Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i + + 4i 7 + Sice their imagiary part has to be equal, 4i 7 64i (64)(7) 4 + 4i + 697 697.. (985 AIME Problem 3) Fid c if a, b, ad c are positive itegers which satisfy c (a + bi) 3 07i, where i. [Solutio: c 98, where a 6 ad b ] Expadig out both sides of the give equatio we have c + 07i (a 3 3ab ) + (3a b b 3 )i. Two complex umbers are equal if ad oly if their real parts ad imagiary parts are equal, so c a 3 3ab ad 07 3a b b 3 (3a b )b. Sice a, b are itegers, this meas b is a divisor of 07, which is a prime umber. Thus either b or b 07. If b 07, 3a 07 so 3a 07 +, but 07 + is ot divisible by 3, a cotradictio. Thus we must have b, 3a 08 so a 36 ad a 6 (sice we kow a is positive). Thus c 6 3 3 6 98. 3. (995 AIME Problem 5) For certai real values of a, b, c, ad d, the equatio x 4 + ax 3 + bx + cx + d 0 has four o-real roots. The product of two of these roots is 3 + i ad the sum of the other two roots is 3 + 4i, where i. Fid b. [Solutio: b 05] Sice the coefficiets of the polyomial are real, it follows that the o-real roots must come i complex cojugate pairs. Let the first two roots be m,. Sice m + is ot real, m, are ot cojugates, so the other pair of roots must be the cojugates of m,. Let m be the cojugate of m, ad be the cojugate of. The, m 3 + i, m + 3 + 4i m 3 i, m + 3 4i.
By Vieta s formulas, we have that b mm + + m + m + m + m (m + )(m + ) + m + m 05. 4. (984 AIME Problem 8) The equatio 6 + 3 + 0 has complex roots with argumet θ betwee 90 ad 80 i the complex plae. Determie the degree measure of θ. [Solutio: θ 60 ] We shall itroduce aother factor to make the equatio easier to solve. If r is a root of 6 + 3 +, the 0 (r 3 )(r 6 + r 3 + ) r 9. Thus, the root we wat is also a 9th root of uity. This reduces θ to either 0 or 60. But θ ca t be 0 because if r cos 0 + i si 0, the r 6 + r 3 + 3. This leaves θ 60. 5. (994 AIME Problem 8) The poits (0, 0), (a, ), ad (b, 37) are the vertices of a equilateral triagle. Fid the value of ab. [Solutio: ab 35] Cosider the poits o the complex plae. The poit b+37i is the a rotatio by 60 of a+i about the origi, so (a + i) (cis 60 ) (a + i) Equatig the real ad imagiary parts, we have: b a 3 37 + a 3 Thus, the aswer is ab ( 3)(5 3) 35. ( + ) 3i b + 37i. a 3 b 5 3 Note: There is aother solutio where the poit b + 37i is a rotatio of 60 degrees of a + i; however, this triagle is just a reflectio of the first triagle by the y-axis, ad the sigs of a ad b are flipped. However, the product ab is uchaged. 6. (999 AIME Problem 9) A fuctio f is defied o the complex umbers by f() (a + bi), where a ad b are positive umbers. This fuctio has the property that the image of each poit i the complex plae is equidistat from that poit ad the origi. Give that a+bi 8 ad that b m/, where m ad are relatively prime positive itegers, fid m +. [Solutio: m + 59, where m 55 ad 4] Pluggig i yields f() a + bi. This implies that a + bi must fall o the lie Re() a, give the equidistat rule. By a + bi 8, we get a + b 64, ad pluggig i a yields b 55 4. The aswer is thus 59. 7. (000 AIME II Problem 9) Give that is a complex umber such that + cos 3, fid the least iteger that is greater tha 000 + 000. [Solutio: 000] Usig the quadratic equatio o ( cos 3) + 0, we have cos 3± 4 cos 3 4 cos 3 ± i si 3 cis 3. Usig De Moivre s Theorem we have 000 cos 6000 + i si 6000, 6000 6(360) + 40, so 000 cos 40 + i si 40. We wat 000 + 000 cos 40. Fially, the least iteger greater tha is 000.
