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HOMEWORK ASSIGNMENT 3 ACCELERATED PROOFS AND PROBLEM SOLVING [MATH08071] Each problem will be marked out of 4 points. Exercise 1 ([1, Exercise 6.2]). Let u and v be two complex numbers. Prove that |u + v| 6 |u| + |v|. Solution. We have 2 |u + v| 6 |u| + |v| ⇐⇒ |u + v|2 6 |u| + |v| , since |u + v| and |u| + |v| are positive. On the other hand, we have |u + v|2 = (u + v)(ū + v̄) = uū + vv̄ + ūv + uv̄ = |u|2 + |v|2 + ūv + uv̄, and (|u| + |v|)2 = |u|2 + |v|2 + 2|u||v|. Thus, to prove that |u + v| 6 |u| + |v|, it is enough to prove that ūv + uv̄ 6 2|u||v| holds. But ūv + uv̄ = 2Re(uv̄). We know that Re(uv̄) = |uv̄| cos θ 6 |uv̄|| cos θ| 6 |uv̄|, where θ is an argument of uv̄. Then ūv + uv̄ = 2Re(uv̄) 6 2|u||v|, which exactly what we had to prove to prove that |u + v| 6 |u| + |v|. √ Exercise 2 ([1, Exercise 6.4]). (a) What is i? (b) Find all the tenth roots of i. Which one is nearest to i in the Agrand diagram. √ 7 (c) Find the seven roots of the equation x − 3 + i = 0. Which one of these roots is closest to the imaginary axis. √ √ Solution. What is i? We know that i is a complex number z such that z 2 = i. There are two such complex numbers. Let us find them. Put z = r(cos θ + i sin θ), where r = |z| and θ ∈ R such that 0 6 θ < 2π. Then z 2 = r2 cos(2θ) + i sin(2θ) = i = cos(π/2) + sin(π/2)i, which implies that r = 1 and θ = (π/2 + 2πk)/2 for any integer k by [1, Proposition 6.2]. Since we assume that 0 6 θ < 2π, we see that k is either 0 or 1. Then cos(π/4) + i sin(π/4) and cos(5π/4) + i sin(5π/4) are two possible solutions of z 2 = i. These are the same complex numbers as √ √ √ √ 2 2 2 2 +i and − −i , 2 2 2 2 √ √ which we can shortcut as ±( 2/2 + i 2/2). Now let us find all the tenth roots of i and find the one that is closest to i in the Agrand diagram. Let z be a tenth roots of i. Then z 10 = i. There are 10 such complex numbers. This assignment is due on Thursday 15th October 2015. 1 Let us find them. Put z = r(cos θ + i sin θ), where r = |z| and θ ∈ R such that 0 6 θ < 2π. Then z 10 = r10 cos(10θ) + i sin(10θ) = i = cos(π/2) + sin(π/2)i, which implies that r = 1 and π + 2πk π 2πk π + 4πk 1 + 4k θ= 2 = + = = π 10 20 10 20 20 for any integer k by [1, Proposition 6.2]. Since we assume that 0 6 θ < 2π, we see that k ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then ! ! ! ! ! ! 1 1 5 5 37 37 cos π + i sin π , cos π + i sin π , . . . , cos π + i sin π 20 20 20 20 20 20 are all possible solutions of z 10 = i (all tenth root of i). Which one of them is is closest to i in the Agrand diagram? The one that minimize the number |z − i|. This is the same root that minimize |z − i|2 . But |z − i|2 = (z − i)(z̄ + i) = z z̄ + zi − z̄i = 2 + 2Re(zi) = 2 − 2 sin θ, since z z̄ = 1 and Re(zi) = −Im(z) = − sin θ. But we have ) ( 5 9 37 1 π, π, π, . . . , π , θ∈ 20 20 20 20 so we can find the right θ just by plugging in all of them into 2 − 2 sin θ and choosing the one that makes 2 − 2 sin θ to be as small as possible. But one can find θ without doing this. The function sin(x) is increasing x ∈ (0, π/2), the function sin(x) is decreasing for x ∈ (π/2, π), we have sin(x) < 0 for x ∈ (π, 2π), and sin(π/2 + x) = sin(π/2 − x). Moreover, sin(x) = 1 only for x = π/2 if x ∈ [0, 2π]. The graph of 2 − 2 cos(x) is shown on the picture . Thus, to get the θ we want, we should minimize 1 + 4k π |θ − π/2| = π − , 20 2 2 which is quite easy. This gives us the answer 9 π, 20 