Chapter 6 Complex Numbers ad Polyomals pages 7 7... INVESTIGATION 6A 6. Gettg Started GRAPHING COMPLEX NUMBERS For You to Explore. a + + + + + + + b + + + + c + + + + + 6 + + + d + + + + + 9 + 6 8 + 6 e + + + + + + + + f + + + +. a + + 6, or 6 + b + 9 +, or + Alteratvely, + 9. c 7 + + 7 d 7 + 7 9 + Alteratvely, 7 + 7 7 9. e + + f + + 6 6 69 Alteratvely, + 69.. a The cojugate of + s. b The cojugate of s +. c If z x + y, the z + z x + y+ x y x. d If z x + y, the z z x + yx y x y x + y.. I geeral, f A x, y the A y,x. To see why ths s the case, otce that the rotated pot, a, b, wll be the same dstace from the org as x, y whch s x + y ad the segmets coectg the two pots to the org wll be perpedcular to each other. That s, the segmet coectg the ew pot to the org wll have a slope that s the opposte of the recprocal of the segmet coectg the org to the pot x, y. Sce the slope betwee, ad a, b s b, the slope from a the org to the rotated pot s x. Also, f x, y s y Quadrat I, the x ad y are oegatve, so a, b wll le Quadrat II. If x, y s Quadrat II, the ts rotato wll be Quadrat III, f x, y s Quadrat III, the ts rotato wll le Quadrat IV, ad f x, y s Quadrat IV, the ts rotato wll le Quadrat I. The two pots that are 9 rotatos of x, y are y,x ad y, x, but y,x s the correct quadrat each case. Therefore, a If A,, the A,. b If A,, the A,. c If A,, the A,. d If A 6,, the A, 6.. a If z, the z. b If z, the z. c If z +, the z + + d If z 6, the z 6 6 + + 6. 6. Gve that z x + y, a z x + y x + xy + y x y + xy. b Therefore, the real part of z s x y. c The magary part of z s xy. 7. Expadg the left sde of the proposed detty, x y + xy x x y + y + x y x + x y + y x + x y + y x + y for all x ad y, so the equato s a detty. O Your Ow 8. We are gve fx x x +. a fx x x + + x + b Sce x for all x, we see that fx x + Therefore, the graph of y fxalways les above the x-axs, so t wll have o x-tercepts. c f + + + + d By the Quadratc Formula, f x x +, the x ± ± ± ± Therefore, aother soluto to fx sx. Mathematcs III Solutos Maual Chapter 6, page 7
9. To fd the legth of the le segmet betwee pots a, b ad c, d, you ca use the dstace formula, a c + b d or draw the horzotal ad vertcal compoets of the le segmet ad use the Pythagorea Theorem: a The legs are ad, so the hypoteuse s + 7. b The legs are ad, so the hypoteuse s +. c The legs are 7 ad 6, so the hypoteuse s 7 + 6 8. d The legs are ad, so the hypoteuse s +. e The legs are 6 ad 8, so the hypoteuse s 6 + 8. f The legs are ad, so the hypoteuse s +. g The legs are x ad y, so the hypoteuse s x + y.. You showed a prevous exercse that f z x + y, the z z x + y. Alteratvely, use the fact that a + ba b a b to compute each product: a + 6 + 7 b + + c 7 + 67 6 7 6 9 + 6 8 d + 9 + 6 e 6 + 86 8 6 8 6 + 6 f + +. a x y 9, xy, x + y 9 + b x y x y xy x + y 8 6 7 8 7 9 The table produces Pythagorea trples.. a y O x b The ampltude of ths fucto s ad ts perod s π. c The graph s the same as part a. d The two graphs appear to be the same. I fact, the ext secto, you wll prove that the equato cos x cos x s x s a detty.. a x + y x + x y + xy + y x + x y xy y x xy + x y The real part s x xy. b c The graph of y cos x cos x s x. d Ths appears to be the same as the graph of y cos x. I fact, the ext secto, you wll show that the equato cos x cos x cos x s x s a detty. It s also terestg to otce that f you replace x + y wth cos x + s x part a, the aswer wll be cos x cos x s x. Mata Your Sklls. Use dfferece-of-squares factorg for each. a a b b a b c a b d a cb e x + yx y x y x + y 6. The Complex Plae Check Your Uderstadg. a The magtude s + 7. b The magtude s +. c The magtude s +. d The magtude s 6 + 6.. Each of these s a specal case of x + yx y x y x + y. a + 6 + 7 b + + c + 9 + d 6 + 6 6 + 6 Mathematcs III Solutos Maual Chapter 6, page 7
. a The umbers are z + ad z + : c The umbers are z ad z : + + argz.9 ad argz.96 d The umbers are z ad z + : argz.88 ad argz.9 b The umbers are z + ad z : + + argz.86 ad argz.87 argz. ad argz.88. Frst, the magtude of each of these complex umbers s. Sce these complex umbers are equal except for the postve ad egatve values of a ad b, the value of a + b wll always equal + 8 To fd the drecto, cosder where each umber s located. a Note that ths complex umber forms a sosceles rght tragle whe a alttude s added, so the agle Mathematcs III Solutos Maual Chapter 6, page 7
at the org s exactly π. Ths s the drecto agle for +. s Quadrat II, the drecto agle for + s π. + + b Ths complex umber also forms a sosceles rght tragle, so the agle at the org s exactly π. Sce the complex umber s Quadrat IV, the drecto agle for s 7π. d Sce the complex umber forms a sosceles rght tragle Quadrat III, the drecto agle for s π. c We aga have a complex umber that forms a sosceles rght tragle whe a alttude s added, so the agle at the org s exactly π. Sce the umber. Specfc values of z wll be dfferet, but results should be smlar ad results mght ot be exact, depedg o the choce of z ad the related approxmato error. The bg dea here s that multplcato by s rotato by π couterclockwse, multplcato by s rotato by π, ad multplcato by s rotato by π also, ote that. Multplcato by doubles the magtude ad leaves the drecto uchaged. The effect o the drecto of takg the recprocal of z, arg arg z π argz z Mathematcs III Solutos Maual Chapter 6, page 7
s ot so obvous. The magtude of the recprocal s the recprocal of the magtude. Cosder several dfferet choces for z: z + z z a I each case, z z ad argz argz + π or argz π, depedg o the quadrat z les. If z +, the z +,so z + 8 z Also, argz π ad argz π as see Exercse. If z, the z +,so z + z Sce ta π ad z s Quadrat 6 IV, argz π 6. Sce ta π ad z s Quadrat IV, argz π. If z, the z, so z z. We also kow that argz π, sce les o the postve magary vertcal axs, ad argz π, sce les o the egatve real axs. b I each case, z z ad arg z argz + π or argz π, depedg o the quadrat z les. If z +, the z z,so z + 8 z Also, argz π ad arg z π as see Exercse. If z, the z z +,so z + z We saw part a that argz π 6. Sce ta ad z s Quadrat II, argz π π π. If z, the z z,so z z. We also kow that argz π, sce les o the postve magary vertcal axs, ad arg z π, sce les o the egatve magary axs. c I each case, z z ad arg z argz + π or argz π, depedg o the quadrat z les. If z +, the z z,so z + 8 z Also, argz π ad arg z 7π Exercse. as see If z, the z, so z + z We saw part a that argz π 6. Sce ta π ad z s Quadrat III, argz π + π π. If z, the z z, so z z. We also kow that argz π, sce les o the postve magary vertcal axs, ad arg z, sce les o the postve real axs. d I each case, z z ad argz arg z. e I each case, z z ad the rule for argumet s stated above. O Your Ow 6. If you trple a complex umber, you trple ts legth, but the agle stays the same. So z z, ad argz argz. 7. For all real umbers c ad all complex umbers z, cz c z. If c, the cz, so there s o argumet. If c>, argcz argz. If c<, argcz argz + π. No matter what c s, cz c z. Ths rule for the product of absolute values that you saw a earler course holds for complex umbers. If c>, argcz argz, sce multplyg by a postve scalar wo t chage the agle. If c<, the the agle wll be reflected over the org, that s, t wll rotate π radas aroud the ut crcle. 