Lecture 11 Complex Numbers

Lecture 11 Complex Numbers Holde Lee /19/11 1 Itroductio 1.1 Defiitios The set of real umbers R from the familiar itegers 0, 1, 1,,,... to ratioal umbers 1, 3,... to irratioal umbers, π, e are familiar to us ad have a clear place i the world aroud us. But math is ot quite complete without the complex umbers. I the reals there is o solutio to the polyomial equatio x = 1, as the square of ay real umber is positive. However, we ca create a solutio: let s adjoi a umber to R whose square is 1. Call it i, ad let the resultig umber system be C. Sice we wat additio ad multiplicatio to work out, we have to iclude i C all umbers of the form z = a + bi, where a ad b are real. Defiitio 1.1: The set of umbers i the form a + bi, where a ad b are real, is called the complex umbers C. We call a the real part of z ad b the imagiary part, deoted by R(z = Re(z ad I(z = Im(z, respectively. A good way to imagie complex umbers is to thik of them as poits o a plae (the complex plae, by graphig a + bi as the poit (a, b. The x-axis cosists of the real umbers a R, while the y-axis cosists of the imagiary umbers bi, b R. However, the complex umbers have much more structure tha just ordered pairs of real umbers for istace, we ca multiply complex umbers as we show below. Let s see how basic operatios work i C. To add or subtract complex umbers, we simply add the real ad imagiary compoets: (1+i (3 5i = (1 3i+( ( 5i = + 7i. We multiply complex umbers by the distributive law, ad recall that i = 1: for example (1 + i(3 5i = 1 3 + 1 5i + i 3 + i 5i = 3 + 6i 10i = 13 + 6i. To defie divisio, we eed the followig: Defiitio 1.: The cojugate of a complex umber z = a + bi, where a, b are real, is z = a bi. Note that the product of a complex umber ad its cojugate is always real: (a + bi(a bi = a (bi = a + b. This allows us to divide complex umbers: to evaluate a+bi we multiply both the umerator ad the deomiator by the complex cojugate of c + di, c di. For c+di example, 1 + i 3 5i = (1 + i(3 + 5i (3 5i(3 + 5i = 3 + 11i + 10i 3 + 5 = 7 34 + 11 34 i.

Note that complex cojugatio preserves additio ad multiplicatio, i.e. z 1 + z = z 1 + z ad z 1 z = z 1 z. Ideed, let z 1 = a + bi ad z = c + di. The z 1 + z = (a bi + (c di = (a + c (b + di = z 1 + z. z 1 z = (a bi(c di = (ac bd (ad + bci = (ac bd + (ad + bci = z 1 z. The absolute value of z = a + bi is defied by z = a + b. It is the distace from z to 0 whe we plot z i the complex plae. Now egative umbers have square roots too, sice ( ai = a. With this, we ca solve ay quadratic equatio by completig the square or usig the quadratic formula. But what about geeral polyomials equatios? Do we have to adjoi more elemets to C so that every cubic polyomial ax 3 + bx + cx + d = 0 where a, b, c, d C, has a solutio? Due to the followig miracle, we do ot: Theorem 1.3 (Fudametal Theorem of Algebra: Every ocostat polyomial with coefficiets i C has a zero i C. (See lecture 8 for a proof. This shows that C is a very atural umber system to do algebra over. 1. Motivatios So what are complex umbers good for? Who cares about polyomial equatios that ca t be solved over the reals after all, complex solutios do t make sese i real life, right? Complex umbers were first developed to solve cubic polyomial equatios the formula higed o the use of complex umbers eve whe the solutios of the equatio were real. At first mathematicias were hesitat to adopt complex umbers, but complex umbers soo proved their use i a variety of other ways. A math problem (for example, a differetial equatio might require as a itermediate step to solve a polyomial equatio that may ot have real solutios. But if we kow how to work with complex umbers, we ca proceed just as we would i the real case, ad the simplified aswer may i fact ot have complex umbers at all, but would be hard to obtai otherwise. Calculus over the complex umbers turs be much icer so that it is ofte advatageous to exted the domai of real fuctios to the complex umbers, ad the look at their properties. This is the case with the ifamous zeta fuctio, which has applicatios to umber theory. May problems that do t look like they ivolve complex umbers tur out to be much easier if we use them. For istace, to fid Pythagorea triples we wat to solve a + b = c over the itegers. We may rewrite this as c b = a ad factor as (c b(c + b = a, ad the use umber theory. Suppose we wat to do the same with a 3 + b 3 = c 3.