θ= and the corresponding tenth root of i is cos(9π/20) + i sin(9π/20). √ Now let us find the seven roots of the equation x7 − 3 + i = 0 and to determine which√one of these roots is closest to the imaginary axis. Let z be a root of the equation x7 − 3 + i = 0. Put z = r(cos θ + i sin θ), where r = |z| and θ ∈ R such that 0 6 θ < 2π. Then √ z 7 = r7 cos(7θ) + i sin(7θ) = 3 − i = 2 cos(11π/6) + sin(11π/6)i , which implies that r = √ 7 2 and 11π 6 θ= + 2πk 11π + 12πk 11 + 12k = = π 7 42 42 for any integer k [1, Proposition 6.2]. Since we assumed earlier that 0 6 θ < 2π, we see that k ∈ {0, 1, 2, 3, 4, 5, 6}. Then √ 7 2 cos ! 11 π +i sin 42 11 π 42 √ 7 2 cos 47 π 42 ! √ 7 2 cos 71 π 42 ! !! , √ 7 ! 23 π +i sin 42 2 cos !! + i sin 47 π 42 !! + i sin 71 π 42 , , √ 7 √ 7 23 π 42 ! 2 cos 59 π 42 ! 2 cos 83 π 42 !! , √ 7 2 cos ! 35 π +i sin 42 !! + i sin 59 π 42 !! + i sin 83 π 42 , , √ are all possible solutions of x7 − 3 + i = 0. Which one of these complex numbers is closest to the imaginary axis. The one that minimizes the number | cos θ|. But we have ( θ∈ ) 11 23 35 47 59 71 83 π, π, π, π, π, π, π , 42 42 42 42 42 42 42 so we can find the θ with smallest | cos θ| just by plugging in all of them into | cos θ| and choosing the one that makes | cos θ| to be as small as possible. But one can find θ without doing this. The function | cos(x)| is decreasing for x ∈ (0, π/2), the function | cos(x)| is increasing for x ∈ (π/2, π), the function | cos(x)| is increasing for x ∈ (π/2, π), the function | cos(x)| is decreasing for x ∈ (π, 3π/2), the function | cos(x)| is increasing for x ∈ (3π/2, 2π), and we have two types of symmetries: | cos(π/2 + x)| = | cos(π/2 − x)| and | cos(3π/2 + x)| = | cos(3π/2 − x)|. Moreover, we known that cos(x) = 0 only for x = π/2 and x = 3π/2 if x ∈ [0, 2π]. The graph of | cos(x)| is shown on the picture 3 35 π 42 !! , . Thus, to find θ we should minimize |θ − π/2|, minimize |θ − 3π/2|, then chose the smallest between these two values and this gives us our θ. Minimizing π 12k − 10 π π 11 + 12k π− = π = 12k − 10, θ − = 42 2 42 2 42 we get k = 1. This gives us the first candidate θ1 = 23π/42. Minimizing 11 + 12k 12k − 52 3π 3π π π− π = 12k − 52, θ − = = 42 2 42 2 42 we get k = 4. This gives us the second candidate θ4 = 59π/42. Since π π 2π 3π < = θ4 − θ1 − = , 2 21 21 2 we see that the final answer here is given by θ1 . Thus, we see that ! !! √ 23 23 7 2 cos π + i sin π 42 42 √ is the solution of x7 − 3 + i = 0 that is closest to the imaginary axis. Exercise 3 √ ([1, Exercise 6.10]). Prove that there is no complex number z such that |z| = |z + i 5| = 1. √ Solution. Let z be any complex number. Applying Exercise 1 to −z and z + i 5, we see that √ √ √ √ √ √ |z| + |z + i 5| = | − z| + |z + i 5| > | − z + z + i 5| = |i 5| = 5 > 4 = 2 = 1 + 1, √ which implies that the equalities |z| = 1 and |z + i 5| = 1 can not √ be satisfied simultaneously. Thus, there is no complex number z such that |z| = |z + i 5| = 1. Exercise 4 ([1, Exercise 7.2]). Use the method for solving cubic equations given in [1, Chapter 7] to find the roots of the equation x3 − 6x2 + 13x − 12 = 0. 