8. a The umber x + y wll be Quadrat I ad x y wll be Quadrat IV: z x + y z x y Mathematcs III Solutos Maual Chapter 6, page 76
b The sum of z ad z s x + y + x y x, whch les o the postve real axs: z x + y z + z x Other solutos are,,, ad +,, +, ad +, +,, ad There are may others.. You ca fd the solutos by factorg: x 6 x + x x x x + x x + x Thus, the solutos are,,, ad. z x y O c The four pots deftely form a quadrlateral. What are the legths of the sdes? The legth of the segmet betwee ad z x + y s x + y,ass the legth of the segmet betwee ad z x y. The legth of the segmet betwee z + z x ad z s x x + y x + y, whch s the legth of the segmet betwee z + z ad z. Therefore, the four pots form a rhombus. 9. a z cos t + s t cos t + s t + cos ts t cos t s t + s t cos t b The magary part s s t cos t.. There s more tha oe possble aswer. Oe aswer s +,, +, ad : + +. a + ; drecto s. b + + ; drecto s ta.69. c + 9 + +,so + +. Sce + s Quadrat I, ts drecto s ta 67.8. d + + + 9 + 6,so + 9 + 6 97. Sce + s Quadrat II, ts drecto s.7. 8 + ta 6 9 e + + + + 9 +, so + 9 + 8,6 69. Sce + s Quadrat II, ts drecto s 8 + ta 9.76.. Oe way to do ths exercse s to use z w zw, but aother s to calculate the squares ad the the magtude. a +, ad +. b + +, ad +. c + 8 + 6, ad 8 + 6. d 7 + 8 +, ad 8 +. e + +, ad + 69. f I each case, the magtude s a teger, ad the square of the orgal umber s magtude. So, thk of a umber wth magtude 9. Oe s + others are +,, ad, to ame a few. The square t: + +. Ths umber has magtude 9.. a The solutos are x ad x. Ther magtudes are both ad ther drectos are ad π radas, respectvely. b The solutos are x ad x ±. The complex solutos drectos, determed usg Mathematcs III Solutos Maual Chapter 6, page 77
-6-9 tragles, are π ad π radas. Each of ther magtudes s. c The solutos are x, x, ad x ±. Ther magtudes are all, ad ther drectos, π, π, ad π radas.. Expad the left sde ad the factor by groupg: ac bd + bc + ad a c abcd + b d + b c + abcd + a d a c + b d + b c + a d a c + a d + b c + b d a c + d + b c + d a + b c + d Other methods are possble, cludg expadg each sde utl the expressos are equal. 6. + +. The correct aswer choce s D. Mata Your Sklls 7. ω ω 8 ω 6 ω ω ω 9 ω ω ω ω 7 ω ω 6. Aother Form for Complex Numbers r cs θ Check Your Uderstadg. a The magtude s ad the drecto s 6. b The magtude s ad the drecto s.. a Sce z +, z 7 + + The z + 6 + 87 6 Notce that z s Quadrat II ad z + cos + s so argz. Note that, ths case at least, z z ad argz argz. b Sce z z z, z + + + + z z ad argz 8 argz. c zw + + + + so ad zw z w argzw 6 argz + argw That s, o the crcle, the drecto of zw s the same as the drecto of z plus the drecto of w.. a Aswers wll vary. Sce cos ad s, all aswers should be of the form z r cs r r, where r s a postve real umber. Here are pots wth drecto the pots,,, ad : Mathematcs III Solutos Maual Chapter 6, page 78
b The set s a ray from the org potg the drecto of : c Sce the magtude of the complex umber r cs s r, the umber we are lookg for s cs. Alteratvely, draw a -6-9 rght tragle wth hypoteuse of legth alog the ray from part b: The the legs are ad, so the complex umber s.. a Sce q ad argq π, π π π q cs cos + s b q cos π + s π + +.. a The smplest way s to use r cs θ form: a cs 6 ad b cs. You ca also wrte them x + y form: a + ad b +. b If you wrote a ad b usg r cs θ form, you could calculate the product ths way: ab cos 6 + s 6 cos + s cos 6 + s 6 cos + s cos 6 cos + s 6 s + cos 6 s + cos s 6 + + + + If you wrote a ad b usg x + y form, you could calculate the product ths way: ab + + + + + + + + Ether way, the product s., ad arg 9. 6. a Aswers wll vary. The pots plotted below are a cs, b cs +, c cs +, d cs 6 +, e cs 9, f cs 8, ad g cs 7 f b If you let θ vary but keep the magtude at, each umber wll be uts away from +. Therefore, all possble z values wll le o the crcle wth radus cetered at the org. e g d c b a Mathematcs III Solutos Maual Chapter 6, page 79
7. a If z csθ, the z cosθ+ sθ b r cs θ r cs θ r r cos θ + s θ O Your Ow 8. a Oe way to fd the magtude ad drecto of s to express the form r cs θ: cos + s Therefore, cos + s ad arg. Note that you could also compute the magtude ad drecto the old way : + ad ta, so sce s Quadrat IV, arg. b Sce we just showed that cs, we kow that ths umber also has a magtude of ad a drecto of. c From our earler work, we kow that the magtude of cs π s ad ts drecto s π. d Sce cos π + s π π cs the magtude of s ad ts drecto s π. 9. Note that z cs π cos π + s π. a z z. b We ve see before that multplyg a complex umber by z has the same effect as rotatg the umber about the org 9. To see why, otce that whe you multply a + b by you get b + a. That s, the magtude s uchaged, the real part of the product s the opposte of the orgal umber s magary part, ad the magary part of the product s the same as the orgal umber s real part. Ths meas that a + b s always the ext quadrat couterclockwse from the quadrat a + b s ad sce b a s the opposte of the recprocal of a b, the umbers a + b ad b + a are perpedcular. Therefore, b + a s the rotato of a + b aroud the org by 9 couterclockwse.. We kow w cs cos + s.68 +.8798 ad z cs 6 cos6 + s6.6986 +.7,so zw.68 +.8798.6986 +..68.6986.8798.7 +.68.7 +.8798.6986.9.9 +.6978 8.89 9.99997 so zw, ad argzw 7.. a We kow that w csα cosα + sα ad z csβ cosβ + sβ. b wz cosα + sαcosβ + sβ cosα cosβ sα sβ + cos α s β + s α cos β. a Both z ad w are Quadrat I. z + ad argz 6.6 ta w + ad argw ta 8.99 zw + s also Quadrat I, so zw + ad argzw ta b Aga, z ad w are both Quadrat I. z ad argz 6.6 as part a w + ad argw ta.6968 zw + 7 s Quadrat I, so zw + 7 6 ad argzw ta 7 6.9. c z + ad argz 9 w + 9 ad argw 9 zw, so zw ad argz 8 d As part a, z ad argz 6.6. Notce that w. The w +. Sce ta ta 6.6, argw 6 6.6.99. zw that s z ad w are recprocals, so zw ad argzw.. a The examples suggest that z w zw. b argzw argz + argw c If z a + b ad w c + d, the zw ac bd + ad + bc. The z w a + b c + d a + b c + d a c + a d + b c + b d zw ac bd + ad + bc a c abcd + b d + a d + abcd + b c a c + a d + b c + b d The expressos are equal for ay choce of the varables, so z w s always equal to zw. Mathematcs III Solutos Maual Chapter 6, page 8
. Ths s the complex umber π cs cos π + s π cs π The correct aswer choce s A. Mata Your Sklls cos π s π. a If w +, the w + ad argw ta π. wz + + ++ whch has magtude ad drecto π. b If w, the w ad argw π. wz + + so wz w z by the prevous exercse. Sce ta ta π ad wz s Quadrat II, argwz π π π. c If w, the w ad argw. wz +,so wz w z ad argwz ta ta π 6. d If w +, the w ad argw ta π. The wz + + + + so wz w z ad argwz ta +.8997 or 7. e If w +, the w +. Sce ta.668 or about 6.6 ad w s Quadrat II, argw π.668.6779 or about 8 6.6.99. The wz + + +.6.6799 so wz + + + 7 ote we could have also used the fact that wz w z. Sce ta.6799 ta.96.6 ad wz s Quadrat III, argwz 8 +.96 8.96. b The drecto of z s + + 6,so t s a postve real umber. The magtude s 7, so z 7.. For z, square the magtude ad double the argumet. For z, cube the magtude ad trple the argumet. I geeral for z, rase the magtude to the th power ad multply the argumet by. For example, f z has magtude ad argumet, the z 6 has magtude 6 6 ad argumet 6.. a If z has magtude 9 ad argumet, the z has magtude 9 79 ad argumet. That does t match the magtude crtero for z. b If z has magtude ad argumet, the z has magtude 7 ad argumet, whch matches both crtera for z. c If z has magtude ad argumet, the z has magtude 7 ad argumet 6. That does t match the argumet crtero for z. d If z has magtude ad argumet 6, the z has 7 ad argumet 6 8. That s a match, sce 8 s oce aroud a full crcle ad the more, so 8 ad are the same argumet the plae. Therefore, oly b, ad d are possble.. Aswers wll vary depedg o estmates for the magtude ad drecto of z ad w. It appears that z ad w have magtudes approxmately. ad., ad drectos approxmately ad, so oe estmate for the magtude s.. 6 ad a estmate for the drecto s + 6.. To fd the magtude of the product, multply the magtudes as you ormally would. To fd the argumet of the product, add the argumets ad the subtract 6, so that the argumet of the product s betwee ad 6. 6. a z z z z z 6. The Multplcato Law Check Your Uderstadg. a The magtude of z s z z 9, ad the drecto s argz + argz +. The powers of rotate 9 couterclockwse, ad. Mathematcs III Solutos Maual Chapter 6, page 8
b d z z z z z z z z z z c The powers of rotate 9 clockwse, ad. Oe reaso ths happes s that s the cojugate of. e The powers of rotate clockwse, ad each power has tmes the magtude of the oe before t. Ths occurs sce s the cojugate of +. z z z z + z z 8 6 z z + z 6 8 The powers of + rotate couterclockwse, ad each power has tmes the magtude of the oe before t. A full rotato happes at + 8 6, ad ths ca be foud usg the fact that 8 6. The powers of + rotate about 6.6 couterclockwse, ad each has tmes the magtude of the oe before t. Mathematcs III Solutos Maual Chapter 6, page 8
f b z z 6 8 6 8 z z The powers of rotate about 6.6 clockwse, ad each has tmes the magtude of the oe before t. Ths occurs sce s the cojugate of +. O Your Ow 7. a The product has magtude. b Ths sum has ukow magtude. The magtude could be aythg from to. c The magtude of the cojugate x y s always the same as that of x + y, so the magtude of w must be. d The magtude of the recprocal s the recprocal of the magtude, whch s stll. e The magtude of z s. f The magtude of z s, ot. 8. a 8 + 6 6 + 8 6 6 8 7 6 8 c 8 + 7 6 + 6 + 8 6 6 8 6 8 Ths tragle s rotated 9. + 8 6 8 + 6 + 8 6 6 8 6 8 Ths tragle s rotated ad each sde s tmes as log. 9. a Aswers vary. Oe possble aswer s + ad +. I geeral, the product must have magtude ad argumet 9, so you eed to pck umbers wth magtudes whose product s ad argumets whose sum s 9 or 9 more tha a multple of 6. There are a lot of possbltes. b If zw, the w. To dvde, multply by the z cojugate dvded by tself: w z z z z z z z z Alteratvely, fd the magtude ad argumet of w. The magtude would be dvded by the magtude of z, ad the argumet would be 9 mus the argumet of z.. Aswers vary, depedg o the estmates for the magtude ad argumet for z ad w. Oe estmate s that z has magtude ad argumet ad w has magtude. ad argumet. The the estmate for zw gves ts magtude as. 7. ad ts argumet as +. These are estmates, so aswers wll vary qute a bt. Mathematcs III Solutos Maual Chapter 6, page 8
. a b z s the crcle wth radus cetered at the org the ut crcle: w + z The magtude of z s ad the argumet s about 6. sce ta.7 ad z s Quadrat IV. The magtude of w s ad the argumet s about 7.6 sce ta 7.6. b Add the argumet agles: 7.6 + 6. 97.9. Ths agle s Quadrat I, equvalet to 97.9 6 7.9. c The magtude of zw. The argumet s 7.9 see prevous part.. a + + 7 + + b The magtude of z s gve by w 7 9 + + To fd the argumet, use verse taget: z arg ta ta 8. w 7 7 c The magtude of z s the magtude of z dvded by w the magtude of w. The argumet of z s the w argumet of z mus the argumet of w.. a z s the crcle wth radus cetered at the org: c z < s the rego sde the ut crcle: d z > s the rego outsde the ut crcle: e Let z a + b. The z a + b ad z. If z, the a +b z a + b, so a +b a + b, whch Mathematcs III Solutos Maual Chapter 6, page 8
