MATH : HONORS CALCULUS-3 HOMEWORK 6: SOLUTIONS

Transcription

1 MATH : HONORS CALCULUS-3 HOMEWORK 6: SOLUTIONS 25-1 Find the absolute value and argument(s) of each of the following. (ii) (3 + 4i) 1 (iv) i (ii) Put z = 3 + 4i. From z 1 z = 1, we have z 1 z = 1 and arg(z 1 ) + arg(z) = 0: z 1 = 1 z = = 1 5. arg(z 1 ) = arg(z) = arctan 4 3 (iv) Let ω be the real seventh root of 3 + 4i. From ω 7 = 3 + 4i, we have ω 7 = 5 and 7 arg(ω) = tan : ω = 7 5, arg(ω) = 1 7 arctan 4 3. There are seven 7th roots of 3 + 4i, each separated by θ = 2π/7, so for = 0, 1,..., 6. arg(z) = 1 7 arctan π Describe the set of all complex numbers z such that (ii) z = z 1 (iv) z a + z b = c (ii) Note that z 0, so z = z 1 if and only if zz = z 1 z if and only if z = 1. Hence the set is the complex unit circle. (iv) If c / R 0, there is no solution. If b a > c, then c < b a b z + z a, so there is no solution. If c = b a, then the set is the line segment between a and b. If c = 0, there is a solution if and only if a = b, in which case, the only solution is z = a. Now assume b a < c. The curve formed by the set is such that the sum of the distances to a and to b is constant for every point on the curve, so by definition the curve is an ellipse with focal points a and b Prove that if a 0,..., a n 1 are real and a + bi (for a and b real) satisfies the equation z n + a n 1 z n a 0 = 0, then a bi also satisfies this equation. (Thus the nonreal roots of such an equation always occur in pairs, and the number of such roots is even). Conclude that z n + a n 1 z n a 0 is divisible by z 2 2az + (a 2 + b 2 ). 1

2 Let σ : C C be defined by x + iy x iy. Put α = a + ib and β = x + iy, and note that σ(α + β) = σ((a + x) + i(b + y)) = (a + x) i(b + y) = (a ib) + (x iy) = σ(α) + σ(β), and σ(αβ) = σ((ax by) + i(bx + ay)) = (ax by) i(bx + ay) = (a ib)(x iy) = σ(α)σ(β). It follows that if z = a + bi satisfies z n + a n 1 z n a 0 = 0, then 0 = σ(0) = σ(z n + a n 1 z n a 0 ) = σ(z n ) + σ(a n 1 z n 1 ) + + σ(a 0 ) = σ(z) n + a n 1 σ(z) n a 0, as desired. The polynomial has roots a + bi and a bi and hence has factor (z (a + bi))(z + (a bi)) = z 2 2az + (a 2 + b 2 ) Prove that if ω is an nth root of 1, then so is ω. A number ω is called a primitive nth root of 1 if {1, ω, ω 2,..., ω n 1 } is the set of all nth roots of 1. How many primitive nth roots of 1 are there for n = 3, 4, 5, 9? Let ω be an nth root of 1, with ω 1. Prove that n 1 =0 ω = 0. If ω n = 1, then (ω ) n = (ω n ) = 1 = 1. Let ζ n = cos 2π n + i sin 2π n ; the n roots of unity are {1, ζ n,..., ζn n 1 }. We prove that ζn is a primitive nth root of 1 if and only if gcd(, n) = 1. Note that a root of unity appears in {1, ζ n, ζ 2 n, ζ 3 n,..., ζ (n 1) } twice if and only if m 1 m 2 (mod n) for some 0 m 1 < m 2 n 1. That is, if and only if (m 1 m 2 ) n for some 0 m 1 < m 2 n 1. This is equivalent to m n for some 1 < m < n 1. Since m < 1, this is only possible if and only if and n share some factor. This completes the proof. Let ϕ(n) be the number of positive integers less than n that are co-prime to n. Then ϕ(3) = 2, ϕ(4) = 3, ϕ(5) = 4, and ϕ(9) = 6. Z Z 2

