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Oxford University Mathematical Institute Introduction to Complex Numbers, Michaelmas Term 2016: Specimen Solutions to Exercise Sheet Q1. Which of the following quadratic equations have real roots, which do not? 3x2 + 2x − 1 = 0; 2x2 − 6x + 9 = 0; −4x2 + 7x − 9 = 0; 4x2 − 9x + 5 = 0. A1. Let D be the discriminant of the equation. For the first equation D = 22 − 4 × 3 × −1 = 16. Since 16 > 0 the equation has real roots. In fact , of course, 3x2 + 2x − 1 = (3x − 1)(x + 1) and so the roots are −1, 1/3. For the second equation D = (−6)2 − 4 × 2 × 9 = −36, which is negative, so 2x2 − 6x + 9 = 0 has no real roots. For the third equation D = 49 − 144 < 0 so there are no real roots. And for the fourth equation D = 1, so it has two real roots: and in fact 4x2 − 9x + 5 = (x − 1)(4x − 5) so the roots are 1, 5/4. Q2. Write a careful proof of the theorem that if z, w ∈ C then z + w = z + w and z w = z w. A2. Write z = a + bi and w = c + di. Then z + w = (a + c) + (b + d)i = (a + c) − (b + d)i = (a − bi) + (c − di) = z + w and z w = (ac − bd) + (ad + bc)i = (ac − bd) − (ad + bc)i; z w = (a − bi)(c − di) = (ac − bd) − (ad + bc)i , and so z w = z w, as the theorem states. Q3. Put each of the following complex numbers into standard form a + bi: √ 7 − 2i (1 + 2i)(3 − i); (2 + i)(1 − 2i); (1 + i)4 ; (1 − 3i)3 ; ; 5 + 12i A3. i . 1−i (1 + 2i)(3 − i) = 5 + 5i ; (2 + i)(1 − 2i) = 4 − 3i ; (1 + i)4 √ (1 − 3i)3 7 − 2i 5 + 12i i 1−i = ((1 + i)2 )2 = (2i)2 = −4 ; √ √ √ √ = (1 − 3i)2 (1 − 3i) = (−2 − 2 3i)(1 − 3i) = −8 ; (7 − 2i)(5 − 12i) 11 94 = = − i; 169 169 169 i(1 + i) 1 1 = = − + i. 2 2 2 Q4. Find the modulus and argument of each of the following complex numbers: √ 1 + 3i; (2 + i)(3 − i); (1 + i)5 . A4. √ √ |1 + 3i| = 1 + 3 = 2 , p √ |(2 + i)(3 − i)| = (4 + 1)(9 + 1) = 5 2 , √ √ |(1 + i)5 | = ( 2)5 = 4 2 , arg(1 + √ 3i) = tan−1 √ 3 = π/3 ; arg((2 + i)(3 − i)) = arg(7 + i) = tan−1 17 ; arg(1 + i)5 = 5 arg(1 + i) = 1 5π 4 . Q5. On separate Argand diagrams sketch each of the following subsets of C: A := {z |z| < 1}; B := {z Rez = 3}; C := {z − 14 π < arg z < 14 π}; D := {z arg(z − i) = 12 π}; E := {z |z − 3 − 4i| = 5}; F := {z |z − 1| = |z − i|} . A5. Figure A: circular disc, not including boundary circle Figure B: line “x = 3” Figure C: right-angled sector not including 0 or boundary rays Figure D: “vertical” ray from i, not including point i Figure E: circle with centre at 3 + 4i, radius 5 Figure F : line “x = y” Q6. For each of the following complex numbers w, what transformation of the Argand diagram does multiplication by w represent? i; (1 + i); (1 − i); (3 + 4i). A6. Multiplication by i is a rotation about 0 anticlockwise through angle multiplication by (1 + i) is scaling (magnification) by a factor through angle 14 π about 0; √ 1 2 π; 2 and rotation anticlockwise multiplication by (1 − i) is scaling (magnification) by a factor through angle 14 π about 0; √ 2 and rotation clockwise multiplication by (3 + 4i) is scaling (magnification) by a factor 5 and rotation anticlockwise through angle tan−1 34 about 0. 