PUTNAM TRAINING, 2008 COMPLEX NUMBERS

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1 PUTNAM TRAINING, 008 COMPLEX NUMBERS (Last updated: December 11, 017) Remark. This is a list of exercises o Complex Numbers Miguel A. Lerma Exercises 1. Let m ad two itegers such that each ca be expressed as the sum of two perfect squares. Prove that m has this property as well. For istace 17 = 4 + 1, 13 = + 3, ad = 1 = Prove that si k = si si 1 si Show that if z is a complex umber such that z + 1/z = cos a, the for ay iteger, z + 1/z = cos a. 4. Factor p(z) = z 5 + z Fid a close-form expressio for si kπ. 6. Cosider a regular -go which is iscribed i a circle with radius 1. What is the product of the legths of all ( 1)/ diagoals of the polygo (this icludes the sides of the -go). 7. (Putam 1991, B) Suppose f ad g are o-costat, differetiable, real-valued fuctios o R. Furthermore, suppose that for each pair of real umbers x ad y f(x + y) = f(x) f(y) g(x) g(y) g(x + y) = f(x) g(y) + g(x) f(y) If f (0) = 0 prove that f(x) + g(x) = 1 for all x. 8. Give a circle of lights, exactly oe of which is iitially o, it is permitted to chage the state of a bulb provided that oe also chages the state of every dth bulb after it (where d is a divisor of strictly less tha ), provided that all /d bulbs were origially i the same state as oe aother. For what values of is it possible to tur all the bulbs o by makig a sequece of moves of this kid? 1

2 PUTNAM TRAINING, 008 COMPLEX NUMBERS 9. Suppose that a, b, u, v are real umbers for which av bu = 1. Prove that a + b + u + v + au + bv Let P be a regular polygo iscribed i a uit circle. Deote by S the set of all chords whose edpoits are vertices of P, ad let A be the average legth of all chords i S. Write a closed form expressio for A i terms of, ad compute lim A.

3 PUTNAM TRAINING, 008 COMPLEX NUMBERS 3 Hits 1. If m = a + b ad = c + d, the cosider the product z = (a + bi)(c + di) = (ac bd) + (ad + bc)i.. The left had side of the equality is the imagiary part of 3. What are the possible values of z? 4. If ω = e πi/3 the ω ad ω are two roots of p(z). 5. Write si t = (e ti e ti )/i. e ik. 6. Assume the vertices of the -go placed o the complex plae at the th roots of uity. 7. Look at the fuctio h(x) = f(x) + ig(x). 8. Assume the lights placed o the complex plae at the th roots of uity 1, ζ, ζ,..., ζ, where ζ = e πi/. 9. Let z 1 = a bi, z = u + vi. We have z 1 = a + b, z = u + v, R(z 1 z ) = au + bv, I(z 1 z ) = 1, ad must prove z 1 + z + R(z 1 z ) 3. { 10. The legth of a chord ca be expressed as si πk = I e πk }, i where k is a iteger, 0 k 1, ad I{z} = imagiary part of the complex umber z.

4 PUTNAM TRAINING, 008 COMPLEX NUMBERS 4 Solutios 1. If m = a + b ad = c + d, the cosider the product z = (a + bi)(c + di) = (ac bd) + (ad + bc)i. We have ad z = a + bi c + di = (a + b )(c + d ) = m, z = (ac bd) + (ad + bc), so m is also i fact a sum of two perfect squares.. The left had side of the equality is the imagiary part of e ik = ei(+1) 1 = ei(+1/) e i/ = cos ( + 1) cos 1 + i{si ( + 1) + si 1} e i 1 e i/ e i/ i si 1. The imagiary part of that expressio is cos 1 cos ( + 1 ) si 1 = si si 1 si We have that z = e ±ia, so z + 1/z = e ia + e ia = cos a, hece: z + 1/z = e ia + e ia = cos a. 4. Factorig a polyomial is easier to accomplish if we ca fid its roots. I this case we will look for roots that are roots of uity e kπi/ : p(e kπi/ ) = e 10kπi/ + e kπi/ + 1. The three terms of that expressio are complex umbers placed o the uit circle at the vertices of a equilateral triagle for = 3 ad k = 1,, so if ω = e kπi/3, the ω ad ω are roots of p(z), hece p(z) is divisible by (z ω)(z ω ) = z + z + 1. By log divisio we fid that the other factor is z 3 z + 1, hece: p(z) = (z + z + 1)(z 3 z + 1). 5. Write si t = (e ti e ti )/i ad cosider the polyomial We have: p(x) = (x e πik/ ). P = si kπ = e πik/ e πik/ i = e πi()/ (i) (e πik/ 1) = p(1). O the other had the roots of p(x) are all th roots of 1 except 1, so (x 1)p(x) = x 1, ad p(x) = x 1 x 1 = 1 + x + x + + x.