characterzes all complex umbers z a + b so that z. As show part b, ths s the ut crcle: f If z z, the z z zz, so ether z or z : g If z z, the z z z z, so ether z or z. The oly pot where z s the org the umber +, ad z descrbes the ut crcle.. The umber cs π. Multplyg by ths adds π to the argumet. The correct aswer choce s C. Mata Your Sklls. a + + + + b + + + + + + c + + + + + d + + + + + + + + e + + + + + + 6 + + + 6 + + f Each set of four + + +, + + 6 + 7,..., 6 + 6 + 6 + 6, 6 + 6 + 66 + 67 sums to, ad ths sum eds wth a complete set of four, so the sum s zero. 6. a The sum s + + +, ad all terms cacel out to leave. b Sce + ω + ω, the sum s ω, whch equals. c The sum s ω + ω + + +. d The sum s zero, sce each set of three terms adds to. e The sum s, sce ω 6 ad the other terms add to. f Use the patter establshed the earler parts. The sum to ω 67 wll be the same as the sum to ω, whch s +. Each set of three terms cacels, ad there are a total of 68 terms. 6A MATHEMATICAL REFLECTIONS. Sce ad arg 9, multplcato by results a scale factor of ad a rotato of 9 couterclockwse aroud the org.. The complex umbers z x + y wth magtude are those such that x + y, whch s equvalet to x + y, whch descrbes the crcle wth radus cetered at the org.. Aswers wll vary. I geeral, you wat umbers of the form r cs rcos + s r + r Three examples wth magtudes,, ad are +, +, ad +.. Let z a + b. The z z a + ba b a b a + b z Mathematcs III Solutos Maual Chapter 6, page 8
. We have z z z z z z z z z z z 6. If z r ad argz θ, the z r cs θ r cos θ + r s θ 7. If z a + b, the the magtude of z s z a + b ad the argumet of z, deoted argz, s the agle the approprate quadrat havg a referece agle of ta b. a 8. + ad ta 6,so arg 6 6. + 9 + 9 6, ad ta ta,so arg + 8. Therefore, usg the Multplcato Law, ad + 6 argz + 6 9, so + s the complex umber wth magtude ad argumet 9. That s, +. O The solutos are the four vertces of a square.. a z, z 6, z.8.6.. z, z, z 9... z, z, z 8 b..6.8 z, z 7 z, z, z, z, z, z, z 6, z 7, z 8, z 9, z INVESTIGATION 6B DE MOIVRE S THEOREM 6. Gettg Started z, z 8.8.6 z, z 7 For You to Explore. a Aswers may vary. b I geeral, pck large umbers wth the fewest factors. Some strateges clude Oly the frst player who pcks a prme umber wll score, sce after that move, wll be off the table ad the player wll lose a tur. So 9 s a good frst choce. Perfect squares wll always gve hgher totals to the player tha to the oppoet. So s a good choce.. You ca solve the equato by factorg over C. x x x + x + x x + x The four solutos are thus,,, ad... z, z 9.. z, z 6...6.8 z, z z z, z +, z +, z, z, z, z6, z 7 +, z8 +, z9, z Mathematcs III Solutos Maual Chapter 6, page 86
c e z z z 9 z.8.6.. z 8 z z 7 z z 9.. z.. z z z z z z z 8 z.6.8 z 6 z 6 z 7 z d z, z +, z 7 +, z 7 +, z 7 6 6 6, z.78.997, z 6.7.689, z 7.978 +.6, z 8. +.966, z 9.7 +.88, z.988 +. z z.8. z z, z... z 6.6...6.8 z z 7 z 8 z z 9 z f z.... z. z, z +, z, z +, z, z, z 6 8, z 7 8 8, z 8 6, z 9 6 + 6, z z z. z z 9 z 8 z 6...6.8 z, z.6 +.6, z, z.88 +.88, z 6, z.., z 6 6, z 7.., z 8 6, z 9. +., z.. a If z, the powers fall alog a crcle. b If z >, the powers spral out away from the org. c If < z <, the powers spral toward the org.. z 7. z z, z cs 6, z cs 7, z cs 8, z cs, z cs 8, z 6 cs 6, z 7 cs, z 8 cs 88, z 9 cs, z cos πk cos + cos π + cos π + cos π + + + Mathematcs III Solutos Maual Chapter 6, page 87
6. 7. 8. x x x + x + Oe root s. To fd the other two, use the quadratc formula. ± ± So the other two are + ad. O ± The pots are the three vertces of a equlateral tragle. s πk s + s π + s π + + 6 k cos πk 7 cos π 7 + cos π 7 + cos 6π 7 + cos 8π π π + cos + cos 7 7 7.6..99.9. +.6 O Your Ow 9. a b 6 c 88 d e f 7 or, more accurately, 7 mod 6. a cosx + 9 cos x cos 9 s x s 9 s x s x b cosx+x cos x cos x s x s x cos x s x c Depedg o how you go about smplfyg the expresso, you may ed up wth dfferet results. Here s oe way: cos x cos x s x cos x s x s x cos x cos x + s x s x cos x s x cos x cos x + s x 6 s x cos x A secod way starts smlarly, but eds wth a smpler expresso at the ed. cos x cos x s x cos x s x s x cos x cos x s x cos x cos x + cos x s x cos x cos x cos x s x cos x cos x s x s x cos x s x cos x s x cos x 8 s x cos x d Use your results from part c. For ether result, replace s x wth cos x. If you got the frst result, you have to make two replacemets: cos x + s x 6 s x cos x cos x + s x 6 s x cos x cos x + cos x 6 cos xcos x cos x + + cos x cos x 6 cos x + 6 cos x 8 cos x 8 cos x + If you got the secod expresso, you have to make just oe replacemet: cos x 8 s x cos x 8 cos xcos x 8 cos x + 8 cos x Ether way, you ed up wth the same expresso.. a Yes, sce x x + x, ad x + x appears the gve factorzato of x 6. b Yes, sce x 6 x + x. Also, x x x + x +, ad x x + x + appears the gve factorzato of x 6. c No. x x + x x + x + x. So order for x tobea factor of x 6, x + would have to be a factor of x x + x + x +, but t s ot. Thk about t a dfferet way: f x s a factor of x 6, the ay soluto for x would also be a soluto for x 6. s a soluto for x, sce, ad. But s ot a soluto for x 6, sce 6, ad.. x orx ; + 8;. Use the quadratc formula. x + or x ; + + 8; + 6 7. x,x.9 +.9, x.9.9, x.89 +.878, or x.89.878 Mathematcs III Solutos Maual Chapter 6, page 88
. e 7 cos πk 8 s + s π + s π + s π + s π + s π + O s π + s 7π +.77 + +.77 + +.77 + +.77 6.6 Powers of Complex Numbers The pots appear to be the fve vertces of a regular polygo. 6. a.9 b.89 c.89 Mata Your Sklls 7. a b c cos πk 6 d s πk s πk 6 7 s k s + s π + s π + s π + + + s + s π + s π + s π + s π + s π +.866 +.866 + +.866 +.866 cos + cos π + cos π + cos π + cos π + cos π +. +. + +. +. s + s + s 9 + s + s 8 + s + s 7 + s +.77 + +.77 + +.77 + +.77 Check Your Uderstadg. a z cs π cs π 9 cs π,so z 9 ad argz π. b z cs π cs π 7 cs π,so. z 7 ad argz π. c z cs π cs π z ad argz π. cs π,so d z cs π cs π,so z ad argz argz π. e z cs π cs π cs, so z ad argz. Note that ths preserves the dea that aythg to the power other tha s equal to, so z cs. cs π, f z cs π cs π so z ad argz π π π. a cs π r cs θ cs π r cs θ cs π So r ad θ π + πk for teger values of k. If r, the r ±. If θ π + πk, the θ π + πk for teger values 6 of k. The two values for θ the terval [, π are π 6 ad 7π 6. So ow you have two possble values for r, ad two possble values for θ. The combatos are thus cs π 6, cs π 6, cs 7π 6, ad cs 7π 6. Does that mea you have possble solutos? Well, o. Notce that sce π 6 s 8 from 7π 6, the uts the drecto of π wll lad you the same spot 6 as uts the drecto of 7π 6 sce uts dcates that you move backward. Thus, cs π 6 cs 7π 6, ad cs 7π 6 cs π. So, as expected, there are oly two dstct 6 solutos.. a cs, so z 6, ad argz. b z 6, ad argz 6 or argz. Mathematcs III Solutos Maual Chapter 6, page 89
