sols8 - MATH 135 Solutions to Assignment 8 W10 This...

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Unformatted text preview: MATH 135 Solutions to Assignment 8 W10 This assignment is due at 8:30am on Wednesday March 17, in the drop box outside the Tutorial Centre, MC 4067 Remember to show all of your computational work. 1. [8 marks, distributed as (a) 2, (b) 2, (c) 2 (d) 2] Express each of the following complex numbers in the standard form x + iy , where x, y ∈ R. (a) (3 + 7i)2 Answer: (3 + 7i)2 = (32 + 42i + (7i)2 ) = −40 + 42i. √ (b) (2 + 5 i + 5i)3 Answer: √ √ √ (2 + 5 i + 5i)3 = (2 + 5(1 + 5) i)3 √ √ √ √ √ √ = 23 + 3 · 22 · 5(1 + 5) i + 3 · 2 · ( 5(1 + 5) i)2 + ( 5(1 + 5) i)3 √ √ √ √ √ = 8 + 12 5(1 + 5) i − 30(6 + 2 5) − 5 5(16 + 8 5) i √ √ = (−172 − 60 5) − (140 + 68 5) i. (c) 1 i101 Answer: By Proposition 8.23, i101 = i because 101 ≡ 1 (mod 4). Hence, 1 i101 = 1 i = = −i = 0 + (−1)i. i (−1) √ ( 7 − i)2 √ (d) √ ( 3 − i)(1 + 12 i) Answer: √ √ ( 7 − i)2 (6 − 2 7i) √ √ =√ ( 3 − i)(1 + 12 i) (3 3 + 5i) √ √ 6 − 2 7i 3 3 − 5i =√ ·√ 3 3 + 5 i 3 3 − 5i √ √ √ (18 3 − 10 7) − (5 + 21)6i = 27 − 25i2 √ √ √ 9 3 − 5 7 3(5 + 21) − i. = 26 26 2. [9 marks, distributed as (a) 3, (b) 3, (c) 3] Solve each of the following equations for z ∈ C. Express your answers in cartesian form. (a) 3z + 1 = 2 + 8i z − 5i Answer: We have 3z + 1 = 2 + 8i ⇐⇒ (3z + 1) = (2 + 8i)(z − 5i) ⇐⇒ 3z + 1 = 2z + 8iz − 10i + 40 z − 5i −39 + 10i −39 + 10i −1 − 8i 119 + 302i ⇐⇒ −39 + 10i = (−1 + 8i)z ⇐⇒ z = = · = −1 + 8i −1 + 8i −1 − 8i 65 119 302 ⇐⇒ z = + i. 65 65 (b) z 2 = 15z Answer: Write z = x + iy with x, y ∈ R. Then we have z 2 = 15z ⇐⇒ (x + iy )2 = 15(x − iy ) ⇐⇒ (x2 − y 2 ) + i(2xy ) = 15(x − iy ) ⇐⇒ x2 − y 2 = 15x and 2xy = −15y . From 2xy = −15y , we get y (2x + 15) = 0, thus y = 0 or x = − 15 . 2 When y = 0, the equation x2 − y 2 = 15x gives x2 = 15x so x = 0 or x = 15. When x = − 15 , the equation x2 − y 2 = 15x gives y 2 = x2 − 15x so y 2 = 2 y= √ ± 152 3 . (3)·(152 ) , 4 that is Thus z = x + iy = 0, 15, − 15 + 2 √ (c) z 2 = 15 + 31 i √ 15 3 2i or − 15 − 2 √ 15 3 2 i. Answer: Write z = x + yi with x, y ∈ R. Then we have √ √ √ z 2 = 15 + 31 i ⇐⇒ (x + yi)2 = 15 + 31 i ⇐⇒ (x2 − y 2 ) + (2xy )i = 15 + 31 i √ ⇐⇒ x2 − y 2 = 15 and 2xy = 31 . From 2xy = x2 − √ √ 31 we get y = 31 2x . Put this into the equation x2 − y 2 = 15 to get 31 = 15 =⇒ 4x4 − 31 = 60x2 =⇒ 4x4 − 60x2 − 31 = 0 =⇒ (2x2 − 31)(2x2 + 1) = 0 4x2 31 1 =⇒ x2 = or x2 = − . 2 2 31 2 √ 31 2x Since x ∈ R we must have x2 = When x = y= 1 − √2 . 31 2 31 2 so x = ± √ 31 2. 2 31 we have y = = √ 31 2 · = 1 √, 2 and when x = − 31 2 we have Thus z = ± + 1 √ 2 i =± 1 √ 2 31 + i . 