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Unformatted text preview: Student Solutions Manual to Accompany Introduction to Complex Analysis and Its Applications Donald Trim The University of Manitoba c 2017 Donald W. Trim PREFACE This manual contains solutions to even-numbered exercises in the text. Solutions are sufficiently detailed that the reader should have no difficulty following the logic. Mechanical calculations, such as integrations and algebraic simplifications, are sometimes omitted to conserve space. The author would appreciate being informed of any errors, typographical or otherwise, or suggestions for improvement. He may be contacted by e-mail at [email protected] c 2017 Donald W. Trim All rights reserved. No part of this manual may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission of the author. SECTION 1.1 EXERCISES 1.1 √ √ √ 2. (a + b 3) + (c + d 3) = (a + c) + (b + d) 3 √ √ √ 4. (a + b 3)(c + d 3) = (ac + 3bd) + (ad + bc) 3 1 2 SECTION 1.2 EXERCISES 1.2 2. (2 + 4i) − (3 − 2i) = (2 − 3) + (4 + 2)i = −1 + 6i 4. (−2 + i)(3 − 4i) = (−6 + 4) + (3 + 8)i = −2 + 11i 6. i3 − 3i2 + 2i + 4 = −i + 3 + 2i + 4 = 7 + i 1−i 1 − i 3 − 2i 1 − 5i 1 5 8. = = = − i 3 + 2i 3 + 2i 3 − 2i 13 13 13 10. i24 − 3i13 + 4 = (i4 )6 − 3i(i4 )3 + 4 = 1 − 3i + 4 = 5 − 3i         i−4 1+i 2+i i − 4 1 − 2i 1+i +3 = 6i +3 12. 6i 2−i 2i + 1 1 + 2i 1 − 2i  2 − i 2+ i  −2 + 9i −18 + 6i −6 + 27i 1 + 3i + +3 = = 6i 5 5 5 5 24 33 =− + i 5 5 2 2 2 14. 1 + i + (1 + i) = (1 − i) + 2i = −2i − 2i = −4i 16. If we set z1 = x1 + y1 i and z2 = x2 + y2 i, then (a) z1 + z2 = (x1 + x2 ) + (y1 + y2 )i = (x1 + x2 ) − (y1 + y2 )i = (x1 − y1 i) + (x2 − y2 i) = z1 + z2 (b)z1 − z2 = (x1 − x2 ) + (y1 − y2 )i = (x1 − x2 ) − (y1 − y2 )i = (x1 − y1 i) − (x2 − y2 i) = z1 − z2 (c) z1 z2 = (x1 x2 − y1 y2 ) + (x1 y2 + x2 y1 )i = (x1 x2 − y1 y2 ) − (x1 y2 + x2 y1 )i = (x1 − y1 i)(x2 − y2 i) = z1 z2   x1 + y1 i x2 − y2 i (x1 x2 + y1 y2 ) + (x2 y1 − x1 y2 )i z1 = (d) Since = z2 x2 + y2 i x2 − y2 i x2 2 + y2 2 (x1 x2 + y1 y2 ) − (x2 y1 − x1 y2 )i = , x2 2 + y2 2 x1 − y1 i x2 + y2 i (x1 x2 + y1 y2 ) + (x1 y2 − x2 y1 )i x1 − y1 i z1 = = = , and z2 x2 − y2 i x2 − y2 i x2 + y2 i x2 2 + y2 2   z1 z1 = . it follows that z2 z2 18. 20. 22. 24. 26. 