(Thik Fermat s Last Theorem! We rewrite this as c 3 b 3 = a 3 ad factor (c b(c + cb + b = a 3. However, the left side is ot completely factored. If, however, we were workig over the complex umbers, we could factor this further: (c b(c ωb(c ω b = a 3 for some complex umber ω. The we could solve this by usig umber theory but over complex umbers istead (this is part of what is called algebraic umber theory. 1.3 Examples Let s warm up with some problems. Example 1.4: Fid the sum Solutio. We calculate 1 + i + i + i 3 + + i 011. 1 = 1, i = i, i = 1, i 3 = i, i 4 = 1. Hece the powers of i cycle through 1, i, 1, ad i. So we group the sum ito four terms at a time (1 + i + i + i 3 + (i 4 + i 5 + i 6 + i 7 + + (i 008 + i 009 + i 010 + i 011 ad ote that the sum i each group is 1 + i 1 i = 0. Hece the aswer is 0. Example 1.5: Show that if f is a polyomial with real coefficiets ad f(z = 0, the f( z = 0. Coclude that the roots of f (with multiplicity ca be grouped ito complex cojugate pairs. Proof. Let f(z = a z + +a 1 z+a 0. Usig the fact that cojugatio preserves additio ad multiplicatio, f( z = a z + + a 1 z + a 0 = a z + + a 1 z + a 0 = a z + + a 1 z + a 0 sice a m real = a z + + a 1 z + a 0 = f(z. Thus if f(z = 0 the f( z = 0. Thus the oreal roots of f come i complex cojugate pairs. Example 1.6 (AIME1 009/: There is a complex umber z with imagiary part 164 ad a positive iteger such that Fid. z z + = 4i. 3

Solutio. We ca write z = a + 164i where a is real. The Clearig the deomiator gives a + 164i (a + + 164i = 4i. a + 164i = 656 + 4(a + i. Hece matchig the real ad imagiary parts, a = 656 ad 164 = 4(a+ = 4( 656+. We get 41 = 656 + or = 697. 1.4 Problem Set 1 1. [1] Simplify a+bi, where a, b, c, d are real ad c + di 0. c+di. [1] Factor x + y ad x 3 + y 3 over the complex umbers. 3. [1] Fid i 4017. 4. [1.5] (AMC 1B, 004/16 A fuctio f is defied by f(z = i z. How may values of z satisfy z = 5 ad f(z = z? 5. [1.5] a ad b are complex umbers such that x a + b = x a + x b is a idetity, true for all values of x. Fid all possible values of a b. 6. [1.5] (AMC 1A, 007/18 The polyomial f(x = x 4 + ax 3 + bx + cx + d has real coefficiets, ad f(i = f( + i = 0. What is a + b + c + d? 7. [1.5] (AMC 1B, 008/19 A fuctio f is defied by f(z = (4 + iz + αz + γ for all complex umbers z, where α ad γ are complex umbers ad i = 1. Suppose that f(1 ad f(i are both real. What is the smallest possible value of α + γ? 8. [1.5] (AIME1 007/3 The complex umber z is equal to 9+bi, where b is a positive real umber. Give that the imagiary parts of z ad z 3 are equal, fid b. 9. [] (AIME1 00/1 Let F (z = z+i for all complex umbers z i, ad let z i z = F (z 1 for all positive itegers. Give that z 0 = 1 + i ad z 137 00 = a + bi, where a ad b are real, fid a + b. 10. [.5] Let Λ = {a + bi a, b Z, (a, b 0}. Evaluate λ Λ 1 λ 6. (Note: You may assume the sum is absolutely coverget, i.e. well-defied. 11. [4] (USAMO 1989/3 Let P (z = z + c 1 z 1 + + c be a polyomial i z, with real coefficiets c k. Suppose that P (i < 1. Prove that there exist real umbers a ad b such that P (a + bi = 0 ad (a + b + a < 4b + 1. 4