8. (005 AIME II Problem 9) For how may positive itegers 000 is (si t + i cos t) si t + i cos t true for all real t? [Solutio: 50, where + 4Z] This problem begs us to use the familiar idetity e it cos(t) + i si(t). Notice that si(t) + i cos(t) i(cos(t) i si(t)) ie it sice si( t) si(t). Usig this, (si(t) + i cos(t)) si(t) + i cos(t) is recast as (ie it ) ie it. Hece we must have i i i mod 4. Thus sice 000 is a multiple of 4 exactly oe quarter of the residues are cogruet to hece we have 50. 9. (990 AIME Problem 0) The sets A { : 8 } ad B {w : w 48 } are both sets of complex roots of uity. The set C {w : A ad w B} is also a set of complex roots of uity. How may distict elemets are i C? [Solutio: C 44] The least commo multiple of 8 ad 48 is 44, so defie e πi/44. We ca write the umbers of set A as { 8, 6,... 44 } ad of set B as { 3, 6,... 44 }. x ca yield at most 44 differet values. All solutios for w will be i the form of 8k+3k. 8 ad 3 are relatively prime, ad it is well kow that for two relatively prime itegers a ad b, the largest umber that caot be expressed as the sum of multiples of a ad b is (ab a b). For 3, 8, this is 3; however, we ca easily see that the umbers 45 to 57 ca be writte i terms of 3 ad 8. Sice the expoets are of roots of uities, they reduce mod 44, so all umbers i the rage are covered. Thus the aswer is 44. 0. (99 AIME Problem 0) Cosider the regio A i the complex plae that cosists of all poits such that both 40 40 ad have real ad imagiary parts betwee 0 ad, iclusive. What is the iteger that is earest the area of A? [Solutio: [A] 57] Let a + bi which is a square of side legth 40. 40 a 40 + b 40 i. Sice 0 a 40, b 40 we have the iequality 0 a, b 40 Also, 40 40 a bi 40a a +b + 40b a +b i so we have 0 a, b a +b 40, which leads to We graph them: (a 0) + b 0 a + (b 0) 0 3
We wat the area outside the two circles but iside the square. Doig a little geometry, the area of the itersectio of those three graphs is 40 40 4 π0 00 00π 57.68. Thus, by roudig to the earest iteger we get 57.. (988 AIME Problem ) Let w, w,..., w be complex umbers. A lie L i the complex plae is called a mea lie for the poits w, w,..., w if L cotais poits (complex umbers),,..., such that ( k w k ) 0. For the umbers w 3 + 70i, w 7 + 64i, w 3 9 + 00i, w 4 + 7i, ad w 5 4 + 43i, there is a uique mea lie with y-itercept 3. Fid the slope of this mea lie. [Solutio: 63] k w k 0 k 3 + 504i Each k x k + y k i lies o the complex lie y mx + 3, so we ca rewrite this as k x k + y k i 3 + 504i x k + i (mx k + 3) Matchig the real parts ad the imagiary parts, we get that 5 x k 3 ad 5 (mx k + 3) 504. Simplifyig the secod summatio, we fid that m 5 x k 504 3 5 489, ad substitutig, the aswer is m 3 489 m 63.. (996 AIME Problem ) Let P be the product of the roots of 6 + 4 + 3 + + 0 that have a positive imagiary part, ad suppose that P r(cos θ + i si θ ), where 0 < r ad 0 θ < 360. Fid θ. [Solutio: ] Let w the 5th roots of uity, except for. The w 6 +w 4 +w 3 +w + w 4 +w 3 +w +w+ 0, ad sice both sides have the fifth roots of uity as roots, we have that 4 + 3 + + + 6 + 4 + 3 + +. Log divisio quickly gives the other factor to be +. Thus, + 0 ± 3 cis 60, cis 300 Discardig the roots with egative imagiary parts (leavig us with cisθ, 0 < θ < 80), we are left with cis 60, 7, 44; their product is P cis(60 + 7 + 44) cis 76. 3. (997 AIME Problem ) cos Let x. What is the greatest iteger that does ot exceed 00x? si [Solutio: 4] Usig the idetity si a + si b si a+b a b cos si x + cos x si x + si(90 x) si 45 cos(45 x) cos(45 x), ote that cos + si cos(45 ) cos 4
si ( ) cos x Thus, 00x 00( + ) 4. cos si 44 + 4. (00 AIME I Problem ) Let F () +i i for all complex umbers i, ad let F ( ) for all positive itegers. Give that 0 37 + i ad 00 a + bi, where a ad b are real umbers, fid a + b. [Solutio: a + b 75, where a ad b 74] Iteratig F we get: F () + i i F (F ()) F (F (F ())) +i i + i ( + i) + i( i) i ( + i) i( i) + i + i + + i i +i i ( + )(i + ) ( + )(i) ( )( + ) ( )() + i + i + i + i i + + ( + ) + ( ) ( + ) ( ). + From this, it follows that k+3 k, for all k. Thus, ( + )(i + ) ( )( i) 00 3 667+ 0 + i 0 i ( 37 + i) + i ( 37 + i) i 37 + i + 74i 37 Thus a + b + 74 75. 