28. (e) With part (c) we can write that z n = z z n−1 = z z n−1 . Another application of this same property gives z n = z (z z n−2 ) = z (z z n−2 ) = z 2 z n−2 . Continuation leads to z n = z n . If z = x + yi, then z z = r 2 implies that r 2 = (x + yi)(x − yi) = x2 + y 2 . This is the equation of a circle with centre z = 0 and radius r. p √ √ then z z = (−1)(−1) = 1 = 1. On the other hand, If z1 = z2 √= −1, 1 2 √ √ √ z1 z2 = −1 −1 = (i)(i) = −1. √ √ −5 ± 13 −5 ± 25 − 12 = x= 2 2 Since x2 + 8x + 16 = (x + 4)2 , the equation has a double root x = −4. √ √ −2 ± 4 − 28 = −1 ± 6i x= 2 p p √ p √ √ −5 ± 25 − 12 5 5 12 5 − 25 −5 ± 25 − 4 45 √ √ √ = =− √ ± i x= 2 3 2 3 2 3 2 3 3 SECTION 1.2 4 2 2 2 2 2 30. Since x √ + 4x + 3 = (x + 3)(x + 1), it follows that x = −3 or x = −1. Hence x = ± 3i or x = ±i. 32. If we let the numbers be z and w, then z + w = 6 and zw = 10. Substitution from the√first into the second gives z(6 − z) = 10, or z 2 − 6z + 10 = 0. Thus, z = (6 ± 36 − 40)/2 = 3 ± i, and the numbers are z = 3 + i and w = 6 − z = 3 − i. Choosing z = 3 − i gives the same pair of numbers. 34. If we set z = x + yi in the equation z 2 = 21 − 20i, then (x2 − y 2 ) + 2xyi = 21 − 20i. When we equate real and imaginary parts, x2 − y 2 = 21, 2xy = −20. From the second of these, y = −10/x, which we substitute into the first 21 = x2 − 100 x2 =⇒ 0= x4 − 21x2 − 100 (x2 − 25)(x2 + 4) = . x2 x2 Hence, 0 = (x2 − 25)(x2 + 4), and since x must be real, x = ±5. Correspondingly, y = ∓2, and the required square roots are ±(5 − 2i). 36. If we set z = x + yi in the equation z 2 = −1 + i, then (x2 − y 2 ) + 2xyi = −1 + i. When we equate real and imaginary parts, x2 − y 2 = −1, 2xy = 1. From the second of these, y = 1/(2x), which we substitute into the first −1 = x2 − 1 4x2 =⇒ 0= 4x4 + 4x2 − 1 . 4x2 √ √ Thus, 0 = 4x4 + 4x2 − 1,qfrom which x2 = (−4 ± 16 + 16)/8 = (−1 ± 2)/2. Since √ √ p√ 2 − 1) = x must be real, x = ± ( 2 − 1)/2. Correspondingly, y = ±1/( 2 q  q q √ √ √ ( 2 − 1)/2 + ( 2 + 1)/2i . ± ( 2 + 1)/2. The required square roots are ± p √ −4 ± 16 − 4(2i)(3 − 2i) −4 ± −24i = To find the square roots of −24i 38. z = 4i 4i we set z = x + yi in z 2 = −24i, the result being (x2 − y 2 ) + 2xyi = −24i. When we equate real and imaginary parts, x2 − y 2 = 0, 2xy = −24. From the second, y = −12/x, and when this is substituted into the first, 144 x4 − 144 = . x2 x2 √ √ Hence, x2 = ±12, from which √ x = ±2 3. Correspondingly, y = ∓2 3, and the square roots of −24i are ±2 3(1 − i). Solutions of the original equation are √ √ ! √ √ ! √ −4 ± 2 3 3 3 3 −4 ± 2 3(1 − i) =∓ − i=∓ + 1∓ i 4i 2 4 2 2 √ ! √ 3 3 + 1± i. =± 2 2 0 = x2 − 4 SECTION 1.2 √ √ 40. z = (−2i± −4 − 12i)/(2i) = −1± −1 − 3i i To find the square roots of −1−3i we set z = x + yi in z 2 = −1 − 3i, the result being (x2 − y 2 ) + 2xyi = −1 − 3i. When we equate real and imaginary parts, x2 − y 2 = −1, 2xy = −3. From the second, y = −3/(2x), and when this is substituted into the first, −1 = x2 − 9 4x2 =⇒ 0= 4x4 + 4x2 − 9 . 