Polar form.1 Basic Facts ad Examples Cosider the complex plae. Whe we write a complex umber as a + bi, we graph the poit by goig horizotally a uits ad vertically b uits. This is called the rectagular form of the complex umber. Additio of complex umbers correspods icely to additio of vectors. Cojugatio correspods to reflectio over the x-axis (real axis. But what does multiplicatio ad divisio? To aswer this questio, we eed to write our complex umbers i a differet form. We could istead specify a locatio z o the complex plae by the distace r from 0, ad the agle θ made with the positive real axis. Drawig a triagle, we see that z = r(cos θ + i si θ which we abbreviate as z = r cis θ. This is called the polar form, z is called the modulus, ad θ is called the argumet. Let s try to multiply complex umbers i this form. Let z 1 = r 1 cis α ad z = r cis β. The usig the additio idetities for sie ad cosie, z 1 z = r 1 r (cos α + i si α(cos β + i si β We state this as a theorem. Theorem.1 (De Moivre: I particular, = r 1 r [(cos α cos β si α si β + i(cos α si β + cos β si α] = r 1 r [cos(α + β + i si(α + β]. (r 1 cis α(r cis β = r 1 r cis(α + β. (r cis θ = r cis(θ. Usig this theorem we ca fid th roots of a complex umber z, i.e. fid the solutios to x = z. Writig z = r cis θ ad x = s cis φ, this is equivalet to (s cis φ = r cis θ. Usig De Moivre s Theorem, this is equivalet to s cis φ = r cis θ. Hece we eed s = r ad φ equal to θ as agles. This meas they are allowed to differ by a multiple of πi. Hece r = s ad φ = θ + πik φ = θ + πik for some iteger k. Takig k = 0, 1,..., 1 the possibilities for k modulo give distict possible values of φ (other values differ from oe of these by a multiple of πi. Thus each ozero complex umber has th roots. For example, the third roots of i = 8 cis π are cis ( π + πk 6 3 for k = 0, 1,, i.e. cis π 6 = 3 + 1 i cis 5π 6 = 3 + 1 i cis 3π = i. 5

Example.: Let z 0 ad > 1. Show that the sum of the th roots of z equals 0. Solutio 1. Let x be a th root of z, ad ω = cis π. The 1, ω,..., ω 1 are the th roots of 1 so z, zω,..., zω 1 are the th roots of z. So the sum is z(1 + ω + + ω 1 = z ω 1 ω 1 = 0 sice ω = 1. Solutio. Let x 1,..., x be the th roots of z ad ω = cis π above. The x 1 ω,..., x ω are also the th roots of uity, so are equal to x 1,..., x i some order. Thus the sum s equals both x 1 + + x ad ω(x 1 + + x. Sice s = sω, s = 0. Example.3 (AIME 008/9: A particle is located o the coordiate plae at (5, 0. Defie a move for the particle as a couterclockwise rotatio of π radias about the origi 4 followed by a traslatio of 10 uits i the positive x-directio. Give that the particle s positio after 150 moves is (p, q, fid the greatest iteger less tha or equal to p + q. Solutio. Thik of the coordiate plae as the complex plae, ad let z be the particle s positio after moves, as a complex umber. A couterclockwise rotatio of π 4 correspods to multiplicatio by t = cis π, ad traslatio by 10 uits i the positive 4 x-directio correspods to addig 10. Thus Calculatig the first few terms, z 0 = 5 z 1 = 5t + 10 z = tz 1 + 10. z = (5t + 10t + 10 = 5t + 10t + 10 so we see that z = 5t + 10t 1 + + 10, easily proved by iductio. Sice t is a eighth root of 1, by Example (. the sum of the eighth roots 1 + t + t + + t 7 equals 0. Sice t 8 = 1, Hece 0 = 1 + t + + t 7 = t 8 + + t 15 = = t 144 + + t 151. z 150 = 5t 150 + 10t 149 + + 10 = 10[(1 + t + + t 7 + + (t 144 + t 15 ] 10t 151 5t 150 = 10t 7 5t 6 = 10 cis 7π 4 5 cis 3π ( = 10 i + 5i. Thus p + q = 10 + 5 10 1.41 + 5 = 19.1. The aswer is 19. Example.4: Let P (x = (x + (x + 1 (x + 1 (x 1(x (x. Show that all the zeros of P (x are purely imagiary, i.e. have real part 0. 6