2 Q7. Use De Moivre’s Theorem to show that if θ ∈ R then cos 5θ = 16 cos5 θ − 20 cos3 θ + 5 cos θ; sin 5θ = (16 cos4 θ − 12 cos2 θ + 1) sin θ . A7. By De Moivre’s Theorem and a small case of the Binomial Theorem, cos 5θ + i sin 5θ = (cos θ + i sin θ)5 = cos5 θ + 5i cos4 θ sin θ − 10 cos3 θ sin2 θ − 10i cos2 θ sin3 θ + 5 cos θ sin4 θ + i sin5 θ . Focussing on real parts we see that cos 5θ = cos5 θ − 10 cos3 θ sin2 θ + 5 cos θ sin4 θ = cos5 θ − 10 cos3 θ (1 − cos2 θ) + 5 cos θ (1 − cos2 θ)2 = 16 cos5 θ − 20 cos3 θ + 5 cos θ , as required. And from the imaginary parts we find that sin 5θ = 5 cos4 θ sin θ − 10 cos2 θ sin3 θ + sin5 θ = 5 cos4 θ − 10 cos2 θ (1 − cos2 θ) + (1 − cos2 θ)2 sin θ = (16 cos4 θ − 12 cos2 θ + 1) sin θ , also as required. Q8. Write down the primitive 6th roots of unity and the primitive 8th roots of unity in standard form a + bi. 2 (which is −ω ), where ω is a primitive A8. The primitive 6th roots of unity √ are −ω and −ω √ cube-root of 1. Therefore they are 12 + 23 i and 12 − 23 i. The primitive 8th roots of unity are the four complex numbers ± √12 ± √12 i. (Note that these all have absolute value 1 and their arguments are π/4, 3π/4, 5π/4, 7π/4, respectively.) Q9. Is 3 + 4i a root of unity? What about 3 5 + 45 i ? A9. No, 3 + 4i is not a root of unity. Its absolute value is 5, so it hasn’t a hope. And no, 35 + 45 i is not a root of unity either, even although its absolute value is 1. There are many ways to see this. Here is one. Consider complex numbers of the form (3 + 4i) + 5(a + bi), where a, b are integers. Notice that ((3 + 4i) + 5(a + bi)) × ((3 + 4i) + 5(c + di)) = ((3 + 5a)(3 + 5c) − (4 + 5b)(4 + 5d)) + ((3 + 5a)(4 + 5d) + (4 + 5b)(3 + 5c))i = (3 + 4i) + 5(A + B i) , where A := −2 + 3a + 3c + 5ac − 4b − 4d − 5bd and B := 4 + 4a + 3d + 5ad + 3b + 4c + 5bc . Thus the product of two numbers of this special form is again of our special form. Therefore 3 4 n + i = 51n ((3 + 4i) + 5(an + bn i)) , 5 5 where an , bn are integers. Clearly, 5n never divides 3 + 5an if n > 1, and therefore no power of 3 4 3 4 5 + 5 i is 1. That is, 5 + 5 i is not a root of unity. 3 Q10. Let φ := cos(2π/5) + i sin(2π/5), a primitive 5th root of 1. Define α := φ + φ4 , β := φ2 + φ3 . (i) Show that α and β are real numbers. (ii) Show that α + β = −1 and αβ = −1 (so that α, β are the roots of x2 + x − 1 = 0). √ (iii) Deduce that cos(2π/5) = 14 ( 5 − 1). A10. (i) Since cos(2π − x) = cos x and sin(2π − x) = − sin x we find that φ = cos(2π/5) − i sin(2π/5) = cos(8π/5) + i sin(8π/5) = φ4 . 