5 PUTNAM TRAINING, 008 COMPLEX NUMBERS 5 Cosequetly p(1) =, ad P =. 6. Assume the vertices of the -go placed o the complex plae at the th roots of uity 1, ζ, ζ,..., ζ, where ζ = e πi/. The the legth of the diagoal coectig vertices j ad k is ζ i ζ k, ad the desired product ca be writte P = ζ j ζ k. 0 j<k< By symmetry we obtai the same product if we replace the coditio j < k with k < j, ad multiplyig both expressios together we get: P = ζ j ζ k = ζ j 1 ζ k j. 0 j,k< j k 0 j,k< j k Note that ζ j = 1, ad for each k, r = k j takes all o-zero values from k + 1 to k. Sice ζ r = ζ r+ we may assume that r rages from 1 to 1, so we ca rewrite the product like this: ( P = 1 ζ ) r. Next cosider the polyomial r=1 p(x) = (x ζ r ). r=1 Its roots are the same roots of x 1 except 1, hece x 1 = (x 1)p(x) ad hece p(x) = x 1 x 1 = 1 + x + x + + x, (1 ζ r ) = p(1) =. r=1 cosequetly P =, ad P = /. 7. Defie h(x) = f(x) + ig(x). The h is differetiable ad h (0) = bi for some b R. The give equatios ca be reiterpret as h(x +y) = h(x)h(y). Differetiatig respect to y ad substitutig y = 0 we get h (x) = h(x)h (0) = bi h(x), so h(x) = Ce bix for some C C. From h(0 + 0) = h(0)h(0) we get C = C. If C = 0 the h = 0 ad f ad g would be costat, cotradictig the hypothesis. Thus C = 1. Fially, for ay x R, f(x) + g(x) = h(x) = e bix = Assume the lights placed o the complex plae at the th roots of uity 1, ζ, ζ,..., ζ, where ζ = e πi/. Without loss of geerality we may assume that the light at 1 is iitially o. Now, if d < is a divisor of ad the lights ζ a, ζ a+d, ζ a+d,..., ζ a+( 1)d

6 PUTNAM TRAINING, 008 COMPLEX NUMBERS 6 have the same state, the we ca chage the state of this /d lights. The sum of these is ( ) ( ) ζ a + ζ a+d + ζ a+d + + ζ a+( 1 ζ 1)d = ζ a 1 1 = ζ a = 0. 1 ζ d 1 ζ d So if we add up all the roots that are o, the sum will ever chage. The origial sum was 1, ad the goal is to get all the lights tured o. That sum will be 1 + ζ + ζ + + ζ = 1 ζ 1 ζ = 0 1. Hece we ca ever tur o all the lights. 9. Let z 1 = a bi, z = u + vi. The z 1 = a + b, z = u + v, R(z 1 z ) = au + bv, I(z 1 z ) = 1. O the other had: Now for ay real t, Hece z 1 z = R(z 1 z ) + I(z 1 z ) = R(z 1 z ) + 1. (t 3 + 1) 0 = 3t + 1 t 3 = 4t + 4 ( 3 t). ( z 1 + z ) 4 z 1 z = 4 ( R(z 1 z ) + 1 ) ( 3 R(z 1 z ) ). So, z 1 + z 3 R(z 1 z ). Or z 1 + z + R(z 1 z ) 3, as required. 10. Name the vertices cosecutively V 0,..., V. The legth of the chord joiig vertices V 0 ad V k is si πk { } = I e πk i, where I{z} = imagiary part of the complex umber z. So, the sum of all chords with a edpoit i vertex V 0 is (icludig the zero term k = 0 for coveiece): { { } } I e πk i = I e πk i { } e πi 1 = I e π i 1 { } = I e π i 1 { } e π i = I e π i e π i { (cos π = I + i si π ) } i si π = cos π si π.

7 PUTNAM TRAINING, 008 COMPLEX NUMBERS 7 Repeatig the same computatio for all chords with a edpoit i V 1, the V, ad so o util V, we get the same expressio times. Addig ad takig ito accout that each chord has bee couted twice, we get that the sum of the legths of all chords is cos π si π. Sice there are () chords, the average is A = cos π 1 si π. For the limit we ca use L Hôpital after replacig with x = 1 0: lim A cos π = lim 1 si π x = lim x 0 = lim x 0 = lim 1 x x si πx x 0 π = 4 π. cos πx cos πx si πx

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