ANSWERS SOLUTIONS iiii i. and 1. Thus, we have. i i i. i, A.

Size: px

Start display at page:

Download "ANSWERS SOLUTIONS iiii i. and 1. Thus, we have. i i i. i, A."

Shannon Snow
6 years ago
Views:

1 013 ΜΑΘ Natioal Covetio ANSWERS (1) C A A A B (6) B D D A B (11) C D D A A (16) D B A A C (1) D B C B C (6) D C B C C 1. We have SOLUTIONS iiii i , C.. The powers of i cycle betwee i, 1, i, ad 1. Thus, we have i i i i, A. 3. The absolute value of the etire fractio is the ratio of the absolute values of the umerator ad deomiator. Usig this, we have 3 i 3 i 5 1i 5 1i 5, A. 13 Page 1 of 13

2 013 ΜΑΘ Natioal Covetio. We let x equal the expressio we wish to evaluate. With a substitutio, we obtai i i i i x x. Solvig this equatio with the quadratic formula gives i 1 1i x x 0 x. Now, we must evaluate 1 i. We ca write this i cis form as 1i cis. the square root of this, we utilize de Moivre s Theorem to obtai cis cis, 0,1 8 Combiig this with the rest of the solutio gives x 1 cis, 0, cis 8, 0,1 1 3 cis, 0,1, A. 8 To tae 5. Note that Re(cis ) cos ad Im(cis ) si. Thus, we have Re cis(( 1) ) Im cis(( 1) ) cos1 cos 3 cos89 si si 6 si178 cos1 cos 3 cos89 (si1cos1 )(si 3cos3 ) (si89cos89 ) si1si 3 si89 The bottom expressio ca be writte as Page of 13

3 013 ΜΑΘ Natioal Covetio si1si si 3 si89 si1si 3 si89 si si si88 si1si si 3 si89 (si1cos1 )(si cos ) (si cos ) 1 si 6si 7 si89 si 5 cos1 cos cos si 6si 7 si89 i89si88 si 6 5 s 89, where we have used the fact that si(90 ) cos. Our aswer is the 5 1 1, B The powers of i cotai two sets of umbers that are additive iverses of each other, amely (1, 1) ad ( i, i). Thus the oly sets of four umbers that will satisfy a 0 are permutatios of either (1,1, 1, 1), ( i, i, i, i), ad ( i, i,1, 1). The first two have 6 distict arragemets each, while the last has! total arragemets, givig (6) 36 overall. There are 56 possibilities, givig a probability of 36 9, B The solutios to the equatio z form a hexago i the complex plae, similar to the 6 th roots of uity, except the side legth of the hexago is Thus z3 z6 is equal to the distace betwee two diagoally opposite poits o the hexago. This is simply (3) 6, D. 8. We have v1 a, b ad v c, d, givig v1 v ac bd. Ituitio would lead us to try Re z w ac bd. This, however, is the cojugate of what we wish to obtai. Naturally, we would the tae the cojugate of either z or w. This gives us Re( z w) Re(( ac bd) i( ad bc)) ac bd, D. Page 3 of 13

4 013 ΜΑΘ Natioal Covetio 9. Goig by the defiitio, we have i i( i 1)( i )( i 3)! 10 5, A We have (cis3 cis35 ) cos 35cis3 cos35 cos 3 cos35 cos 3 i si 3 i cos35 si 3 cos(3 35 ) cos(3 35 ) i si(3 35 ) si(3 35 ) cos 78 i si 78 cos8 i si 8 cis78 cis8. Thus, we have (78)(8) 6, B. 11. Note that we ca rewrite the equatio as ( a6) ( b3) 6, or the equatio for a circle. If we were to covert z to the Cartesia plae, we would simply write z ( x, y) ( a, b). Hece, R is a circle with radius 8, ad thus has a area of 8 6 C. 1. Let z a bi. The we have z z a bi a bi a bi a b a a b bi a a a b a b b a b a a b. Now, sice z a b, this becomes Page of 13

5 013 ΜΑΘ Natioal Covetio z a z z a z 1 0. Usig the quadratic formula, we solve for z as a tae z a, D. z a a. Sice a, we 13. We have x 3y i x y v1 v x 1 i y 3 i 5 6 i. This gives us the systems of equatios x3y 5 xy 6 which we solve as ( xy, ) (8, 1), which gives xy 7, D. 1. We have B A I Thus, 0. i Solvig gives 1 i, A. det( B) Page 5 of 13

6 013 ΜΑΘ Natioal Covetio 15. It is clear that each f ( x ) will be a polyomial of degree, sice the roots are the vertices of a square. Now, ote that each set of roots is a rotatio of radias couterclocwise from the previous set of roots ad furthermore, each set of roots has 1 times the amplitude of the previous set of roots. We bega with the fourth roots of uity, which are cis, 0 3. This meas the th set of roots are cis ( 1) cis ( 1). Of course, we ca write this as x cis ( 1) ( 1). This implies that 1 1 f( x) x. Thus we have f (0), A The fuctio will ot itersect the x -axis whe it has imagiary roots. This requires that the discrimiat be less tha 0. We have ()( )(9) 0,,, D Let the roots be r 1, r,, r 013, where r 1 1. The sum of the roots tae two at a time ca be writte as rr i j, 0 i, j 013, i j. This ca be writte as cyc rr i j rr 1 a rb rc ra rb rc, cyc cyc cyc cyc cyc Sice r 1 1. We ca see that this summatio cotais both the sum of the roots ad the sum of the roots tae two at a time of 01 g( x) 1 x. This is just , B Page 6 of 13