4x2 q√ √ √ Hence, x2 = (−4± 16 + 144)/8 = (−1± 10)/2, from which x = ± ( 10 − 1)/2. q √ Correspondingly, y = ∓3/ 2( 10 − 1), and square roots of −1 − 3i are  q q √ √ 10 − 1)/2 − 3i/ 2 10 − 2 . Solutions of the original equation are ± s √ ! s √ 3 3 10 − 1 10 − 1 i i = −1 ± p √ ± −p √ i. 2 2 2 10 − 2 2 10 − 2 √ √ −2 + i ± −1 − 8i −2 + i ± 3 − 4i − 4 − 4i = To find the square roots of 42. z = 2 2 2 −1−8i we set z = x+yi in z = −1−8i, the result being (x2 −y 2 )+2xyi = −1−8i. When we equate real and imaginary parts, z = −1 ± x2 − y 2 = −1, 2xy = −8. From the second, y = −4/x, and when this is substituted into the first −1 = x2 − 2 Hence, x = (−1 ± 16 x2 √ =⇒ 0= √ x4 + x2 − 16 . x2 q √ 65)/2, from which x = ± ( 65 − 1)/2. 1 + 64)/2 = (−1 ± q√ Correspondingly, y = ∓ ( 65 + 1)/2, and the square roots of −1 − 8i are  q q √ √ ( 65 − 1)/2 − i ( 65 + 1)/2 . Solutions of the original equation are ± s s√ √ 1 65 − 1 65 + 1 i − i z = −1 + ± 2 2 2 2 s s √ √ 1 65 − 1 65 + 1 + ∓ i. = −1 ± 8 2 8 44. If we set z = x + yi in 2iz 2 + 4z + 3 − 2i = 0, 0 = 2i(x + yi)2 + 4(x + yi) + 3 − 2i = (−4xy + 4x + 3) + (2x2 − 2y 2 + 4y − 2)i. When we equate real and imaginary parts, −4xy + 4x + 3 = 0, 2x2 − 2y 2 + 4y − 2 = 0. From the first y = (4x + 3)/(4x), and when this is substituted into the second SECTION 1.2  16x4 − 9 4x + 3 −2= +4 . 0 = 2x − 2 4x 8x2 √ √ Hence,√x = ± 3/2, √ and these give y = 1 ± 3/2. The solutions are therefore z = ± 3/2 + (1 ± 3/2)i. 46. If z1 = x1 + y1 i, z2 = x2 + y2 i, z3 = x3 + y3 i, and z4 = x4 + y4 i, then the slope of the line joining z1 and z2 is (y2 − y1 )/(x2 − x1 ), and the slope of the line joining z3 and z4 is (y4 − y3 )/(x4 − x3 ). The lines are parallel if and only if their slopes are equal; that is, 2 y2 − y1 y4 − y3 = x4 − x3 x2 − x1  4x + 3 4x ⇐⇒ 2 5  (y4 − y3 )(x2 − x1 ) − (x4 − x3 )(y2 − y1 ) = 0. But, Im [(z1 − z2 )(z3 − z4 )] = Im {[(x1 − x2 ) + (y1 − y2 )i][(x3 − x4 ) − (y3 − y4 )i]} = (y1 − y2 )(x3 − x4 ) − (x1 − x2 )(y3 − y4 ). The lines are therefore parallel if and only if Im [(z1 − z2 )(z3 − z4 )] = 0. 6 SECTION 1.3 EXERCISES 1.3 2. Since r = 2, and an argument for −2i is −π/2, a polar representation for −2i is 2[cos (−π/2) + sin (−π/2) i]. √ 4. Since r = 9 + 16 = 5, and an argument for 3 + 4i is Tan−1 (4/3) ≈ 0.927, a polar representation is 5[cos (0.927) + sin (0.927) i]. √ √ 6. −2[cos (π/3)−sin (π/3) i] = −1+ 3 i Since r = 1 + 3 = 2, and an argument for the complex number is 2π/3, a polar representation is 2[cos (2π/3) + sin (2π/3) i]. 8. This complex number is essentially in Cartesian form now, but we could evaluate the trigonometric functions to write cos (1.4) + sin (1.4) i ≈ 0.170 + 0.985i 4[cos (2π/3) + sin (2π/3) i] = 4[cos (π/2) + sin (π/2) i] = 4i 10. cos (π/6) + sin (π/6) i 12. [cos (π/5) + sin (π/5) i]10 = cos (2π) + sin (2π) i = 1 −2[cos (−π/3) + sin (−π/3) i]3 −2[cos (π/3) − sin (π/3) i]3 = 14. 3+i 3+i −2[cos (−π) + sin (−π) i] = 3+i 6 − 2i 3 i 2 3−i = = − = 3+√ i 3−i 10 5 5 5(cos θ + sin θ i), where θ = Tan−1 (1/2), and 16. A polar representation for 2 + i is √ −1 one for 4 − i is 17(cos φ + sin φ i), where φ = −Tan (1/4). Hence, √ [ 5(cos θ + sin θ i)]6 125[cos 6θ + sin 6θ i] (2 + i)6 = √ = √ 3 (4 − i)3 [ 17(cos φ + sin φ i)] 17 17[cos 3φ + sin 3φ i] 125 = √ [cos (6θ − 3φ) + sin (6θ − 3φ) i] ≈ −1.66 − 0.654i 17 17 18. (a) If we set z1 = x1 + y1 i and z2 = x2 + y2 i, then |z1 + z2 |2 = |(x1 + x2 ) + (y1 + y2 )i|2 = (x1 + x2 )2 + (y1 + y2 )2 = (x21 + y12 ) + 2(x1 x2 + y1 y2 ) + (x22 + y22 ). Since Re (z1 z2 ) = Re [(x1 + y1 i)(x2 − y2 i)] = x1 x2 + y1 y2 , we have |z1 + z2 |2 = |z1 |2 + 2Re (z1 z2 ) + |z2 |2 . (b) With z1 = x1 + y1 i and z2 = x2 + y2 i, |z1 + z2 |2 + |z1 − z2 |2 = |(x1 + x2 ) + (y1 + y2 )i|2 + |(x1 − x2 ) + (y1 − y2 )i|2 = [(x1 + x2 )2 + (y1 + y2 )2 ] + [(x1 − x2 )2 + (y1 − y2 )2 ] = 2(x21 + y12 ) + 2(x22 + y2 )2 = 2(|z1 |2 + |z2 |2 ). (c) Once again with z1 = x1 + y1 i and z2 = x2 + y2 i, |z1 − z2 |2 − |z1 + z2 |2 = |(x1 − x2 ) + (y1 − y2 )i|2 − |(x1 + x2 ) + (y1 − y2 )i|2 = [(x1 − x2 )2 + (y1 − y2 )2 ] − [(x1 + x2 )2 + (y1 − y2 )2 ] = −4x1 x2 = −4(Re z1 )(Re z2 ). SECTION 1.4 7 EXERCISES 1.4 2. Since r = 2, and an argument for the complex number is −π/2, an exponential form is 2e−πi/2 . 4. Since r = 5, and an argument for the complex number is Tan−1 (4/3), an exponential −1 form is 5eiTan (4/3) . 6. −2[cos (π/3) − sin (π/3)i] = 2[− cos (π/3) + sin (π/3)i] = 2[cos (2π/3) + sin (2π/3)i] = 2e2πi/3 8. e−πi = cos (−π) + sin (−π) i = −1 4e2πi/3 = 4eπi/2 = 4[cos (π/2) + sin (π/2) i] = 4i eπi/6 12. [cos (π/5) + sin (π/5) i]10 = (eπi/5 )10 = e2πi = cos 2π + sin 2π i = 1 10. −2[cos (−π/3) + sin (−π/3) i]3 −2(e−πi/3 )3 −2[cos (π/3) − sin (π/3) i]3 = = 3+i 3+i 3+i 2 3−i 6 − 2i 3 i −2e−πi = = = − = 3+i 3+i3−i 10 5 5 √ φi √ θi −1 16. Since 2 + i = 5e , where θ = Tan (1/2), and 4 − i = 17e , where φ = −Tan−1 (1/4), √ ( 5eθi )6 125e6θi 125 (2 + i)6 √ √ = = = √ e(6θ−3φ)i 3 φi 3 3φi (4 − i) ( 17e ) 17 17e 17 17 125 = √ [cos (6θ − 3φ) + sin (6θ − 3φ) i] ≈ −1.66 − 0.654i. 17 17 14. √ 18. Since 3 − i = 2e−πi/6 and √ 3 + i = 2eπi/6 , !4  √ 5  −πi/6 4  5  5 2e 3−i 1+i 1+i1+i 2i −πi/3 4 √ = = (e ) = ie−4πi/3 πi/6 1−i 1−i1+i 2 2e 3+i √ ! √ i 1 3i 3 =− − = i[cos (−4π/3) + sin (−4π/3) i] = i − + 2 2 2 2 20. When the complex number in DeMoivre’s theorem 1.32 has modulus r = 1, z n = (eθi )n = enθi . For n = 4, this becomes (eθi )4 = e4θi =⇒ (cos θ + sin θ i)4 = cos 4θ + sin 4θ i. When the left side is expanded with the binomial theorem, the result is cos4 θ + 4 cos3 θ sin θ i − 6 cos2 θ sin2 θ − 4 cos θ sin3 θ i + sin4 θ = cos 4θ + sin 4θ i. When we equate real parts, we obtain cos4 θ − 6 cos2 θ sin2 θ + sin4 θ = cos 4θ. 22. When the complex number in DeMoivre’s theorem 1.32 has modulus r = 1, z n = (eθi )n = enθi . For n = 5, this becomes 8 SECTION 1.4 (eθi )5 = e5θi =⇒ (cos θ + sin θ i)5 = cos 5θ + sin 5θ i. When the left side is expanded with the binomial theorem, the result is cos5 θ + 5 cos4 θ sin θ i − 10 cos3 θ sin2 θ − 10 cos2 θ sin3 θ i + 5 cos θ sin4 θ + sin5 θ i = cos 5θ + sin 5θ i. When we equate real parts, we obtain cos5 θ−10 cos3 θ sin2 θ+5 cos θ sin4 θ = cos 5θ. 24. When we raise both sides of equation 1.35a to power three, 1 θi 1 1 (e + e−θi )3 = (e3θi + 3eθi + 3e−θi + e−3θi ) = (2 cos 3θ + 6 cos θ) 8 8 8 1 = (3 cos θ + cos 3θ). 4 cos3 θ = 26. When we raise both sides of equation 1.35b to power four, 1 θi 1 4θi (e − e−θi )4 = (e − 4e2θi + 6 − 4e−2θi + e−4θi ) 16 16 1 1 (6 − 8 cos 2θ + 2 cos 4θ) = (3 − 4 cos 2θ + cos 4θ). = 16 8 sin4 θ = 28. The left side of the equation can be written in the form i sin θ n  n n  θi n  1 + cos θ + sin θ i e 1 + i tan θ cos θ = = = = (e2θi )n = e2nθi . i sin θ 1 − i tan θ cos θ − sin θ i e−θi 1− cos θ A similar calculation on the right gives i sin nθ 1+ nθi 1 + i tan nθ cos nθ = cos nθ + i sin nθ = e = = e2nθi . i sin nθ 1 − i tan nθ cos nθ − i sin nθ e−nθi 1− cos nθ 30. When we raise both sides of equation 1.35b to power n, n  θi n n e − e−θi 1 X n = nn (−1)n−k ekθi e−(n−k)θi sin θ = 2i 2 i k k=0 = 1 2n in n X k=0 (−1)n−k n k e(2k−n)θi . When n is odd, there is an even number of terms in the summation. We separate them into the first (n + 1)/2 terms and the last (n + 1)/2 terms. In addition, when n is odd, 1/in = (−1)(n+1)/2 i, so that (n−1)/2 n n n (n+1)/2 X X (−1) i (−1)k+1 (−1)k+1 e(2k−n)θi + e(2k−n)θi . sinn θ = 2n k k k=0 k=(n+1)/2 In the first line below, we change variables in the second summation by setting m = n − k, and in the second line we change back to k, 9 SECTION 1.4 (−1)(n+1)/2 i sin θ = 2n n "(n−1)/2 X (−1)k+1 n k k=0 0 X + e(2k−n)θi (−1)n−m+1 m=(n−1)/2 = "(n−1)/2 (−1)(n+1)/2 i X 2n (−1)k+1 n k k=0 + (−1)(n+1)/2 i = 2n "(n−1)/2 X n−k+1 (−1) (−1) n k k=0 + X k+1 −(−1)  k=0 Since k =   e[2(n−m)−n]θi # n n−k  (n−2k)θi e # e(2k−n)θi (n−1)/2 n  k=0 k+1 n n−m e(2k−n)θi (n−1)/2 X  n n−k  (n−2k)θi e # .  n , n−k sinn θ = = (−1)(n+1)/2 i 2n X i 2n (−1)(n+1)/2 = 2n−1 n [e(2k−n)θi − e(n−2k)θi] (−1)k+1 n [−2i sin (n − 2k)θ] k=0 (n+1)/2 (n−1)/2 X (−1) (−1)k+1 (n−1)/2 k=0 (n−1)/2 X k=0 (−1)k+1 k k n k sin (n − 2k)θ. When n is even, there is an odd number of terms in the summation. We separate them into the first n/2 terms, the (n + 2)/2nd term, and the last n/2 terms. In addition, when n is even, 1/in = (−1)n/2 , so that "     n/2 (n−2)/2 X (−1) n n k n (2k−n)θi n/2 (−1) + (−1) e sin θ = 2n k n/2 k=0 # n   X k n (2k−n)θi (−1) e . + k k=(n+2)/2 In the first line below, we change variables in the second summation by setting m = n − k, and in the second line we change back to k, "(n−2)/2     X (−1)n/2 n n k n (2k−n)θi n/2 (−1) + (−1) e sin θ = 2n k n/2 k=0 10 SECTION 1.4 0 X + n−m (−1)  m=(n−2)/2 = "(n−2)/2 X (−1)n/2 2n k (−1) k=0 n k (2k−n)θi e n/2 + (−1) (n−2)/2 + = "(n−2)/2 X (−1)n/2 2n X (−1)n−k (−1) k=0 n k (2k−n)θi e (n−2)/2 + X (−1)k k=0  n Since = , k n−k   (−1)n n (−1)n/2 n + sin θ = 2n 2n n/2 n =  k=0 k n n−m n n−k  n/2 + (−1)  n n−k    n n/2 [2(n−m)−n]θi e  e(n−2k)θi  n n/2 # #  # e(n−2k)θi .  (n−2)/2 X n k=0 n/2 (n−2)/2 X (−1) (−1)n n! + n 2 2 [(n/2)!] 2n−1 (−1)k k=0 (−1)k n k k [e(2k−n)θi + e(n−2k)θi ] cos (n − 2k)θ. 11 SECTION 1.5 Im z EXERCISES 1.5 2. If we set z = reθi in z 2 = 21 − 20i, r 2 e2θi = 21 − 20i = 29eΘi , Q Re z where Θ = −Tan−1 (20/21). Conditions 1.21 require 21-20i 2θ = Θ + 2kπ, k an integer. r 2 = 29 √ and Hence r = 29 and θ = Θ/2 + kπ. For k = 0 and k = 1 we obtain √ √ √ √ z1 = 29e(Θ/2+π)i = 29eΘi/2 eπi = − 29eΘi/2 = −z0 . z0 = 29eΘi/2 , √ The square roots are ± 29eΘi/2 . We can use a calculator to approximate real and imaginary parts, or we can use trigonometry to find exact values. Because cos Θ = 21/29 and Θ is an angle in the fourth quadrant (right figure above), it follows that r r   1 − cos Θ 1 − 21/29 2 Θ =− =− = −√ , sin 2 2 2 29 r   r 1 + cos Θ 1 + 21/29 Θ 5 cos = = =√ . 