Proof. Sice a polyomial of degree at most 1 has at most 1 zeros, it suffices to show that there are 1 pure imagiary roots. Let x = yi ad θ k (y = ta 1 y k. Makig this substitutio ad rewritig i polar form, P (x = 0 is equivalet to (iy + (iy + 1 (iy + 1 = (iy 1(iy (iy cis θ (y cis θ 1 (y = cis(π θ 1 (y cis(π θ (y cis(θ 1 (y + θ (y = cis(π (θ 1 (y + + θ (y (θ 1 (y + + θ (y π (mod π. (1 Note that each θ k is a cotiuous fuctio of y, ad that (θ 1 (y + + θ (y π = π as y (θ 1 (y + + θ (y π = π as y Thus (θ 1 (y + + θ (y attais the value π kπ for k = 1,,..., 1. Hece there are 1 values of y such that (1 holds ad hece there are 1 purely imagiary roots of P, as eeded.. Trigoometry The polar form of complex umbers ca be used to simplify trig expressios ad prove trig idetities. Example.5: Show that cos 3θ = 4 cos 3 θ 3 cos θ ad si 3θ = 3 si θ 4 si 3 θ. Solutio. By De Moivre s Theorem, cos 3θ + i si 3θ = (cos θ + i si θ 3 Matchig the real ad imagiary parts, ad = cos 3 θ + 3 cos θ(i si θ + 3 cos θ(i si θ + (i si θ 3 = (cos 3 θ 3 cos θ si (θ + (3 cos θ si θ si 3 θi cos 3θ = cos 3 θ 3 cos θ si (θ = cos 3 θ 3 cos θ(1 cos (θ = 4 cos 3 θ 3 cos θ si 3θ = 3 cos θ si θ si 3 θ = 3(1 si θ si θ si 3 θ = 3 si θ 4 si 3 θ. Example.6: Show that cos 0 + cos 1 + cos + + cos 89 = 1+cot.5. Solutio. Let ω = cos 1 + i si 1. The ω = cos + i si. Hece the desired sum equals the real part of 1 + ω + ω + + ω 89. Usig the geometric series formula, 1 + ω + ω + + ω 89 = ω90 1 ω 1 = i 1 (cos 1 1 + i si 1 = (i 1((cos 1 1 i si 1 (cos 1 1 + si 1 = (i 1((cos 1 1 i si 1 cos 1 = 1 cos 1 + si 1 + (cos 1 1 + si 1 i cos 1. 7

This has real part 1 cos 1 + si 1 = 1 si 1 (1 + = cos 1 1 cos 1 where we used the trig idetity cot θ =.3 Calculus viewpoit si θ. 1 cos θ 1 + cot.5 Usig calculus we ca write cos θ+i si θ i a more suggestive form. The Taylor expasios of cos ad si are Thus cos x = 1 x! + x4 4! x6 6! + si x = x x3 3! + x5 5! x7 7! + cos x + i si x = 1 + ix x! ix3 3! + x4 4! + ix5 5! + = 1 + (ix + (ix! = e ix. + (ix3 3! + (ix4 4! + (ix5 5! Therefore we ca write cis θ as e iθ. Thus De Moivre s Theorem simply correspods to the fact that e iθ 1 e iθ = e i(θ 1+θ the familiar rule for expoets..4 Problem set 1. [1] Show that cos θ = eiθ + e iθ si θ = eiθ e iθ. i. [] Prove that cos θ = cos θ si θ = / ( = ( 1 k k k=0 cos 1 θ = ( 1 ( 1/ k=0 ( cos θ si θ + cos k θ si k θ. ( ( 1 k k + 1 ( cos 3 θ si 3 θ + 3 cos k 1 θ si k+1 θ. 8

3. [] Evaluate si π + si π ( 1π + + si. 4. [] (AIME 000/9 Give that z is a complex umber such that z + 1 z = cos 3, fid the least iteger that is greater tha z 000 + 1 z 000. 5. [] (AMC 1B, 005/ sequece of complex umbers z 0, z 1,... is defied by the rule z +1 = iz z. Suppose that z 0 = 1 ad z 005 = 1. How may possible values are there for z 0? 6. [] (AMC 1A 008/5 A sequece (a 1, b 1, (a, b, (a 3, b 3, of poits i the coordiate plae satisfies (a +1, b +1 = ( 3a b, 3b + a. Suppose that (a 100, b 100 = (, 4. What is a 1 + b 1? 7. [] (AIME 005/9 For how may positive itegers less tha or equal to 1000 is true for all real t? (si t + i cos t = si t + i cos t 8. [.5] (AIME 1994/11 The equatio x 10 + (13x 1 10 = 0 has 10 complex roots r 1, r 1, r, r,..., r 5, r 5. Fid 1 + + 1. r 1 r 1 r 5 r 5 9. [.5] (AIME 001/14 There are complex umbers that satisfy both z 8 z 8 1 = 0 ad z = 1. These umbers have the form z m = cos θ m + i si θ m where 0 θ 1 < < θ < 360 ad agles are measured i degrees. Fid the value of θ + θ 4 + + θ. 10. [3] Fid a closed form for the sum ( + 0 ( + 3 ( + 6 Hit: Oe way to evaluate ( ( 0 + + is the followig. By the biomial formula, we kow ( ( ( ( ( = (1 + 1 = + + + + + 0 = (1 1 = 0 ( 0 1 + ( 1 ( 3 + ( 3 Addig these two gives (( ( ( = + + + 0 1 ( ( ( 1 = + + + 0 1 How ca you geeralize this? 9 4 + ( 4

