Complex Analysis Homework 1

Transcription

1 Comlex Analysis Homework 1 Steve Clanton Sarah Crimi January 27, 2009 Problem Claim. If two integers can be exressed as the sum of two squares, then so can their roduct. Proof. Call the two squares that sum to the first integer a 2 and b 2 and the two squares that sum to the second c 2 and d 2. We rove that the roduct of these two integers is the sum of (ac bd) 2 and (ad + bc) 2 by considering the quantity (a + ib) (c + id) 2. First, we evaluate this quantity directly by erforming the multilication and taking the modulus of the resulting roduct: (a + ib) (c + id) 2 = (ac + aid + ibc + ibid) 2 = (ac bd) + i (ad + bc) 2 = (ac bd) 2 + (ad + bc) 2 We can also recall that the modulus of the roduct of comlex numbers will be the roduct of the modulus of each comlex number to show (a + ib) (c + id) 2 = (a + ib) 2 (c + id) 2 = ( a 2 + b 2) ( c 2 + d 2). Equating the right hand sides of these two equations gives us ( a 2 + b 2) ( c 2 + d 2) = (ac bd) 2 + (ad + bc) 2. 1

2 Problem 6 Part ii Claim. Given that satisfies the equation + 3 i = 2, the minimum of is 3 and occurs at i and the maximum is 7 and occurs at i. Proof. The set of that satisfy the equation + 3 i = 2 is the set of oints in the comlex lane that are two units distance from the oint 3 + i. The set of modulus is the distance of each of those oints from the origin. We can write 3 + i as 5e iθ for some θ since 3 + i = 5. Intuitively, we minimie by moving all two units directly toward the origin, that is by shifting 2e iθ. Since we start at distance from the origin and can move at most 2 units toward the origin, the minimum is 3. Thus, we can calculate the osition as 3 5 ( 3 + i) = i. Similarly, the maximum of the modulus will be 7 at 7 5 ( 3 + i) = i. 3 i 2 3 i r 2 rojr ' ' roj ' 5 rojr More formally, for contradiction, we assume that there is some other that satisfies the equation + 3 i = 2 that is not a scalar multile of 3 + i with 3. Consider the rojection of the line from 3 + i to onto the line from 3 + i to the origin. Since these two lines cannot be arallel, the length of the rojection is strictly less than 2. For the same reason, the length of the line from the origin to oint, which we know is the definition of, is strictly greater than its rojection onto the line from 3 + i wchich must be strictly greater than 3. 2

3 Problem 11 Question. Exlain geometrically why the locus of such that a arg = const. b is an arc of certain circle assing through the fixed oints a and b. Solution. We will start by finding an arbitrary solution with a geometric construction. Since the construction is not an obvious roof that the oint solves the equation, we must show that the solution is correct. Then, we will show why all solutions are on that articular arc. Afterward, we will generalie the result for all cases. b 0 a Figure 1: A articular solution In Figure 1, we find a solution for the case where b a 0 < arg < π and arg = θ < π. a b One angle is set by the fixed oints a and b, and another is determined by the constant θ. We allow one remaining angle to be any angle α between 0 and 2π arg ( a b θ. The final angle is set to β = 2π arg b a) θ α. We begin to construct the quadrilateral by drawing lines from 0 to oint a and from 0 to oint b. We finish the quadrilateral by drawing lines from a and b at resective interior angles α and β until they meet at some oint we will call. We also add the circle that is determined by the three oints a, b, and. b 0 a a 2 b Figure 2: The symmetry of Figure 1 and its reflection, shifted by We verify that solves our equation by reflecting this figure through the origin and shifting it by as in Figure 2. Now that we understand why solves our equation, we return to Figure 1. If we recall a fact from geometry, we can conclude that the set of for which ab = θ is the set of oints that lie on our circle. Points a and b divide the circle into two arcs, and the solution set is the arc that contains oint. We observe that the remaining arc rovides negative solutions to rovide solutions in the case where θ π. To solve when arg ( b a) π, we can solve the equivalent arg b a = θ. 3

4 Problem 12 1 i 1 i Since the condition Re +1+i = 0 is equivalent to arg +1+i = ± π 2, we use the result of exercise 11 to conclude that all solutions are on the circel formed by 1 + i, 1 i, and the articular solution 1 + i. Thus, the solution is = 1 i 1 i 1 i 2. The condition Im +1+i = 0 imlies that arg +1+i = 0 or arg +1+i = π. This is the secial case where the oints are collinear with the origin. Thus, the locus of is any multile of 1 + i. Problem 16 Question. Starting from the origin, go one unit east, then the same length north, then (1/2) of the revious length west, then (1/3) of the revious length south, then (1/) of the revious length east, and so on. What oint does this siral converge to? Solution. Note that the unit vectors east, north, west, and south corresond to 1, i, 1, and i, resectively. Just as the length is given as as a multile of the revious length, we may exress the direction we go as a 90 rotation left. Equivalently, we use multilication of the unit vector by i to change the direction. The osition after moving 2n times is i + ( i 2 + i 3 + i ! 1 (2n 1)! ) 2n 1 i 2n 1 = Although this looks like Maclaurin series for e i, we do not have the framework to assert that it is and will take one more ste to simlify: 2n 1 i k n 1 k! = ( 1) n ( 1 (2n)! + ) n 1 i ( 1) n = (2n + 1)! (2n)! + i n ( 1) n (2n + 1)! Thus, in the limit, the real art of the siral goes to cos (1) and the imaginary art to sin (1) i k k!

