrounds 0 and not 1 in the ω-plane. For this, we consider the following: 5
6. Then it implies n(C∗, 0) = 1, n(C∗, 1) = 0 and C∗ dω ωn(1 − ω) = 2πiRes 1 ωn(1 − ω) ; 0 . Since 1 ωn(1−ω) = ω−n(1 + ω + ω2 + · · · ) = ∞ k=−n ωk, and hence Res 1 (1 − e−z)n ; 0 = Res 1 ωn(1 − ω) ; 0 = 1. 6