11. [4] (MOSP 007 Let a, b 1,..., b, c 1,..., c be real umbers such that x + ax 1 + + ax + 1 = (x + b 1 x + c 1 (x + b x + c (x + b x + c. Prove that c 1 = c = = c = 1. 1. [5] (APMO 009/5 Larry ad Rob are two robots travelig i oe car from Argovia to Zillis. Both robots have cotrol over the steerig ad steer accordig to the followig algorith: Larry makes a 90 left tur after every l kilometers drivig from the start; Rob makes a 90 right tur after every r kilometers drivig from the start, where l ad r are relatively prime positive itegers. I the evet of both turs occurig simultaeously, the car will keep goig straight without chagig directio. Assume that the groud is flat ad the car ca move i ay directio. Let the car start from Argovia facig towards Zillis. For which choices of the pair (l, r is the car guarateed to reach Zillis, regardless of how far it is from Argovia? 3 Roots of Uity The th roots of uity, i.e. roots of 1, satisfy the equatio x 1 = 0. The th roots of uity, ot coutig 1, satisfy x 1 x 1 = x 1 + + x + 1 = 0. If we plug i x = cis π, the terms of this sum are all the th roots of uity, ad we get Example.. This simple idea the sum of the th roots of uity satisfy the equatios above, ad that their sum is zero ca be very useful. Example 3.1: poits Q 1,..., Q are equally spaced o a circle of radius 1 cetered at O. Poit P is o ray OQ 1 so that OP =. Fid the product P Q k i closed form, i terms of. k=1 Solutio. Lettig ray OQ 1 be the positive real axis, Q i represet the th roots of uity ω i i the complex plae. Hece P Q i equals ω i. The roots of x 1 = 0 are just the th roots of uity, so x 1 = 1 i=0 (x ωi. Pluggig i x = gives k=1 P Q i = 1. Pluggig i roots of uity ca help i factorig polyomials ad establishig divisibility, as the followig example shows: Example 3.: Fid all itegers 1 such that x +1 + x + 1 is divisible by x + x + 1. Solutio. I order for x +1 + x + 1 to be divisible by x ± x + 1, we must have that all zeros of x + x + 1 are zeros of x +1 + x + 1. The zeros are exactly the two third roots of uity ot equal to 1. Let ω be such a root of uity. The we eed ω +1 + ω + 1 = 0. However, we kow ω 3 = 1, so the above equatio is equivalet to ω (+1 mod 3 + ω mod 3 + 1 = 0. For 0 (mod 3 this equals ω +, for 1 (mod 3 this equals ω + ω + 1 = 0, ad for (mod 3 this equals ω +. Hece the aswer is all with (mod 3. 10