c z 8, ad argz π 6 π. d 8 8. 8 8, ad arg8 π. Also, ote that arg π,so arg8 π 9π, or arg8 π.. To add, t s easer to wrte the three umbers usg the form x + y: cs π 6 + cs π 6 + cs π + + + + + To multply, use the r cs θ form. cs π cs π cs π 8 cs 6 6 8 cs π 6 8 cs π 8 π 6 + π 6 + π. a a s a soluto f ad oly f a. By the law of expoets, a a a, ad, sce a s a soluto, a, so a. Thus, sce a, a s a soluto to x. b Follow the same logc as part a: a k s a soluto f ad oly f a k. By the law of expoets, a k a k, ad, sce a s a soluto, a, so a k k. Thus, sce a k, a k s a soluto to x. O Your Ow 6. You ca solve by factorg: x + x 8x 6 x 8x + x 6 xx 8 + x 8 x + x 8 x + x + x 8. Use De Movre s Theorem: cs x cs x, sofyou expad cs, the real part wll be a formula for cos x. cos x + s x cos x + cos x s x + cos x s x + cos x s x + cos x s x + s x cos x + cos x s x cos x s x cos x s x + cos x s x + s x cos x cos x s x + cos x s x + cos x s x cos x s x + s x So cos x s equal to the real part, that s cos x cos x cos x s x + cos x s x 9. Use the facts that cs π ad cs. By De Movre s Theorem, cs π cs π. For eve, k for some teger k. Thus, π πk πk, whch s equvalet to for ay teger k. Sof s eve, cs π cs. For odd, k +. Thus, π πk + π + πk. Ad, from above, πk s equvalet to for ay teger k, soπ + πk s equvalet to π for ay teger k. So, f s odd, cs π cs π.. By De Movre s Theorem, cs x cs x. Frst, to wrte cs form, you eed to kow that ad arg π. So, cs π. The cs π cs π. To see the powers of, frst substtute,,, ad : cs cs π cs π cs π 7. So the solutos are,, ad. The sum of the solutos s + +. The product of the solutos s 6. cos θ + s θ cos θ + s θ + s θ cos θ cos θ s θ+ s θ cos θ Also, by De Movre s Theorem, cs θ cs θ cos θ + s θ Sce both expressos are complex umbers, they are equal f ad oly f both ther real parts are equal ad ther magary parts are equal. Thus, cos θ cos θ s θ, s θ s θ cos θ ad If you crease ay by a multple of four, you get the same result, sce se ad cose are perodc.. a cs π cs π cs b Accordg to part a, you wat cs θ for ay value of θ where θ s equvalet to. Whe θ π k for ay teger k, θ πk, whch s equvalet to. So you wat to test teger values of k. If k, you have cs, whch was already covered the problem statemet. If k, you have cs π, the soluto show part a. For k, you have cs π. Check: cs π cs π cs π cs. k gves the soluto cs 6π. k gves the soluto cs 8π. k gves the soluto cs π cs π, whch s already covered. Addtoal values for k wll oly repeat the solutos already foud. Mathematcs III Solutos Maual Chapter 6, page 9
. a y O b By De Movre s Theorem, cs x cos x + s x cos x + cos x s x + cos x s x + s x cos x + cos x s x cos x s x s x cos x cos x s x + cos x s x s x Equatg real parts, cos x cos x cos x s x. You wat to show that cos x cos x s x cos x cos x The equato above s a detty, sce cos x cos x s x cos x cos x cos x cos x cos x + cos x cos x cos x. a From Exercse, you kow that. cos x cos x cos x Sce cos x, you ca solve the equato cos x cos x to fd the possble values for cos x. You ca save some wrtg by lettg v cos x. v v vv vv v + v, ± So cos x,, or. b The agles betwee ad π are cos, cos, ad cos π 6 π, π x, whch are π, π 6, ad. The correspodg agles betwee π ad π are 6, ad 7π 6. π s ot oe of the fredler agles, but you ca use ts relatoshp to fredler agles to get the soluto. You kow that cos π 6. Ad π 6 π.souse the formulas for double agles: cos x cos x s x cos x Solvg for cos x yelds so cos x + cos x + cos π cos π 6 + +. Sce cs π,fx cs θ, the cs θ cs π cs θ cs π θ π θ π The correct aswer choce s B. Mata Your Sklls 6. a x cs or x cs π b x cs, or x cs π, or x cs π c x cs, or x cs π, or x cs π, or x π d x cs, or x cs π, or x cs π, or x cs 6π, or x cs 8π e x cs, or x cs π, or x cs π, or x cs π, or x cs π, or x cs π f x cs, or x cs π 7, or x cs π 7, 7. a 6 or x cs 6π 7, or x cs 8π 7 or x cs π 7 s πk 7, or x cs π 7, s + s π 7 + s π 7 + s 6π 7 + s 8π 7 + s π π + s 7 7 +.788 +.979 +.9.9.979.788 b 6 πk cos 7 cos + cos π 7 + cos π 7 + cos 6π 7 + cos 8π π π + cos + cos 7 7 7. +.6..9.9. +.6 Mathematcs III Solutos Maual Chapter 6, page 9
c 6 πk cos 7 d 8 πk s 9 cos + cos π 7 + cos 8π 7 cos 6π π + cos 7 7 cos + cos π 7 + cos 8π 7 + cos π 7 + cos π 7 + π + cos 7 + π + cos 7 cos π 7 + cos 6π 7...9 +.6 +.6.9. s + s π 9 + s π 9 + s 6π 9 + s 8π 9 + s π π π 6π + s + s + s 9 9 9 9. +.68 +.988 +.866 +...866.988.68 e 8 cos k cos + cos + cos 8 + cos + cos 6 + cos + cos + cos 8 + cos. +.766 +.76..997.997. +.76 +.766 f By ow, you may have fgured that the patter s that the sum s always equal to. For ths part, though, be careful: the sum oly goes through terms, but the deomator s. So that last term s ot there. But sce ad thus cos πk, cos πk cos πk + cos π cos π 6.866 If you worked t out as the prevous parts have bee worked out, you should have gotte the same result. 6.7 Roots of Uty Check Your Uderstadg. a ω. 8. 6.. ω... ω, ω..6.8 b ω ω ω ω cs π cs π cs π cs π cs π cs π cs π cs π. a + b + ω, ω ω 6 cs π cs π cs 6π cs 8π π cs + π + 6π + 8π π cs cs π + + + + + + + + s πk 6 + s + s π + s π + s π + s π + s π +.866 +.866 + +.866 +.866 c Sce the sum of the roots of uty s, all the magary parts must total, ad the magary parts are s πk 6 for k,,... Mathematcs III Solutos Maual Chapter 6, page 9
. b The fgure below shows the graph from Exercse. Alteratg pots are also roots of x 6..8.8.6.......6.8. a.8.6.......6.8.6.......6.8 So, from the graphs, all solutos to x 6 are also solutos to x. c If ζ s a soluto to x 6, the x ζs a factor of x 6. But, from part b, ζ s also a soluto to x, so x ζs also a factor of x. The same wll be true for all sx roots of x 6. So all the dstct factors of x 6 wll be factors of x, ad thus x 6 must be a factor of x.. As suggested Exercse, 6 dvdes, so ay 6th root of uty wll be a th root of uty. Smlarly, f a umber s a th root of uty, t wll also be a th root of uty, a th root of uty, ad, geeral, a kth root of uty for ay postve teger k. 6. cs. If ths s a th root of uty, the w cs cs Thus, 6k for some atural umber k. Smplfyg yelds 6k. The smallest teger that works s 6. 7. You ca plot the 6th roots of uty by plottg cs for all values of,,...6. So ay teger s a 6th root of uty. To fd a smaller value, compare wth 6. Both are dvsble by, so f you plotted cs for all values of,,...6, you d get cs,so t s a 6th root of uty. Now, sce s prme, you ca t break t dow ay further to equal peces, so ca t be ay smaller. 8. a x s a fourth root of uty, ad t les betwee z ad w. b To fd a root betwee z ad x, thk about how the roots are spread over the ut crcle: z s the dstace aroud the crcle, couterclockwse, ad x s the dstace. So your goal s to fd a measure betwee ad. There are a umber of measures; Mathematcs III Solutos Maual Chapter 6, page 9