5 Problem 2 i To find the radius of convergance,we break into three cases: ii 1. > 1 2. = 1 3. < 1 If > 1, the series will diverge. This can be seen clearly from looking at the sum: Σ n = will get larger and larger as the series continues because > 1. If we think about the series in terms of the sirals that we have been seeing, each eice of the siral will be getting larger not smaller, so there will be more and more room between different arts of the series instead of less. If = 1, there are cases where the series does not diverge or converge. For instance, if = e iπ/, we will obtain a hexagon where the series continuously loos back on itself every 2π. Clearly, there is no single convergance oint here. Also, secifically if = 1, the bottom of the fraction n 1 1 blows u. If < 1, then our series will siral inward tighter and tighter around itself, so that as the series gets infinity large, we get infinitely close to a single oint. Hence, for all < 1 ther is some ositive integer N such that a P n (a) < ɛ for every value of n greater than N where a is what the series converges to. We break u n 1 into two fractions; 1 n of 1 is lim n n We look at the behavoir of 1 n = 0. This can be grased intuitively be realising that since < 1, 1, 2, 3,... n where < 1. The Limit 1 get rogressively smaller. Looking at the series for.5, we see the series is 1,.25,.125,.0625, Thus,as n goes to, n 1 1 = 1 1 which is 1 1. For 1 2 (1 + i), the oint of convergence is 1 1 (1/2 + 1/2i) = 1 + i. iii 5

6 Problem 3 We take advantage of relationshis e iθ e iθ = e i2π = 1 and e iθ + e iθ = 2 cos θ of comlex conjugates throughout. i Figure 3: Primitive 12th Roots of Unity ii Φ 12 () = ( e iπ/6) ( e i5π/6) ( e i7π/6) ( e i11π/6) iii Φ 12 () = ( e iπ/6) ( e i5π/6) ( e i7π/6) ( e i11π/6) = (( e iπ/6) ( e i11π/6)) (( e i5π/6) ( e i7π/6)) ( ( π ) = 2 2 cos + 1) ( ) 5π 2 2 cos ( = 2 ) ( ) = ( ) = iv Φ 8 () = ( e iπ/) ( e i3π/) ( e i5π/) ( e i7π/) = (( e iπ/) ( e i7π/)) (( e i3π/) ( e i5π/)) ( ( π ) = 2 2 cos + 1) ( ) 3π 2 2 cos + 1 ( = 2 ) ( ) = = + 1 6

7 i3 e i e e i5 i7 e Figure : Primitive 8th Roots of Unity v Claim. For any n > 2, the cyclotomic olynomial Φ n () has even degree and real coefficients. Proof. Let ζ denote an arbitrary rimitive n th root of unity. First, we will show that ζ = ζ 1 is also a rimitive root. Then, we will rove a cyclotomic olynomial with even degree has real coefficients. Last, we see that the olynomial with odd degree is a contradiction. We know ζζ = 1 2 = 1, because the modulus of any root of unity is 1. Since this is the definition of multilicative inverse, we have that ζ = ζ 1. By the definition of a rimitive root of unity, every n th root of unity can be written as ζ k for some 0 < k < n and ζ n = 1. Now, we can conclude that ζ is also a rimitive root by showing that an arbitrary root n th can be exressed as a ower of ζ: ζ k = ζ k 1 n k = ζ k ( ζζ ) n k = ζ k ζ n k ζ n k = ζ n ζ n k = ζ n k We choose n > 2 because otherwise we can consider reals number 1 and 1 with trivial conjugation as generators. Thus, the eroes of Φ n must be conjugate airs ζ ζ. Since the conjugates come in twos, we will have an even degree olynomial. For each conjugate air, we have ( (a + bi))( + (a bi))= 2 2a + a 2 + b 2. So, each conjugate air yeilds real numbers. The roduct of real numbers is a real number. So, the roduct of all of the conjugate airs is also a real number. vi Claim. If is a rime number, then Φ () = Proof. Thus far, we have seen that a 12gon insires rimitive roots at π 6, 5π 6, 7π 11π, and. Incidently, 1, 5, 7, are all relatively rime to 12. Similarly, an 8gon insires rimitive roots at π, 3π, 5π, and 7π and 8 is relatively rime to 1,3,5,and 7. So, it seems that the rimitive roots must be comosed of the numbers that are relatively π rime to n for all ngons. A way to see this is to realise that en will revolve around the circle if α is relatively rime to n allowing ζ, ζ 2...ζ n to generate the grou. Any rime number n has n 1 numbers that are relatively rime to kiπ. Thus, there are it; mainly: 1, 2,3...(n 1). These relatively rime numbers corresond to the k in P k+1 = e not simly (n 1) roots as is exected from equation (16), but also (n 1) rimitive roots. So, Φ n () must be of degree (n 1). Also, = 1 cannot be a generator because 1 n = 1 and with n > 2 there is at least one root that is not 1. Since = 1 does not generate all of the elements of the ngon, we must divide out ( 1). This looks a lot like n 1 1. We know that the series for n 1 1 = n 1. Hence, Φ () = n 1. 7