3 Using the geometric sequence formula (since ω 1), n 1 ω = ωn 1 ω 1 = 1 1 ω 1 = 0. = Prove that if z 1,..., z lie on one side of some straight line through 0, then z z 0. Show further that z 1 1,..., z 1 + z 1 0. all lie on one side of a straight line through 0, so that z If the line is the real axis, then either the imaginary parts of the z i are all positive or all negative, so that the sum of the z i also has the property. In particular, z z 0. Now suppose the line is not the real axis, and let θ be the angle between the positive real line and the ray extending into the imaginary part of the plane. Let ω be the complex number with modulus 1 and argument θ. Then {z 1 ω 1,..., z n ω 1 } is a subset of either the upper half or the lower half of the plane. By the previous argument we are done. If the line is the real axis, then I(z i ) and I(z 1 i ) have opposite signs, so in particular, the imaginary part of the sum is not zero. If the line is not the real line, then one can proceed as in part by noting that the inverses all lie on one side of the reflection of the line across the real line Prove that if a 0,..., a n 1 are any complex numbers, then there are complex numbers z 1,..., z n (not necessarily distinct) such that z n + a n 1 z n a 0 = n (z z i ). Prove that if a 0,..., a n 1 are real, then z n + a n 1 z n a 0 can be written as the product of linear factors z + a and quadratic factors z 2 + az + b all of whose coefficients are real. We induct on n. For n = 1 the statement is clear. Suppose it were true for some 1, and let a 0,..., a be any complex numbers. By the Fundamental Theorem of Algebra, the polynomial z +1 + a z + + a 0 has a root, say z +1. But then z +1 + a z + + a 0 z z +1 is a monic polynomial of degree. By the inductive hypothesis, for some z 1,..., z. It follows that as desired. z +1 + a z + + a 0 z z +1 = (z z i ) +1 z +1 + a z + + a 0 = (z z i ), 3

4 By part, the polynomial can be written as n (z z i). If any of the z i have non-zero imaginary part, then Problem 25-7 implies z i {z 1,..., z n }; consequently, the corresponding linear factors can be multiplied to produce a quadratic factor with real coefficients, as desired Let A be a set of complex numbers. A number z is called, as in the real case, a limit point of the set A if for every (real) ε > 0, there is a point a in A with z a < ε but z a. Prove the two-dimensional version of the Bolzano-Weierstrass Theorem: If A is an infinite subset of [a, b] [c, d], then A has a limit point in [a, b] [c, d]. Prove that a continuous (complex-valued) function on [a, b] [c, d] is bounded on [a, b] [c, d]. Prove that if f is a real-valued continuous function on [a, b] [c, d], then f taes on a maximum and minimum value on [a, b] [c, d] Let A 1 = [a, b] [c, d]. For n 1, let A 2n denote either the closed left or closed right half of A 2n 1 that contains infinitely many points, and let A 2n+1 denote either the closed upper half or closed lower half of A 2n that contains infinitely many points. For each positive integer n, let z n be some point in A n A. Then {x n } is Cauchy: for any ε > 0 there is a large N such that the points {z n } n>n = {(x n, y n )} n>n are contained in a rectangle with length and width less than ε. Hence z n z for some z [a, b] [c, d]; by construction, z is a limit point of A. If f is not bounded above, then for each n N there exists a point x n such that f(x n ) > n; the x n may be chosen so that { f(x n ) } n=1 is non-decreasing. By Bolzano-Weierstrass, there is a subsequence {x n } =1 converging to some x [a, b] [c, d], and by continuity, f(x n ) f(x). But { f(x n ) } n=1 has no convergent subsequence by construction, a contradiction. By part, the range of f is non-empty and bounded above; in particular, sup f exists. Call it M. Now for each n N, there exists a point z n [a, b] [c, d] such that M 1/n f(z n ) M. By the Squeeze Theorem, f(z n ) M. The z n are bounded, and by part, contain a convergent subsequence z n z; by continuity, f(z) = f( lim z n ) = lim f(z n ) = M. Hence the maximum is attained. Now let g = f; the minimum of f is the maximum of g, which is attained by the previous argument Let f(z) = (z z 1 ) m1 (z z ) m for m 1,..., m > 0. Show that f (z) = (z z 1 ) m1 (z z ) m m α(z z α ) 1. Let g(z) = m α(z z α ) 1. Show that if g(z) = 0, then z 1,..., z cannot all lie on the same side of a straight line through z. A subset K of the plane is convex if K contains the line segment joining any two points in it. For any set A, there is a smallest convex set containing it, which is called the convex hull of A; if a point P is not in the convex hull of A, then all of A is contained on one side of some straight line through P. Using this information, prove that the roots of f (z) = 0 lie within the convex hull of the set {z 1,..., z }. 4

5 By the product rule, f [ (z) = (z z1 ) m1 m α (z z α ) mα 1 (z z ) m ] = (z z 1 ) m1 (z z ) m m α (z z α ) 1 If z 1,..., z lied on the same side of a straight line through z, then z z 1,..., z z lie on the same side of a straight line through 0. By the contrapositive of Problem 25-11, if z 1,..., z lie on one side of some straight line through 0 and z z 1 = 0, then z1 1,..., z 1 all do not lie on the same side of a straight line through 0. By 25-7, (z z α ) 1 0. Since the m α > 0, we have a contradiction. m α (z z α ) 1 0, Let A be the convex hull of {z 1,..., z }, and note that the z i are in A. If a root z of g was not in the convex hull, then all of A is contained on one side of some straight line through z; this contradicts the result in. 5