oe s. The fgure below shows the values for z, w, 9 x, ad y cs π 9. w.8.6.......6.8 x c Ay ratoal umber betwee ad ca represet a root of uty. Thk about t: f you start at ad dvde the ut crcle evely to parts, the each edpot of the dvsos s a th root of uty. So a pot o the ut crcle that s m of the crcle away from, where <m<, wll always be a root of uty. Ad the complex umber ca be wrtte as cs πm, sce the rada measure of m of the crcle wll be π m. So, the questo ow becomes whether you ca always fd a ratoal umber betwee two other ratoal umbers. You mght recall Farey sequeces: the Farey sequece F for ay postve teger s the set of rreducble ratoal umbers a b wth a b ad gcda,b arraged creasg order. For stace, { F, } { F,, }. F y {,,,,,,,,,, } A key attrbute of members of ay Farey sequece s that you ca fd ay fracto the sequece from the two eghborg fractos: suppose three cosecutve terms the sequece are a b, c d, ad e f.it wll always be true that c d a + e b + f For stace, F, the term ca be foud from ad +, sce + 9. The upshot s that you ca z fd a fracto betwee ay two fractos by addg ther umerators ad addg ther deomators. Sce ay root of uty s a fracto of the crcle, you ca fd a root of uty betwee two others by addg ther umerators ad deomators. If you followed ths approach, the you could say that to fd x, use the fractos ad : + + 8, the value foud above. To fd the root betwee z ad x, follow the same process: + + 9. O Your Ow 9. Let z, z,... represet the 8th roots of uty, ad w, w,... represet the th roots of uty. The fgure below shows all the roots. w w z.8.6.. z... w z..6.8 The oly pot o the graph that s shared by both sets of roots s. If x were a factor of x 8, the every root of x would also be a root of x 8. But, from the graph, oly oe th root of uty,, s also a 8th root of uty.. a b c If you work through several examples, you wll fd that whe s eve, the product s. Whe s odd, the product s.. The roots of x 6 are,, +,, +, ad. a 6 6,sox 6 6. Image factorg the expresso o the left sde. Whe you factor x 6, each factor wll be the form x ζ, where ζ s a sxth root of uty. So whe you factor x 6 6, each factor wll look lke x ζ.so the roots are twce the roots for x 6. That s,,, +,, +, ad. b Sce 6 6, oe of the sx solutos s. But f s a soluto, the ζ s a soluto, where ζ s ay of the sxth roots of uty. That s,,, +,, +, ad. z z 6 w z z 7 Mathematcs III Solutos Maual Chapter 6, page 9
. a The sum of the eghth roots of uty s : cs + cs π + cs π + cs π + cs π + cs π + cs π + cs 7π + + + + b 7 + + cos πk 8 + + + cos + cos π + cos π + cos π + cos π + cos π + cos π + cos 7π + + + + + + + c The sum from part b represets the real part of the sum from part a. Sce the sum part a was, the real part s.. If z s ay root of uty, the ts magtude must be. The roots dvde the crcle evely. Sce the ut crcle has crcumferece π, the each arc legth s π 9,sothe argumets wll be of the form πk 9.. The roots of uty evely dvde a crcle to equal-szed arcs. So ay root of uty wll be a fracto of the crcle, that s, of the form cs k π. Sce k ad are both tegers ad k<, the k s a ratoal umber. Thus, f a s ratoal, cs π a s a root of uty. Ad f a s rratoal, the cs π a s ot a root of uty.. a Let z cos θ. The z cos θ + s θ cos θ s θ cos θ + s θcos θ s θ cos θ s θ cos θ + s θ cos θ s θ z b Recall that cos θ cos θ, ad s θ s θ. Thus, cos θ+ s θ cos θ s θ z. Sce cs θ s a root of uty, the θ s of the form πa, where a s ratoal. But the θ s of the form π a, ad f a s ratoal, so s a. Thus, z s a root of uty. 6. a Aother value of for whch ths works s + + +, ad the product of the frst teth roots of uty s also. b If ω s a th root of uty, the ω. The product of the frst k roots of uty wll be ω ω ω ω k ω ++++ +k so the product of a subset of cosecutve roots of uty wll equal f the sum of the frst k roots of uty equals. You may recall that the sum of the tegers from to k s equal to kk+. So you wat to fd stuatos where kk + For k or, you do t get a that satsfes the restrcto k<. If k, the kk+ 6, ad the product of the frst sxth roots of uty s, as stated the problem. If k, the kk+, ad the product of the frst teth roots of uty s : ω ω ω ω ω +++ ω, sce ω s a teth root of uty. That s the soluto from part a. So for ay value of, where s a tragular umber a sum of the frst k postve tegers, the product of the frst kth roots s. I fact, there are more choces: f ω, the ω a for ay teger a, so you also have to cosder cases where kk+ a. If you are usure why ths s true, cosder ths: let k m Now, suppose m a, where a ad are tegers, ad suppose ζ s a th root of uty. The the product of the frst kth roots of uty s ζ ζ ζ ζ k ζ k ζ m ζ a Now, sce ζ s a th root of uty, the ζ. Thus, t s certaly true that ζ a, whch meas that ζ m. There may be other cases that do ot follow ths patter. The soluto to the problem s curretly beg studed by researchers umber theory. 7. a Let be a odd teger. The the th roots of uty are cs πk πk, for k tok. Let a cs The oe of the two roots of x a s cs πk The other root s cos πk πk + s cos πk πk s cos π + πk + s cs If k s eve, the cs πk t s of the form cs πj + kπ.. π + πk s a th root of uty, sce.ifk s odd, the + k s Mathematcs III Solutos Maual Chapter 6, page 9
8. a eve, ad thus +kπ s of the form πj. So, as tally stated, f a s a root of uty, the ether a or a s a root of uty. b Suppose s eve. The f you follow the same process, you stll ed up wth choosg betwee πk. As before, f k s eve, the cs πk s a root of uty. But f k s odd, the ether form s a root of uty: + k s also odd, so ether fracto s the form πj. ad +kπ c d s πk 6 k s π cos πk cos Equato # New Roots x x x x x x 6 x 7 6 x 8 x 9 6 x b There are may aswers here. Frst, the umber of ew roots s always eve begg wth x. Ths ca be prove: f z s a ew root, the so s z, ad as log as z s oreal whch ca oly happe whe z orz, z ad z must come pars. Secod, f p s prme, the umber of ew roots of x p sp. Oe strog clam related to the clam about prmes s that the umber of ew roots of x s the umber of values {,,,..., } that do ot have commo factors wth. For example, cosder the solutos to x. If w s the frst prcpal root of uty foud, the ew roots are w, w,w 7, ad w 9. These umbers,, 7, ad 9 have o commo factors wth other tha. Numbers that share commo factors are t ew roots: for example, w 6 teth roots s the same as w ffth roots. 