4 Now notice that 3 is one of the roots. Reconcile this with the roots you have found. Solution. Let us solve this problem other way around. We do not need to notice that 3 is one of the roots. Then we see that x3 − 6x2 + 13x − 12 = (x − 3)(x2 − 3x + 4) by dividing x3 − 6x2 + 13x − 12 by x − 3. Solving x2 − 3x + 4 = 0, √ √ we get two other roots, which are 3/2 + 7i/2 and 3/2 − 7i/2. Now let us apply the method for solving cubic equations given in [1, Chapter 7]. Put y = x − 2. Then x = y + 2 and y must be a root of the equation (y + 2)3 − 6(y + 2)2 + 13(y + 2) − 12 = 0, which can be rewritten as y 3 + y − 2 = 0. Then the formula in [1, Chapter 7] gives us r r 2√ 2√ 3 3 y = 1+ 21 + 1 − 21, 9 9 where we consider y is a real number here (and all cubes roots are real as well). Then The other two solutions are given by ! ! r r r r 2√ 2√ 2√ 2√ 3 3 3 3 2 2 1+ 21ω + 1 − 21 ω and 1+ 21ω + 1 − 21 ω, 9 9 9 9 where ω = cos(2π/3) + i sin(2π/3). Then x is one of these three complex numbers: r r 2√ 2√ 3 3 2+ 1+ 21 + 1 − 21, 9 9 ! ! r r √ √ 2 2 3 3 2+ 1+ 21 ω + 1− 21 ω 2 , 9 9 ! ! r r √ 2√ 2 3 3 2+ 1+ 21 ω 2 + 1− 21 ω, 9 9 √ √ which do not looks like 3, 3/2 + 7i/2, and 3/2 − 7i/2. But this is OK. Keeping in mind that √ √ 1 3 3 1 ω=− + i and ω = − − i, 2 2 2 2 one can show (similar to [1, Example 7.1]) that r r 2√ 2√ 3 3 2+ 1+ 21 + 1 − 21 = 3, 9 9 ! ! r r √ 2√ 2√ 3 7 3 3 2 2+ 1+ 21 ω + 1− 21 ω = + i , 9 9 2 2 ! ! r r √ √ 2√ 2 3 7 3 3 2 2+ 1+ 21 ω + 1− 21 ω = − i , 9 9 2 2 but this is tricky to see (it is easy to check this approximately using calculator). Let us use the method suggested by Tony Gilbert. Namely, arguing as in [1, Example 7.1], we may guess that s √ 21 3 √ 1±2 =p± q 9 5 for some rational numbers p and q (it is not obvious that such p and q exist). Since we believe that r r 2√ 2√ 3 3 2+ 1+ 21 + 1 − 21 = 3, 9 9 we see that p = 1/2. Now we can find q by taking cube of s √ 21 1 √ 3 1+2 = + q, 9 2 which gives us q = 7/12. Then we can deduce our reconciliation. Exercise 5 ([1, Exercise 7.6]). Let n be a positive integer. (a) Factorize x2n+1 − 1 as a product of real linear and quadratic polynomials. (b) Factorize x2n + x2n−1 + · · · + x2 + x + 1 as a product of real quadratic polynomials. (c) Let ω = e2πi/(2n+1) Show that 2n+1 X ω i+j = 0. i=1,j=1,i<j Solution. It follows from [1, Proposition 6.3] that x2n+1 − 1 = (x − ω)(x − ω 2 ) · · · (x − ω 2n+1 ) where ω = e2πi/(2n+1) . Since ω 2n+1 = 1, we see that ω 2n+1−k ω k = 1 for every k ∈ {1, . . . , n}. This implies that ω 2n+1−k is a complex conjugate of ω k for every k ∈ {1, . . . , n}. Then (x−ω 2n+1−k )(x−ω k ) = x2 −(ω 2n+1−k +ω k )x+ω 2n+1−k ω i = x2 −2Re(ω k )x+1 = x2 −2 cos(2πk/(2n+1))x+1) for every for every k ∈ {1, . . . , n}. Then x2n+1 − 1 can be factorized as (x−1) x2 −2 cos(2π/(2n+1))x+1) x2 −2 cos(4π/(2n+1))x+1) · · · x2 −2 cos(2nπ/(2n+1))x+1) , because ω 2n+1 = 1. Since x2n+1 − 1 = (x − 1) x2n + x2n−1 + · · · + x2 + x + 1 and we know the how to factorize x2n+1 − 1 as a product of real linear and quadratic polynomials, we see that x2n + x2n−1 · · · + x2 + x + 1 can be factorized as x2 −2 cos(2π/(2n+1))x+1) x2 −2 cos(4π/(2n+1))x+1) · · · x2 −2 cos(2nπ/(2n+1))x+1) . Finally, we have to show that 2n+1 X ω i+j = 0 i=1,j=1,i<j to complete the solution. Note that ω 1 , ω 2 , ω 3 , . . . , ω 2n+1 are all roots of the polynomial x2n+1 − 1 by [1, Proposition 6.3]. Since ω i+j = ω i ω j , we see that 2n+1 2n+1 X X ω i+j = ωiωj , i=1,j=1,i<j i=1,j=1,i<j 6 where the later sum is just the sum of all products of pairs of roots of the polynomial x2n+1 − 1. By [1, Proposition 7.1] this sum must be equal to the coefficients in front of x2n−1 in x2n+1 − 1, which is zero. Thus, the sum is zero. References [1] M. Liebeck, A concise introduction to pure mathematics Third edition (2010), CRC Press 7