2 2 2 29 √ √ √ Thus, the square roots are ± 29(5/ 29 − 2i/ 29) = ±(5 − 2i). 4. If we set z = reθi in z 3 = 1, r 3 e3θi = 1 = 1e0i . Conditions 1.21 require r3 = 1 and 3θ = 0 + 2kπ = 2kπ, k an integer. Hence r = 1 and θ = 2kπ/3, and for k = 0,1,2, we obtain √ √ 3i 3i 1 1 0i 2πi/3 4πi/3 , z2 = 1e . =− + =− − z0 = 1e = 1, z1 = 1e 2 2 2 2 √ 6. If we set z = reθi in z 2 = 2 + 2 3i, √ r 2 e2θi = 2 + 2 3i = 4eπi/3 . Conditions 1.21 require r2 = 4 and 2θ = π/3 + 2kπ, k an integer. Hence r = 2 and θ = π/6 + kπ. For k = 0, we obtain 2eπi/6 = 2[cos (π/6) + sin (π/6) i] = √ The other square root is − 3 − i. √ 3 + i. 8. If we set z = reθi in z 3 = 4 − 3i, r 3 e3θi = 4 − 3i = 5eΘi where Θ = −Tan−1(3/4). Conditions 1.21 require r3 = 5 and 3θ = Θ + 2kπ, k an integer. 12 SECTION 1.5 Hence r = 51/3 and θ = Θ/3 + 2kπ/3, and for k = 0,1,2, we obtain z0 = 51/3 eΘi/3 = 51/3 [cos (Θ/3) + sin (Θ/3) i] ≈ 1.671 − 0.364i, z1 = 51/3 e(Θ/3+2π/3)i = 51/3 [cos (Θ/3 + 2π/3) + sin (Θ/3 + 2π/3) i] ≈ −0.520 + 1.629i, z2 = 51/3 e(Θ/3+4π/3)i = 51/3 [cos (Θ/3 + 4π/3) + sin (Θ/3 + 4π/3) i] ≈ −1.151 − 1.265i. √ √ 3 11 =− ± i To find the square roots of −3/2 + 11i/2, we 10. z = 2 2 √2 set z = reθi in z 2 = −3/2 + 11i/2, √ 3 11i √ Θi 2 2θi = 5e r e =− + 2 2 √ where Θ = π − Tan−1 ( 11/3). Conditions 1.21 require √ and 2θ = Θ + 2kπ, k an integer. r2 = 5 2 −3 ± √ 9 − 20 Hence r = 51/4 and θ = Θ/2 + kπ. For k = 0, we obtain z = 51/4 eΘi/2 = 51/4 [cos (Θ/2) + sin (Θ/2) i] ≈ 0.607 + 1.367i. √ The other square root is −0.607−1.367i. The square roots of −3/2− 11i/2 lead to the solutions z ≈ ±(0.607 − 1.367i). Alternatively, because this is a real polynomial equation, the remaining roots must be the complex conjugates of ±(0.607 + 1.367i). 12. (a) If we set z = reθi in z n = 1, r n enθi = 1 = 1e0i . Conditions 1.21 require rn = 1 and nθ = 0 + 2kπ = 2kπ, k an integer. Hence r = 1 and θ = 2kπ/n, and for k = 0, . . ., n − 1, we obtain     2kπ 2kπ 2kπi/n + sin i. = cos ωk = e n n (b) Using part (a), ωk ωj = e2kπi/n e2jπi/n = e2(k+j)πi/n . If r is the remainder when k + j is divided by n, then k + j = mn + r, where m is either 0 or 1. Hence, ωk ωj = e2(mn+r)πi/n = e2mπi e2rπi/n = e2rπi/n = ωr . (c) If we set S = ω0 + ω1 + · · · + ωn−1 , and multiply this by ω1 , then using part (b), ω1 S = ω1 ω0 + ω1 ω1 + · · · + ω1 ωn−1 = ω1 + ω2 + · · · + ωn . Since ωn = ω0 , we have ω1 S = ω0 + ω1 + · · · + ωn−1 = S. Consequently, (1 − ω1 )S = 0, from which S = 0. (d) If we set S = 1 + ωk + ωk2 + · · · + ωkn−1 , and multiply this by ωk , we obtain ωk S = ωk + ωk2 + ωk3 + · · · + ωkn . 