7 013 ΜΑΘ Natioal Covetio 18. We proceed by casewor. Our first case, a real result, ca be achieved by rollig both real umbers or both imagiary umbers. Note that both the first ad secod subcases are symmetric so the total expected value is Our secod case, a imagiary result, is achieved whe we multiply a imagiary umber by a real umber. The expected value of this is The total expected value is 9 5 $.5, A. 19. Writig the expressio i cis form gives us i 3 i 013 cis cis 013 cis 013, A Let the first term be a ad a commo ratio be r. If, at some poit i the series, the th 1 term i the series equals the first, we have a ar r 1, 1. Thus the possible ratios are the th roots of uity. There must be 50 of these roots i the secod quadrat, or betwee 90 ad 180. Sice the roots of uity are cis, for some, x we must have 360x 90 x 360( x 50) 180 x 50 Page 7 of 13

8 013 ΜΑΘ Natioal Covetio Subtractig the secod from the first gives Thus the smallest value of is 01, which gives , C. 1. This is just i i f( i) 1!! , D.. Note that cis cis cis. Usig this, we have cis cis1cis cis (01) cis cis , B. 3. The plotted poits form a spiral shape, composed of segmets which we ca treat as hypoteuses of right triagles for our purposes of calculatig distace. Sice the powers of i traverse the axes couterclocwise, each two set of cosecutive poits alog with the origi form a right triagle. For example, z 3 i 3, givig 1 z1 i i zz I geeral, we have zz , ad Page 8 of 13

9 013 ΜΑΘ Natioal Covetio Thus, z1z zz3 z01z Therefore, our aswer is 07090(mod100) 90, C Note that A is a 60 couterclocwise rotatio matrix. So every apply it, we simply retur to the same vector. This meas that times we 37 6(6) 1 A z A z Az i 1 3 i 3, B. 3 3 i 5. We have (!) ( 11)(!) 1 1 ( 1)(!)! 1 ( 1)!! 1 ( 1)! 1 Thus, the sum becomes ( 1)!!! ( 1)!! 1! ad we have 1! 1 ( 1)! 11 ( 1)!, Page 9 of 13

10 013 ΜΑΘ Natioal Covetio 1 i 1 i ( 1)! i ( 1)! i i 1! i i e 1 i ie i, C. 6. We ca write f( x ) as f x x x x ( ) x x x x x x x ( 1) ( 1) ( 1) ( 1) ( x 1)( x x x 1) 010 ( x 1) x ( x 1) ( x 1) ( x 1)( x 1)( x x 1). Thus, we ow that f( x ) has 1, i, ad i as roots. Sice the powers of i cycle, we are oly worried about the powers of i that come out to 1, or every fourth power. Note that R(1) f(1) 01, by the Remaider theorem. Sice we begi at 0, our aswer is (mod100) (mod100) 56, D. 7. The fuctio i this problem is similar to the fuctio give i Problem 6. We ca write f ( x ) as 1 f ( x) x j0 1 1 j0 (1 x) j x j1 j 1 1 j0 1 j0 x j (1 x) x x x j j 1 j (1 x)(1 x ) x j0 1 x x x Page 10 of 13

11 013 ΜΑΘ Natioal Covetio Solvig 1 x 1 gives us that the etire set is a uio of all. i x cis,,, The sum for a give is Note Thus the etire sum (while accommodatig for x1 0 x cis ) is Fially, we must fid such that log 809. We ca easily verify that the smallest such is 1, C. 8. We ow that z m 9. Cosider this modulo 8. Sice the quadratic residues mod8 are 0,1, ad, the possible values of z (mod8) are 0 0 0(mod 8) 0 1 1(mod 8) 0 (mod 8) 11 (mod 8) 1 5(mod 8) The aswer choices, mod8, are Page 11 of 13

12 013 ΜΑΘ Natioal Covetio 010 (mod 8) 011 3(mod 8) 01 (mod 8) 013 5(mod 8) Thus our aswer is 011, B. 9. Our ituitio for a ew set of axes is based o the fact that the Eisestei itegers have a argumet of 60. Through some playig aroud, we ca fid the set of axes as show below: As we ca see, the plotted poits form a equilateral triagle with a side legth of 3. Thus, the area is , C. 30. We have z a b 1 a b ( 1 i 3) b b 3 a i. Thus, Page 1 of 13

13 013 ΜΑΘ Natioal Covetio z b b 3 a a ab a ab b b 3b a b ab, C. Page 13 of 13

APPENDIX F Complex Numbers

APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios