De Moivre s Theorem - ALL. Let x ad y be real umbers, ad be oe of the complex solutios of the equatio =. Evaluate: (a) + + ; (b) ( x + y)( x + y). [6]. (a) Sice is a complex umber which satisfies = 0,. Hece, + + = = 0. (M)(A) (b) (x + y)( x + y) = x + yx + xy + y. (M) Usig = ad =, we get, (M) (x + y)( x + y) = (x + y ) + ( + )xy (M) = x + y xy, (Sice l + + = 0) (A). (a) Express as a product of two factors, oe of which is liear. (b) Fid the eros of, givig your aswers i the form r(cos θ + i si θ) where r > 0 ad < θ. (c) Express + + + + as a product of two real quadratic factors. [0]. (a) = ( )( + + + + ) (C) (b) = 0 = cos 0 + i si 0 (accept = ). cos i si,cos i si (C) Accept cos i si,cos i si (c) cos cos i si cos i si i si cos i si Thus, + + + + = cos (M)(C) cos (M)(C) cos cos (C) + + + + = cos cos + + + + = ( 0.68 + )( +.68 + ) (C) (C)
. (a) Express the complex umber 8i i polar form. (b) The cube root of 8i which lies i the first quadrat is deoted by. Express (i) i polar form; (ii) i cartesia form. [6]. (a) 8i = 8, arg 8i = (or.7, 90 ) (A)(A) 8i = 8 cos i si (C) 6 (b) (i) =, arg = (or 0., 0 ) (A)(A) = cos i si (C) 6 6 (ii) = +i (or.7 + i) (A)(A) (C) [6] cos i si cos i si. Cosider the complex umber =. cos i si (a) (i) Fid the modulus of. (ii) Fid the argumet of, givig your aswer i radias. (b) Usig De Moivre s theorem, show that is a cube root of oe, ie =. (c) Simplify (l + )( + ), expressig your aswer i the form a + bi, where a ad b are exact real umbers. []. = cos isi cos cos isi isi (a) (i) = (A) cos isi cos isi (ii) = (M) cos isi arg = (M) = 6 = (A)
arg = or.09 radias (G) (b) = cos isi (M) = cos + i si (M) = l + 0i = (AG) (c) ( + ) ( + ) = + + + = + + + (sice = ) (Ml) = + + = + + + cos isi cos isi (M)(A) i (A)(A) = + + + + = + (sice + + = 0) (M) = + cos isi (M) i = + (A) = i (A). (a) Prove, usig mathematical iductio, that for a positive iteger, (cos + i si) = cos + i si where i =. (b) The complex umber is defied by = cos + i si. (i) Show that = cos ( ) + i si ( ). (ii) Deduce that + = cos θ. (c) (i) Fid the biomial expasio of ( + l ). (ii) Hece show that cos = (a cos + b cos + c cos ), 6 where a, b, c are positive itegers to be foud. []. (a) The result is true for =, sice LHS = cos θ + i si θ ad RHS = cos θ + i si θ Let the propositio be true for = k. Cosider (cos θ + i si θ) k + = (cos kθ + i si kθ)(cos θ + i si θ) = cos kθ cos θ si kθ si θ + i(si kθ cos θ + cos kθ si θ) (R) (M) (M)
= cos (k + )θ + i si (k + )θ (A) Therefore, true for = k true for = k + ad the propositio is proved by iductio. (R) (b) (i) cos isi cos isi cos isi (M) = (cos θ + i si θ) (A) = cos ( θ) + i si ( θ) (AG) = (M) = cos ( θ) + i si ( θ), by de Moivre s theorem (accept the cis otatio). (A)(AG) Note: Award (M0)(A0) to cadidates who use the result of part (a) with o cosideratio that i this part, < 0. (ii) = ( ) = cos ( θ) + i si ( θ) (A) + = cos θ + i si θ + cos ( θ) + i si ( θ) (M) = cos θ + i si θ + cos (θ) i si (θ) (A) = cos θ (AG) (c) (i) ( + ) = + + 0 + 0 + + (M)(A) (ii) ( cos θ) = cos θ + 0 cos θ + 0 cos θ (M)(A) givig a =, b = ad c = 0(or cos θ = 6 (cos θ + cos θ + 0 cos θ)) (A) 6. (a) Use mathematical iductio to prove De Moivre s theorem (cos + i si) = cos () + i si (), +. (b) Cosider = 0. (i) Show that = cos i si is oe of the complex roots of this equatio. (ii) (iii) Fid,,,, givig your aswer i the modulus argumet form. Plot the poits that represet,,, ad, i the complex plae. (iv) The poit is mapped to + by a compositio of two liear trasformatios, where =,,,. Give a full geometric descriptio of the two trasformatios. [6] 6. (a) (cos θ + i si θ) = cos (θ) + i si (θ), Let = cos + i si = cos which is true. (A) Assume true for = k (cos + i si) k = cos (k) + i si (kθ) (M) Now show = k true implies = k + also true. (M) (cos + i si) k+ = (cos + i si) k (cos + i si ) (M) = (cos (k) + i si (k)(cos + i si) = cos (k) cos si (k)si +i(si (k) cos + cos (k) siθ) (A) = cos (k + ) + i si (kθ + θ) +