13 SECTION 1.5 If this is subtracted from S, the result is S − ωk S = 1 − ωkn = 1 − 1 = 0. Consequently, (1 − ωk )S = 0, from which S = 0. (e) If we multiply S = 1 + 2ωk + 3ωk2 + 4ωk3 + · · · + nωkn−1 by 1 − ωk , the result is (1 − ωk )S = (1 − ωk )(1 + 2ωk + 3ωk2 + 4ωk3 + · · · + nωkn−1 ) = (1 + 2ωk + 3ωk2 + · · · + nωkn−1 ) − (ωk + 2ωk2 + 3ωk3 + · · · + nωkn ) = 1 + ωk + ωk2 + ωk3 + · · · + ωkn−1 − nωkn . Using part (d) and the fact that ωkn = 1, we get (1 − ωk )S = −n =⇒ S= −n . 1 − ωk 14. When we substitute z = x + yi, |x + yi + i| = |x + yi − 2i| 16. 18. 20. 22. 24. =⇒ x2 + (y + 1)2 = x2 + (y − 2)2 =⇒ y = 1/2. This is the horizontal line 1/2 unit above the real axis. This is the circle with centre z = 2 − i and radius 3, and its interior. This is the region between and including two concentric circles with centre z = −3 and radii 1 and 4. This is an ellipse with foci z = ±1 and major axis of length 4, and its interior. This is the region between two ellipses with foci z = ±1 and major axes of lengths 4 and 6, including the ellipses. If we set z = x + yi, then (x + yi)2 + (x + yi) + 2(x − yi) + 4 = 0. When we equate real and imaginary parts x2 − y 2 + 3x + 4 = 0, 2xy − y = 0. The second equation requires that y = 0 or x = 1/2. When y = 0, the first equation is x2 + 3x + 4 = 0 which has no real solutions. When x = 1/2, the first equation gives 3 1 − y2 + + 4 = 0 4 2 26. 28. 30. 32. =⇒ y2 = 23 . 4 √ Hence, only the complex numbers z = 1/2 ± 23i/2 satisfy the given equation. Since the real part of z must be positive, we can write Re z > 0. Since the equation of a circle with centre z = 3 − 4i and radius r is |z − 3 + 4i| = r, the annulus is described by 5 < |z − 3 + 4i| < 8. In the Cartesian plane, the equation of the square is |x| + |y| = 1. In the complex plane, |Re z| + |Im z| = 1. Points inside a circle centred at the origin with radius 2 satisfy |z| < 2. To retain those points above the real axis, we add to this Im z > 0. 14 SECTION 1.5 34. If we substitute z = x + yi, z1 = x1 + y1 i, and z = x2 + y2 i, |(x + yi) − (x1 + y1 i)| = |(x + yi) − (x2 + y2 i)| (x − x1 )2 + (y − y1 )2 = (x − x2 )2 + (y − y2 )2 −2x1 x + x21 − 2y1 y + y12 = −2x2 x + x22 − 2y2 y + y22 2(x2 − x1 )x + 2(y2 − y1 )y = x22 − x21 + y22 − y12 This...
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