3. [9 marks, distributed as (a) 3, (b) 3, (c) 3 ] Draw a picture of each of the following subsets of the plane. Explain your work, and label the pictures with relevant data. (a) z ∈ C and 10 < |z − 10| ≤ √ 480 Answer: We have 10 < |z − 10| when the distance between z and 10 is greater than 10; √ thus z lies outside the circle of radius 10 centered at 10. We have |z − 10| ≤ 480 when √ the distance between z and 10 is less than or equal to 480; thus z lies inside or on the √ circle of radius 480 centered at 10. (b) z ∈ C and |z − (2 + 3 i)| = |z − (4 + 2i)| Answer: We have |z − (2 + 3 i)| = |z − (4 + 2i)| when z is equidistant from the points (2 + 3 i) and (4 + 2i), that is when z lies on the perpendicular bisector of the line segment from (2 + 3 i) to (4 + 2i). (c) z ∈ C and |z | + |z − 4| = 8 Answer: We have |z | + |z − 4| = 8 when the sum of the distance from 0 to z with the distance from z to 4 is equal to 8. This happens when z is on the ellipse, with focii at 0 and 4, which passes through the points −2, ±3i, 4 ± 3i, and 6. Alternatively, writing z = x + iy with x, y ∈ R we can see that this set is the above ellipse as follows. |z | + |z − 4| = 8 ⇐⇒ x2 + y 2 + (x − 4)2 + y 2 = 8 ⇐⇒ x2 − 8x + 16 + y 2 = 8 − x2 + y 2 ⇐⇒ x2 − 8x + 16 + y 2 = 64 − 16 ⇐⇒ 2 x2 + y 2 + x2 + y 2 ⇐⇒ 16 x2 + y 2 = 8x + 48 x2 + y 2 = x + 6 ⇐⇒ 4(x2 + y 2 ) = x2 + 12x + 36 ⇐⇒ 3x2 − 12x + 4y 2 = 36 y2 (x − 2)2 + = 1. 16 12 ⇐⇒ 3(x − 2)2 + 4y 2 = 48 ⇐⇒ #3(a) radius sqrt(480) 111111111111111111111 000000000000000000000 #3(b) 2+3i 4+2i #3(c) 3i 2+sqrt(12)i 4-3i 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 radius 10 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 10+0i 20+0i 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 (10+sqrt(480))+0i 111111111111111111111 000000000000000000000 -2+0i 1.75+0i 4+0i 6+0i -3i 0-3.5i 4-3i 2-sqrt(12)i 4. [4 marks] Let n be a positive integer. Prove that (1 + i2n )(1 + in ) is either 0 or 4. Describe the values of n that give the various answers. Answer: We have (1 + i2n )(1 + in ) = (1 + (−1)n )(1 + in ). By Proposition 8.23, we only need to check the congruence of n modulo 4 in order to determine the value of in . n≡0 n≡1 n≡2 n≡3 (mod 4) =⇒ (−1)n = 1 and in = 1 (mod 4) =⇒ (−1)n = −1 and in = i (mod 4) =⇒ (−1)n = 1 and in = −1 (mod 4) =⇒ (−1)n = −1 and in = −i =⇒ (1 + i2n )(1 + in ) = (2)(2) = 4 =⇒ (1 + i2n )(1 + in ) = (0)(1 + i) = 0 =⇒ (1 + i2n )(1 + in ) = (2)(0) = 0 =⇒ (1 + i2n )(1 + in ) = (0)(1 − i) = 0 Hence, (1 + i2n )(1 + in ) = 0 for n ≡ 1, 2, 3 (mod 4), and (1 + i2n )(1 + in ) = 4 for n ≡ 0 (mod 4). 5. Exercises from the text for practice; NOT to be submitted: Ex.Set.8, pp.218–221: 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 61. ...
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