9. Ths factors as follows: x x x + x + x + x + The correct aswer choce s A. Mata Your Sklls. a b s πk cos πk 6 s π cos π 6.8 Geometry of Roots of Uty Check Your Uderstadg. Each power of w adds to the argumet. For example, w has argumet. To complete the regular polygo, the argumet must be 6 or a teger multple. The equato 6 has the soluto 8, so w 8 completes the regular polygo.. a The pots z,z,...z do ot form a regular polygo because they do ot dvde the crcle to eve peces. The last arc, betwee z ad s ot cogruet to the other arcs. b The powers of z wll come back to, sce z 6 cs 7 6 cs 6 7 cs. It takes 6 tmes aroud, because gcd7, 6. c Ay teger value of θ wll be a vertex o a regular 6-go of course, t could be a vertex o a regular -go for smaller values of. I fact, ay ratoal value of θ wll be a vertex o a regular polygo: f, say, z cs a, where a ad b are tegers, the z s a b vertex o a regular 6b-go, sce cs a 6b b cs 6b a b cs 6a.. a cs π cs π cs 8π cs π b π π z + z cos + s + cos π + s π + cos cos π π + s π s cos π π. a x x x + x + x + x + b z s a root of x. Sce z, z must be a root of the other factor, x + x + x + x +. Ad, sce z s a root, z + z + z + z +. Thus, z + z + z + z. Mathematcs III Solutos Maual Chapter 6, page 96
. a From Exercse, you ca see that a s postve, sce π s the frst quadrat, ad cose s postve Quadrat I. You ca show b s egatve the same way: b z + z cs π + cs π cs π + cs 6π cos π + s π + cos π cos π + s π + cos π s π cos π + s π Sce π s the secod quadrat, cos π s egatve. b Sce a + b z + z + z + z, from Exercse, a + b. c ab z + z z + z z + z + z 6 + z 7. Sce z, z 6 z ad z 7 z. Thus, z + z + z 6 + z 7 z + z + z + z. d From c, ab z + z + z + z whch, from Exercse, equals. 6. Use the sum ad product rule for quadratcs: x ax b x a + bx + ab So, f the sum of two umbers s ad the product s, the a ad b are the roots of the equato x + x Use the quadratc formula to get a + ad b. 7. a From Exercse, a cos π cos 7. But from Exercse 6, you kow that a + sce a s postve. Thus cos 7 a +. b Use the Pythagorea detty: s 7 + cos 7 s 7 + + s 7 8 s 7 + 8 s 7 + 8 s 7 + c You already have two: ad + + +. From the symmetry of the petago, you ca easly get a thrd: + by squarg: + + + 8 + 8 +. You ca fd a fourth + + + + + 6 + + + 8 Smplfyg the magary part ca be trcky. Workg o the umerator, you get + + + + + + + + + + + + Now, plug that back to where you left off: + + + 6 + + 8 Fally, the last root s the cojugate of the oe you just foud, + 8. You kow that x x x + x + x + x +, so you eed to show that x + x + x + x + x + φx + x φ x + Mathematcs III Solutos Maual Chapter 6, page 97
Frst, expad the rght sde: x + φx + x φ x + x φ x + φx + x φ φ x + x + φx φ x + Notce that the coeffcet of x eds up beg : x φ x + x x x + x x φ The coeffcets of x ad x are both φ. You ca φ smplfy by pluggg the value of φ: So x + φ φ + + + + + + + + φ x + x + φ x + x + x + φ φ x + x + O Your Ow 9. Each power of w adds to the argumet. For example, w has argumet 6. To complete the regular polygo, the argumet must be 6 or a teger multple. The equato 6 has the soluto, so w completes the regular polygo.. Drop a alttude from vertex A to the opposte sde, BC. Mark the tersecto pot D. Sce ABC s sosceles, D bsects BC. Sce cos 7 BD AB BD BD, ad, smlarly, cos 7 CD, the BD + CD cos 7 + cos 7 cos 7.. Sce BD bsects ABC, m ABD m DBC,or m DBC m ABC 7 6. Ad, you already kow that m DCB 7,som BDC 8 6 7 7. Thus, by AA, ABC BCD.. a From CPSTP, AB BC q. BC BC CD, thus CD BC BC AB b Sce m ABD 6, DAB s sosceles because DAB DBA. So, AD BD. Also, sce BDC s sosceles because ABC BCD, BD BC. Thus, AD BC. Gve that AC ad BC q, DC AC BC q.. a Rewrte q q as q + q, ad solve usg the quadratc formula: q ± ± b Sce q cos 7, cos 7 q, ad thus cos 7 ± ± Whch of the two values s cos 7? Sce., + s postve, ad s egatve. Sce cos 7 must be postve 7 s the frst quadrat, cos 7 +. Use the formula cos θ cos θ. Sce 7, the cos cos 7 cos 7 + 8. Aswers may vary. The exercses leadg up to ths oe suggest the followg way to justfy the statemet: Let A be oe vertex of petago ABCDE. Draw the two dagoals AC ad AD. ACD s a sosceles tragle wth base agles 7,so ACD s smlar to the tragle from Exercse. I that tragle, the legth of the loger sde was, ad the shorter sde was cos 7 so AC CD φ + +. So the rato AC CD + + + + + + φ Mathematcs III Solutos Maual Chapter 6, page 98
6. If the polygo has sdes, the π πkfor some 7 teger k. Therefore, k. The smallest postve teger ths could be s whe k. The correct aswer choce s C. Mata Your Sklls 7. There are two ways you ca go about solvg ths exercse. Frst, you ca use your calculator to fd a estmate for θ ad the calculate the coses as the problem asks. Or, you ca use the multple-agle formulas. Some you may have already derved, but each s derved aga here. Note, however, that dervg cos θ ad cos θ are pretty trcky, so t s fe f you decded to use estmates o your calculator. a cos θ cos θ.8.6.8 b You ca derve a formula for cosθ: cos θ cos θ + θ cos θ cos θ s θ s θ cos θ cos θ s θ cos θ s θ cos θ cos θ s θ cos θ cos θ cos θ cos θ cos θ cos θ cos θ cos θ + cos θ cos θ cos θ So f cos θ.8, the cos θ.8.8..8. c The formula for cos θ s a lttle more drect: cos θ cos θ cos θ So f cos θ.8, the cos θ cos θ cos θ + 8 cos θ 8 cos θ + cos θ 8.8 8.8 + 8.96 8.6 +.8 d To derve the formula for cos θ, use the fact that θ θ + θ. You wll eed the formula for s θ, so derve that frst: s θ s θ s θ cos θ s θ cos θcos θ s θ8cos θ cos θ So ow, o to cos θ: cos θ cos θ + θ cos θ cos θ s θ s θ 8 cos θ 8 cos θ + cos θ s θ8cos θ cos θs θ 8 cos θ 8 cos θ + cos θ s θ8cos θ cos θ 8 cos θ 8 cos θ + cos θ cos θ8cos θ cos θ 8 cos θ 8 cos θ + cos θ 8 cos θ cos θ+ 8 cos θ cos θ 6 cos θ cos θ + cos θ So, f cos θ.8, the cos θ 6.8.8 +.8.997 8. Let θ cos.8, Quadrat I. The z cs θ.you have already calculated the real part of each umber, sce by De Movre s Theorem, f z cs θ, the z cs θ cs θ. You ca work out all those formulas for se, or you ca just use your calculator to fd the magary part for each calculato. a.8 +.96 b. +.96 c.8 +.76 d.997.78 9. a By the defto, the frst pece defes P,x, ad the secod pece defes P,x x. b P,x x x x c P,x xx x x x x x x d P,x xx x x 8x 6x x + 8x 8x + P,x x8x 8x + x x 6x 6x + x x + x 6x x + x You may otce that these results look famlar, especally f you worked out the multple-agle formulas for cose. Note that P, cos θ cos θ. e P,.8.8.6.8 P,.8.8.8..8. P,.8 8.8 8.8 + 8.96 8.6 +.8 P,.8 6.8.8 +.8.997 Mathematcs III Solutos Maual Chapter 6, page 99