Example 3.3: [MOSP 007] Let be a positive iteger which is ot a prime power. Prove that there exists a equiagular polygo whose side legths are 1,..., i some order. Proof. Sice is ot a prime power, we ca write = pq, where p ad q are relatively prime. Let ω = cis π. The existece of such a equiagular polygo is equivalet to the existece of a permutatio a 1,..., a of the umbers 1,..., such that 1 a k ω k = 0 ( k=0 Ideed, give such a equiagular polygo, place it o the complex plae so that the real axis be parallel to oe of its sides. Thik of the sides of the polygo as vectors formig a loop, ad traslate them to the origi. The we get vectors equally spacced apart; they are i the directio of the th roots of uity ad have legths 1,..., i some order. The sum of the vectors must be 0 because they formed a loop. Hece we get the equatio above. Coversely, we ca easily reverse the above costructio. We use the two factors of to cause double cacelatio. Note that (pr + 1ω rp + (pr + ω rp+q + + (pr + pω rp+(p 1q (3 = prω rp (1 + ω q + + ω (p 1q + ω rp (1 + ω + + pω p 1 = ω rp (1 + ω q + + pω (p 1q. We used the fact 1 + ω q + + ω (p 1q = 0 as the terms are the pth roots of uity. Now addig the above expressio for r = 1,..., q gives q ω rp (1 + ω q + + pω (p 1q = (1 + ω p + + ω (q 1p (1 + ω q + + pω (p 1q = 0 r=1 becuase 1 + ω p + + ω (q 1p = 0 as the terms are the qth roots of uity. However, addig up (3 without simplifyig for r = 1,..., q, we get the pq expoets of ω rage over the umbers rp + kq, where 1 r q ad 0 k q. Sice p, q are relatively prime, these umbers are all distict modulo = pq. (If r 1 p + k 1 q = r p + k q the (r 1 r p = (k k 1 q with r 1 r < q, k k 1 < p so they must equal 0. Thus takig the expoets modulo pq, we get a sum of the form (, as eeded. 3.1 Problems 1. [1] Factor x 5 + x + 1.. [] (AIME 1996/11 Let P be the product of the roots of z 6 + z 4 + z 3 + z + 1 = 0 that have positive imagiary part, ad suppose that P = r(cos θ + i si θ where 0 < r ad 0 θ < 360. Fid θ. 3. [.5] (AIME 1997/14 Let v ad w be distict, radomly chose roots of the equatio z 1997 1 = 0. Let m be the probability that + 3 v + w, where m ad are relatively prime positive itegers. Fid m +. 11

4. [.5] (AIME1 004/13 The polyomial P (x = (1 + x + x + + x 17 x 17 has 34 complex zeros of the form z k = r k [cos(πα k +i si(πα k ], k = 1,, 3,..., 34, with 0 < α 1 α α 34 < 1 ad r k > 0. Give that α 1 + α + α 3 + α 4 + α 5 = m/ where m ad are relatively prime positive itegers, fid m +. 5. [.5] (AIME 003/15 Let 3 P (x = 4x 4 + (4 j(x 4 j + x 4+j. j=1 Let z 1,..., z r be the distict zeros of P (x ad let z k = a k + b k i for k = 1,,..., r, where a k ad b k are real umbers. Let b k = m + p k=1 where m,, ad p are itegers ad p is ot divisible by the square of ay prime. Fid m + + p. 6. [.5] (AIME 009/13 Let A ad B be the edpoits of a semicircular arc of radius. The arc is divided ito seve cogruet arcs by six equally spaced poits C 1, C,..., C 6. All chords of the form AC i or BC i are draw. Let be the product of the legths of these twelve chords. Fid the remaider whe is divided by 1000. 7. [3] Fid all ordered pairs (m, such that x +1 + x + 1 divides x m+1 + x m + 1. 8. [5] (IMO 1990/6 Prove that there exists a covex 1990-go such that all its agles are equal ad the legths of the sides are the umbers 1,,..., 1990 i some order. Ca you geeralize the statemets of this problem ad Example 3.3? 9. [5] (USAMO 1999/3 Let p > be prime ad let a, b, c, d be itegers ot divisible by p, such that {ra/p} + {rb/p} + {rc/p} + {rd/p} = for ay iteger r ot divisible by p. Prove that at least two of the umbers a + b, a + c, a + d, b + c, b + d, c + d are divisible by p. (Note: {x} = x x deotes the fractioal part of x. 4 Complex umbers i combiatorics We give two applicatios of complex umbers to combiatorics problems. beautiful problems of a similar ature, the reader is referred to [1, 8]. For other Example 4.1: Show that a a b rectagle ca be tiled by 1 blocks if ad oly if either a or b. 1