4 Therefore α = φ + φ = φ4 + φ = α, so α is a real number. A similar calculation shows that β ∈ R. (ii) Since φ5 − 1 = 0 and φ5 − 1 = (φ − 1)(φ4 + φ3 + φ2 + φ + 1) we have φ4 + φ3 + φ2 + φ + 1 = 0 (since φ − 1 6= 0). Thus α + β = φ + φ4 + φ2 + φ3 = −1. Similarly, αβ = φ3 + φ4 + φ6 + φ7 = φ3 + φ4 + φ + φ2 = −1, as required. (iii) Now α, β satisfy the equation (x−α)(x−β) = 0, that is, x2 −(α+β)x+αβ √ = 0, which we know to be x2 + x − 1 = 0. Completing the square we see that α and β are 21 (−1 ± 5), only one of √ √ which is positive. Therefore 2 cos(2π/5) = φ+φ = α = 12 (−1+ 5), that is, cos(2π/5) = 14 ( 5−1). Q11. Find the square roots of −7 + 24i. Now solve the equation z 2 − (2 + 2i)z + (7 − 22i) = 0. A11. Let z = reiα and suppose that z 2 = −7 + 24i.√ Then r2 = √ |−7 + 24i| and 2α is one of the possible values of arg(−7 + 24i). Now |−7 + 24i| = 72 + 242 = 625 = 25 and therefore r = 5. 2t Also, tan(2α) = −24/7. If we write t := tan α then − 24 7 = tan(2α) = 1−t2 . This leads to the quadratic equation 24t2 − 24 = 14t or 12t2 − 7t − 12 = 0. Completing the square we find that t = 4/3 or t = −3/4. Thus z = ±5ei τ where τ := tan−1 43 , that is, z = ±(3 + 4i). To solve the equation z 2 − (2 + 2i)z + (7 − 22i) = 0 we again complete the square (rather than citing from memory—imperfect in my case—the standard formula). It may be rewritten (z − (1 + i))2 − (1 + i)2 + (7 − 22i) = 0, that is, (z − (1 + i))2 = −7 + 24i. We have just found the two square roots of −7 + 24i: thus z = (1 + i) ± (3 + 4i), that is, z = 4 + 5i or z = −2 − 3i. Q12. Show that σ 8 = −1 if and only if σ is a primitive 16th root of 1. Find the roots of the equation x8 = −1, and use them to write x8 + 1 as the product of four quadratic factors with real coefficients. A12. Since arg(−1) = π + 2k π for any integer k, from Euler’s formula we know that −1 = cos(π + 2k π) + i sin(π + 2k π) = e(1+2k)π i . Then σ 8 = −1 if and only if |σ| = 1 and 1+2k 8 arg σ = (1 + 2k)π for some integer k, that is, σ = e 8 π i . The primitive 16th roots of 1, however, m are the complex numbers e 16 2π i for integers m that are co-prime with 16 (and, without loss of generality, < 16). These are precisely the odd integers 1 + 2k < 16, and therefore σ 8 = −1 if and only if σ is a primitive 16th root of 1; moreover, the roots of the equation x8 = −1 are ψ m where ψ = eπ i/8 and m is one of 1, 3, 5, 7, 9, 11, 13, 15. From the Factor Theorem it follows that 7 3 Y Y 2k+1 ) , x8 + 1 = (x − ψ 2k+1 ) = (x − ψ 2k+1 )(x − ψ k=0 k=0 m 2k+1 since ψ 16−m = ψ −m = ψ . Now (x − ψ 2k+1 )(x − ψ ) = x2 − 2 cos( 2k+1 8 π)x + 1 and so x8 +1 = x2 − 2 cos( 18 π)x + 1 x2 − 2 cos( 38 π)x + 1 x2 − 2 cos( 58 π)x + 1 x2 − 2 cos( 78 π)x + 1 , a product of four quadratic factors with real coefficients. ΠMN: Queen’s and Mathematical Institute: October 2016 4