5. (004 AIME I Problem 3) The polyomial P (x) ( + x + x + + x 7 ) x 7 has 34 complex roots of the form k r k [cos(πa k ) + i si(πa k )], k,, 3,..., 34, with 0 < a a a 3 a 34 < ad r k > 0. Give that a + a + a 3 + a 4 + a 5 m/, where m ad are relatively prime positive itegers, fid m +. [Solutio: m + 48, where m 59 ad 33] By usig the sum of the geometric series, we see that P (x) ( x 8 ) x 7 x36 x 8 + x x x + x7 x36 x 9 x 7 + (x ) (x9 )(x 7 ) (x ) This expressio has roots at every 7th root ad 9th roots of uity, other tha. Sice 7 ad 9 are relatively prime, this meas there are o duplicate roots. Thus, a, a, a 3, a 4 ad a 5 are the five smallest fractios of the form m 9 or 7 for m, > 0. 3 7 ad 4 9 ca both be see to be larger tha ay of 9, 9, 3 9, 7, 7, so these latter five are the umbers we wat to add. m Thus, 9 + 9 + 3 9 + 7 + 7 6 9 + 3 7 6 7+3 9 7 9 59 33 m + 59 + 33 48. ad so the aswer is 5
6. (994 AIME Problem 3) The equatio x 0 +(3x ) 0 0 has 0 complex roots r, r, r, r, r 3, r 3, r 4, r 4, r 5, r 5, where the bar deotes complex cojugatio. Fid the value of [Solutio: 850] Divide both sides by x 0 to get ( + 3 x) 0 0 + + + + r r r r r 3 r 3 r 4 r 4 r 5 r 5 ( 3 ) 0 3 x x ω where ω cis(π( + )/0) ad 0 9 is a iteger. We see that 3 ω. Thus, x Summig over all terms: (3 ω)(3 ω) 69 3(ω + ω) + ωω 70 3(ω + ω) xx r r + + r 5 r 5 5 70 3(cis(π()/0) + + cis(π(9)/0)) 850 0 850 7. (998 AIME Problem 3) If {a, a, a 3,..., a } is a set of real umbers, idexed so that a < a < a 3 < < a, its complex power sum is defied to be a i + a i + a 3 i 3 + + a i, where i. Let S be the sum of the complex power sums of all oempty subsets of {,,..., }. Give that S 8 76 64i ad S 9 p + qi, where p ad q are itegers, fid p + q. [Solutio: p + q 368, where p 35 ad q 6] We ote that the umber of subsets (for ow, icludig the empty subset, which we will just defie to have a power sum of ero) with 9 i it is equal to the umber of subsets without a 9. To easily see this, take all possible subsets of {,,..., 8}. Sice the sets are ordered, a 9 must go at the ed; hece we ca just apped a 9 to ay of those subsets to get a ew oe. Now that we have draw that bijectio, we ca calculate the complex power sum recursively. Sice appedig a 9 to a subset does t chage aythig about that subset s complex power sum besides addig a additioal term, we have that S 9 S 8 + T 9, where T 9 refers to the sum of all of the 9i x. If a subset of sie ( has ) a 9, the its( power ) sum must be 9i, ad there is oly of these such 8 8 subsets. There are with 9 i, with 9 i 3, ad so forth. So T 9 ( ) 8 8 k0 9 i k+. k This is exactly the biomial expasio of 9i ( + i) 8. We ca use De Moivre s Theorem to calculate the power: ( + i) 8 ( ) 8 cos(8 45) 6. Hece T 9 6 9i 44i, ad S 9 S 8 + 44i ( 76 64i) + 44i 35 + 6i. Thus, p + q 35 + 6 368. 8. (989 AIME Problem 4) Give a positive iteger, it ca be show that every complex umber of the form r + si, where r ad s are itegers, ca be uiquely expressed i the base + i usig the itegers,,..., as digits. That is, the equatio r + si a m ( + i) m + a m ( + i) m + + a ( + i) + a 0 is true for a uique choice of a o-egative iteger m ad digits a 0, a,..., a m chose from the set {0,,,..., }, with a m 0. We write r + si (a m a m... a a 0 ) +i 6
to deote the base + i expasio of r + si. There are oly fiitely may itegers k + 0i that have four-digit expasios Fid the sum of all such k. [Solutio: 490] First, we fid the first three powers of 3 + i: k (a 3 a a a 0 ) 3+i, a 3 0 ( 3 + i) 3 + i; ( 3 + i) 8 6i; ( 3 + i) 3 8 + 6i So we eed to solve the Diophatie equatio a 6a + 6a 3 0 a 6a 6a 3. The miimum the left had side ca go is -54, so a 3, so we try cases: Case : a 3 The oly solutio to that is (a, a, a 3 ) (, 9, ). Case : a 3 The oly solutio to that is (a, a, a 3 ) (4, 5, ). Case 3: a 3 0 a 3 caot be 0, or else we do ot have a four digit umber. So we have the four digit itegers (9a 0 ) 3+i ad (54a 0 ) 3+i, ad we eed to fid the sum of all itegers k that ca be expressed by oe of those. (9a 0 ) 3+i : We plug the first three digits ito base 0 to get 30 + a 0. The sum of the itegers k i that form is 345. (54a 0 ) 3+i : We plug the first three digits ito base 0 to get 0 + a 0. The sum of the itegers k i that form is 45. The aswer is 345 + 45 490. 7