= cos (k + ) + si (k + ) = k + is true. (A) Therefore by mathematical iductio statemet is true for. (R) 7 (b) (i) = cos isi = (cos + si ) (M) = Therefore is a root of = 0 (AG) (ii) cos 8 cos 6 cos 6 8 isi 6 isi 8 isi = (cos + i si )(= (cos 0 + i si 0) = ) (A) Note: Award (A) for all correct, (A) for correct, (A0) otherwise. (iii) Im 6 8 6 6 8 0 6 8 0 6 8 0 Re 6 8 0 6 (A)(A) Note: Award (A) for graph of reasoable sie, scale, axes marked, (A) for all poits correctly plotted, (A) for poits correctly plotted. (A) for poits correctly plotted. (iv) Composite trasformatio is a combiatio of (i ay order) a elargemet scale factor, cetre (0, 0); a rotatio (ati-clockwise) of (7 ), cetre (0, 0) 8 or clockwise (88) (A) (A) 9 Note: Do ot pealie if cetre of elargemet or rotatio ot give. [6] 7. Give that, solve the equatio 8i = 0, givig your aswers i the form = r (cos + i si). [6]
6 7. 8i 8cos i si (A) r (cos i si ) where r 8 ad ( cos i si 6 6 cos i si 6 6 cos isi or ) cos i si (A)(A) (A) (A) (A) (C6) 8. Cosider the complex umber = cos + i si. (a) Usig De Moivre s theorem show that + = cos. (b) By expadig show that (c) Let g (a) = cos d. (i) a 0 Fid g (a). cos = (cos + cos + ). 8 (ii) Solve g (a) = [] 8. (a) = cos + i si = cos ( ) + i si ( ) (M) = cos i si ( ) (A) Therefore cos (AG) (b) 6( ) (M) 6 (M) ( cos ) = cos θ + 8 cos θ + 6 (A) cos 6 (cos 8cos 6) (A)
7 (cos cos ) (AG) 8 a a (c) (i) cos d (cos cos ) d (M) 0 8 0 si si 8 g ( a) si a si a a 8 a 0 (A) (A) (ii) si a si a a 8 a =.96 (A) Sice cos θ 0 the g (a) is a icreasig fuctio so there is oly oe root. (R) 9. Let = cos + i si, for. (a) (i) Fid usig the biomial theorem. (ii) Use de Moivre s theorem to show that cos = cos cos ad si = si si. si θ si θ (b) Hece prove that = ta. cosθ cosθ (c) Give that si =, fid the exact value of ta. [] 9. (a) (i) (cos + i si) = cos + cos i si + cos i si + i si AAAA Note: (= cos + cos si i cos si i si ) (= cos cos si + ( cos si si ) i) Award A for each term i the expasio. (ii) (cos + i si) = cos + i si (A) equatig real ad imagiary parts (M) cos = cos cos si A = cos cos ( cos ) A = cos cos AG N0 ad si = cos si si A = ( si ) si si A = si si AG N0
8 (b) si θ si θ si θ si θ si θ AA cosθ cosθ cos θ cosθ cosθ si θ si θ = AA cosθ cos θ Usig si = cos = cos M si θ si θ si θ cosθ cosθ cosθ A (c) METHOD si θ cosθ si θ 7 = ta AG 7 = (A) A M cosθ = 0 7 METHOD ta θ 0 0 si θ si θ 7 M A M = (A) 7 N0 0 cosθ 7 7 ta θ 7 0 7 MA = A N0 0 0 0. Let y = cos + i si. dy (a) Show that = iy. dθ [You may assume that for the purposes of differetiatio ad itegratio, i may be treated i the
9 same way as a real costat.] (b) Hece show, usig itegratio, that y = e i. (c) Use this result to deduce de Moivre s theorem. si 6θ si θ (d) (i) Give that = a cos + b cos + c cos, where si 0, use de Moivre s theorem with = 6 to fid the values of the costats a, b ad c. si 6θ lim 0 si θ (ii) Hece deduce the value of. [0] 0. (a) dy si θ i cosθ dx A EITHER d y i dθ si θ i cosθ A = i (cos + i si ) A = i y AG N0 i y = i(cos + i si) (= i cos + i si) A = i cos si A dy = dθ AG N0 (b) dy i y dθ MA l y = i + c A Substitutig (0, ) 0 = 0 + c c = 0 A l y = i A y = e i AG N0 (c) cos + i si = e i M = (e i ) A = (cos + i si ) AG N0 Note: Accept this proof i reverse.