Solutio. If a or b the the rectagle ca obviously be tiled. Now suppose that a a b rectagle ca be tiled by 1 blocks. Label the squares of the rectagle (x, y with 0 x a 1, 0 y b 1. Let ω be a primitive th root of uity. Label (x, y with ω x+y. Each 1 tile covers umbers of the form ω t, ω t+1,..., ω t+ 1 for some t; these umbers sum to 0. However the sum of all umbers i the board is (1 + ω + + ω a 1 (1 + ω + + ω b 1 = ωa 1 ω 1 ωb 1 ω 1. This is 0 oly if ω a = 1 or ω b = 1, i.e. a or b. Example 4. (TST 004/: Assume is a positive iteger. Cosider sequeces a 0, a 1,..., a for which a i {1,,..., } for all i ad a = a 0. (a Call a sequece good if for all i = 1,,...,, a i a i 1 i (mod. Suppose that is odd. Fid the umber of good sequeces. (b Call a sequece great if for all i = 1,,...,, a i a i 1 i, i (mod. Suppose that is a odd prime. Fid the umber of great sequeces. Solutio. Let f be a fuctio from {1,,..., } to the set of subsets of {0, 1,..., }. Defie a o-f sequece to be a sequece a 0, a 1,..., a such that 1. a i {1,,..., } for i = 0,...,,. a = a 0, ad 3. a i a i 1 is ot cogruet to a umber i f(i modulo. Claim 4.3: The umber of o-f sequeces is times the umber of sequeces d 1, d,..., d such that 1. d i {0, 1,,..., 1} for each i,. d 1 + d 0 (mod, ad 3. d i f(i. Proof. Give a o-f sequece a i, associate with it a sequece as above, with d i = (a i a i 1 mod. Each sequece d i satisfyig the above is associated with o-f sequeces: a 0 ca be chose arbitrarily (there are choices, ad oce a i 1 has bee defied, a i must be the uique iteger i {0, 1,..., 1} so that a i a i 1 + d i (mod. Claim 4.4: Let P (x = 1 + x + + x 1, ad defie b k such that P (x x j = b k x k. i=1 k 0 j f(i The the umber of valid sequeces {d i } i=1 is j 0 b j. 13

Proof. Note that P (x j f(i xj cotais oly powers of x whose expoets are allowable values for d i. Take a term i the expasio of the left-had-side, suppose it takes the term x d i from P (x j f(i xj. The the term is x d 1+d + +d. Now {d i } i=1 is a valid sequece iff d 1 + d + + d 0 (mod. Hece summig the coefficiets of x k for k gives the umber of valid sequeces. Combiig this with the previous claim, the umber of o-f sequeces i j 0 b j. (a A good sequece is exactly a o-f sequece for f(i = {i mod }. Let Q(x = P (x 1 [ = ] P (x x i. Now i=1 j f(i x j i=0 1 1 Q(1 = (P (1 1 = ( 1 = ( 1 i=0 ad for ay ω 1 a th root of uity, i=0 1 Q(ω = (P (ω ω i i=0 1 = ( ω i i=0 = ω ( 1 = 1 sice the fact that is odd gives ( 1. Now Q(x + 1 is zero for all th roots of uity ω 1, so is divisible by P (x. Writig Q(x + 1 = k 0 c kx k, the fact that P (x Q(x gives that the sums S r = k r (mod c k are all equal (fill i the details basically ay term times P (x is spread amog terms whose expoets make up a complete residue class modulo. Hece, sice the sum of coefficiets of Q(x + 1 is Q(1 + 1, k 0 c k = Q(1 + 1 = ( 1 + 1. The the desired sum for Q(x is ( 1 +1 are ( 1 + 1 good sequeces. 1 ad multiplyig by we get there (b A great sequece is a o-f sequece for f(i = {i mod, i mod }. The calculatios are left to the reader. The followig lemma will be useful below. Lemma 4.5: If ω 1 is pth root of uity, ad a 0,..., a p 1 are ratioal umbers such that a 0 + a 1 ω + + a p 1 ω p 1 = 0 the a 0 = = a p 1. This lemma follows from the fact that 1 + x +... + x p 1 is the irreducible (miimal polyomial of ω, which we will prove i a later lecture. 14

4.1 Problem set 4 1. [3] Cosider a rectagle which ca be tiled with a fiite combiatio of 1 m or 1 rectagles, where m ad are atural umbers. (The tiles caot be rotated. Prove that it is possible to tile this rectagle with oly 1 m or oly with 1 rectagles.. [4] TST 004/b above. 3. [4, with the above preparatio] (IMO 1995/6 Let p > be a prime umber ad A = {1,,..., p}. Fid the umber of subsets of A, havig p elemets ad with sum of elemets divisible by p. Refereces [1] Adreescu, T.; Dospiescu, G.: Problems from The Book. XYZ Press, Alle TX, 008. 15