6.9 Arthmetc Wth Roots of Uty Regular Polygos Check Your Uderstadg. a 6 ζ k + ζ + ζ + ζ + ζ + ζ + ζ b c + ζ + ζ + ζ + ζ 6 + ζ + ζ Or use 6 PolyRemader x k,x 7 whch yelds + x + x + x + x + x + x 6.Oruse PolyRemader 6 x k, + x + x + x + x + x + x 6 whch yelds. 6 ζ k 6 ζ k 6 7 Or use 6 PolyRemader x k,x or 6 PolyRemader x k, both of whch yeld 7. x k 6 ζ k + ζ + ζ + ζ + ζ + ζ + ζ + ζ + + ζ + + ζ + + ζ But you also kow that ζ cs π cs π cs π So + ζ Or you ca use 6 PolyRemader x k,x whch yelds + x.oruse 6 PolyRemader x k, 9 x k whch produces + x, too. How ca we get the CAS to reduce the x? Look at the factorzato of x wth the CAS or by had. x + x + x x + x x + x + x + x + x + x ζ must be a zero of oe of these factors. It s ot a zero of ay of the degree- factors. Ad t s ot a zero of + x + x + x + x, because the zeros of ths are all th roots of uty, ad ζ, ot. So, use 6 PolyRemader x k, x + x x + x whch yelds.. a Start wth the factorzato x 6 x x + x + x + x + x + Put x α. The left sde s, so the rght sde s also. By ZPP, oe of the factors s. But α. So, α + α + α + α + α + b α cs 6π 6, so α +. c α cs π + a cube root of ad α s ts cojugate, so α + α.. If m, every term the sum s, so f.if ad are relatvely prme, the sum wll smplfy to k αk.. a Usg reasog smlar to that Exercse a, z k. b zk z z z z z c Expad the product ad reduce hgher powers to lower oes, or use PolyRemader x k, x k k Alteratvely, we kow the roots of x ad so we kow the roots of + x + x + x + x : they are z, z,z, ad z. So, by the factor theorem, + x + x + x + x x z k k Now put x to fd that the product equals. Mathematcs III Solutos Maual Chapter 6, page
. Oe way to do t s to use the explct formulas for z ad ts powers. They are + + +, + + + + + + +, +, The er sum s ±, so the value sought s. Or use PolyRemader z z z + z, z k 6. a There are several ways to do ths. Oe s to ote that z cs π 6 + ad to do the calculato explctly. Aother s to use the factorzato of x : x + x + x + x x + x + x + x x + x We kow the roots of all the factors but the last oe, ad oe of them s z,soz s a root of So, use x + x PolyRemaderx + x + x 7 + x, x + x The sum s. b Ths s the real part of the above sum, so t s. 7. a The roots of x are the powers of They are ζ cs π cs π 6 + {,ζ,ζ,ζ,ζ,ζ,ζ 6,ζ 7,ζ 8,ζ 9,ζ,ζ } Some of these ca be rased to powers smaller tha to produce : ζ 6, ζ, ζ, ζ 6 ζ 8, ζ 9, ζ 6, ζ That leaves powers of ζ, {ζ,ζ,ζ 7,ζ } The th power of each of these s, ad you ca check by rug aroud the -go that o smaller power produces. So, these are the four prmtve th roots of uty. See Lesso 6.7, Exercse for the graph. b Start wth the factorzato of x. x + x + x + x x + x + x + x x + x Now lst each factor wth the roots of the equato you get by settg that factor equal to : equato Root + x ζ + x ζ 6 + x ± {ζ,ζ 9 } x + x ± {ζ,ζ } + x + x ± {ζ,ζ 8 } x + x????? Now, every power of ζ s a root of oe of these equatos, so the oly oes left for the roots of the last oe are {ζ,ζ,ζ 7,ζ } These are the the roots of the last equato, a fact that you ca check drectly. O Your Ow 8. If 6m, every term the sum s, so f 6. If ad 6 are relatvely prme, you get all the powers of α some order, so f. If s 6m + or6m +, you get the sum of the cube roots of uty twce, whch equals. If s 6m +, you get + +. 9. Oe way to do ths s to calculate PolyRemader x + x x + x x x 6, 6 x k Aother way s to multply out ad the smplfy usg algebra ad the fact that z 7. Ths wll gve you + z + z + z + z + z + z 6 7 7 sce the sum of all the seveth roots s. Mathematcs III Solutos Maual Chapter 6, page
. You eed the sum ad product of α ad α. Sce z s a root of xk, the sum all the powers except for s. For the product, you could use PolyRemader. PolyRemader x + x + x + x + x 9 x + x 6 + x 7 + x 8 + x, Ths produces. The use x the sum of the roots x +the product of the roots x k to obta x + x +.. a Over Z: x x + x + x + x +. The cyclotomc detty gves ths factorzato. The secod factor does t factor ay further: There are o lear factors, because a lear factor over Z would produce a ratoal root, ad all the roots of x + x + x + x + are complex. I fact, they are the frst four powers of ζ cs π. There are o cubc factors because a cubc factor would have a compao lear factor. So, the oly possblty s that there are two quadratc factors wth teger coeffcets. But that ca t happe because each quadratc factor would have to have as roots two of the roots of x + x + x + x +. Ad they d have to be complex cojugates. So, oe par would have to be ζ ad ζ. But the sum of these two s ot a teger -t s cos π +. b Over R: x x + x + x + + x +.We kow from our prevous work that ζ + ζ cos π + ζ ζ ad So, x x x + x + x + x + x x + x + x + + x + Sce each quadratc factor has complex roots, t does t factor ay further over R. c Over C: x x ζ x ζ x ζ x ζ, where ζ cs π + + +. The powers of ζ le o the vertces of a regular -go:, ζ,ζ,ζ,...ζ,ζ Noe of the vertces s utl you get to the th power.. Suppose you dvde t by ad get a quotet q ad remader r, so that r<. I other words, wrte The t q + r r< ζ t ζ q+r ζ q ζ r ζ r But f r s ot, the a ozero power of ζ smaller tha would produce, cotradctg the result of Exercse. So r, ad dvdes.. If z ζ r, z s a th root of uty. We eed to show that z s ot a mth root of uty for ay teger m<. Suppose z m. z m ζ r m ζ rm By Exercse, s a factor of rm. But has o commo factors wth r,so s a factor of m. Ths mples that m, a cotradcto.. a The roots of x 9 are the powers of So, ζ ad ζ are roots of x + x + They are ζ cs π 9 cs π 9 {,ζ,ζ,ζ,ζ,ζ,ζ 6,ζ 7,ζ 8 } Smlarly, ζ ad ζ are roots of x + + x + But ζ, ζ, ζ ad ζ are also the roots of x + x + x + x +. It follows that x + x + x + x + x + x + x + + x + Some of these ca be rased to powers smaller tha 9 to produce : ζ, ζ 6, So,, ζ, ad ζ 6 are cube roots of uty. That leaves 6 powers of ζ, {ζ,ζ,ζ,ζ,ζ 7,ζ 8 } The 9th power of each of these s, ad you ca check by rug aroud the 9-go that o smaller power produces. So, these are the sx prmtve 9th roots of uty. Mathematcs III Solutos Maual Chapter 6, page