0 (d) (i) cos 6 + i si 6 = (cos + i si) 6 M Expadig rhs usig the biomial theorem = cos 6 + 6 cos i si + cos (i si) + 0 cos (i si) + cos (i si) + 6 cos (i si) + (i si) 6 Equatig imagiary parts si 6 = 6 cos si 0 cos si + 6 cos si si 6θ si θ MA (M) A 6 cos 0 cos ( cos ) + 6 cos ( cos ) A (ii) si 6θ lim θ 0 si θ = cos cos + 6 cos (a =, b =, c = 6) A N0 lim θ 0 cos θ cos θ 6cosθ M = + 6 = 6 A N0. Prove by iductio that + ( ) is a multiple of 7 for +. [0]. Let P() be the propositio + ( ) is a multiple of 7 Whe = + ( 0 ) = (which is a multiple of 7) P() is true Assume P(k) is true, ie k + ( k ) = 7m, m + R M For = k + k + + ( k ) = ( k ) + ( k ) M = (7m k ) + k A = (7m k ) + 0 k A = (7m) ( k ) + 0( k ) A = 7(m k ) A k + + ( k ) is a multiple of 7 So P(k) is true P(k + ) is true, ad P() is true. R Hece by iductio P() is true. R N0 Note: Award the fial RRR oly if the previous reasoig is correct. R [0]. Prove that i i is real, where +. [6]. EITHER i cos isi 6 6 i cos isi 6 6 A A
Hece usig De Moivre s Theorem i i cos isi 6 6 cos isi 6 6 MA = cos isi cos isi (A) 6 6 6 6 i i = cos which is real. R 6 i i!!... MA i i i i!!... The terms i odd powers of i are of opposite sig i each series expasio ad hece cacel. MA R Hece + is real. R i i i + i has the form R = MA = Re A expressio is real R a. Express i the form where a, b. [] i b. METHOD r, θ i METHOD cos i si cos isi 8 8 (A)(A) M (M) A ( i )( i ) = i (= i ) (M)A
i i ( )( ) = 8 (M)(A) i 8 METHOD Attempt at Biomial expasio A M i i i i ( ) = + ( ) + ( ) + ( ) (A) i 8 i i = 9 + (A) = 8 A M. Let w = cos isi. (a) Show that w is a root of the equatio = 0. (b) Show that (w ) (w + w + w + w + ) = w ad deduce that w + w + w + w + = 0. (c) Hece show that cos cos. []. (a) EITHER w cos isi (M) = cos + i si A = A Hece w is a root of = 0 AG Solvig = (M) = cos isi, 0,,,,. A = gives cos isi which is w A (b) (w )( + w + w + w + w ) = w + w + w + w + w w w w w M = w A Sice w = 0 ad w, w + w + w + w + = 0. R (c) + w + w + w + w = cos isi cos isi
cos isi cos isi (M) Notes: Award M for attemptig to replace 6 ad 8 by ad. Award A for correct cosie terms ad A for correct sie terms. Note: Note:. = i ad = (a) cos isi cos isi 6 6 8 8 cos isi cos isi cos isi cos isi cos isi cos isi cos cos 0 cos cos m i. Correct methods ivolvig equatig real parts, use of cojugates or reciprocals are also accepted. Use of cis otatio is acceptable throughout this questio. M MAA Fid the modulus ad argumet of ad i terms of m ad, respectively. (b) Hece, fid the smallest positive itegers m ad such that =. [] A AG. (a) i or i (A) 7 arg i or arg ( i) = accept (A) m A A arg ( ) = m arcta m A 7 arg ( ) = arcta () = accept A N (b) m m m k, where k isa iteger (M)A MA
m k m m k m k 6 m k (M) A The smallest value of k such that m is a iteger is, hece m = A =. A N []