= + then for all n N. n= is true, now assume the statement is. ) clearly the base case 1 ( ) ( ) ( )( ) θ θ θ θ ( θ θ θ θ)

Transcription

1 Complex numbers mixed exercise i a We have e cos + isin hence i i ( e + e ) ( cos + isin + cos + isin ) ( cos + isin + cos sin) cos Where we have used the fact that cos cos sin sin b We have ia ia ib ib i( A+ B) i( A B) i( B A) i( A+ B) cosacosb ( e + e ) ( e + e ) e + e + e + e cos + + sin + + cos + sin + cos + sin + cos ( + ) + sin ( + ) cos( A+ B) + cos( A B) ( cos ( A+ B) + cos ( A B) ) ( ( A B) i ( A B) ( A B) i ( A B) ( B A) i ( B A) ( A B ) i ( A B )) We want to prove by induction that if r( cos isin) n n r ( cosn + isinn ) clearly the base case + then for all n N n is true, now assume the statement is true for n k then we have k+ k k k+ r cos + isin r cosk + isink r cos + isin cosk + isink ( ) ( cos cos sin sin cos sin cos sin ) ( cos sin ) r k k i k k k+ + + r k i k k Hence the statement is true for n k+, this completes the induction. (cosx+ isin x) cosx isinx (cosx+ isin x) cos( x) + isin( x) cosx+ isinx cos( x) + isin( x) cos( x x) + isin( x x) cos7x+ isin7x Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

2 a ( + i) If + i,then r + ( ) arg tan So, + i cos + isin ( + i) cos + isin ( ) cos + isin (cos+ isin) Therefore, (+ i()) ( + i) Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

3 b ( i) i Let i, then r + + tan So i cos isin + i cos + isin (cos isin) cos + isin + (+ i()) Therefore, i Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

4 a cos + isin n n (cos + isin ) n cosn + isinn n (cos + isin ) cos( n) + isin( n) cosn isinn n n Therefore + cosn + isinn + cosn isinn n n i.e. + cos n (as required) n de Moivre s Theorem. de Moivre s Theorem. cos( n) cosn sin( n) sinn b + + C + C cos + ()cos cos + cos Hence, + cos cos + c + (cos ) cos cos + cos cos cos + cos Hence, cos cos + cos Soa, b Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

5 d cos d cos + cos d sin + sin sin+ sin sin+ sin () + () So, cos d and k a We wish to show that cos ( cos + cos + cos) i i Starting with cos ( e e ) + we have i i i i i i i i cos ( e + e ) ( e + e + e + e + e + e ) ( cos + cos + cos) ( cos + cos + cos) b The area of the region is given by the integral cos xdx cos xdx Using the first part of the question, this is equivalent to ( cos x+ cos x+ cos x) d x sin x+ sin x+ sinx sin sin sin i i 7 a We have sin ( e e ) sin e e i so that i i ( ) ( e i e i e i + + e i e i + e i ) + + ( cos cos cos ) ( cos cos cos ) Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

6 7 b Let α then recall that we have cosα sin hence we have cos α sin cos cos + cos cos cos cos α α+ α ( cos ( α) cos ( α) + cos ( α) ) ( cos α cos α cosα ) ( cosα cos α cosα ) c We have a a a cos + sin d ( cos ) d [ sin ] + + ( sin a+ a) Hence we require sin a+ a and observe that a solves this and this is the only a cos + sin d using the previous two parts we can evaluate the integral as Solution since the function sin a+ ais increasing. Let e i then we have sin ( ) (( cos i sin ) ( cos i sin ) ) i i + When we expand the right hand side the only terms that survive are odd powers of sin hence we get ( i cos sin i cos sin + i cos sin ) i cos sin cos sin + cos sin ( ) ( ) ( ) sin cos cos cos sin + sin sin cos cos sin + sin sin cos cos cos + cos sin cos cos cos + cos + cos sin cos cos + Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

7 9 a Let e i then we have cos ( + ) (( cos i sin ) ( cos i sin ) ) i i + + Now note the only terms that survive are even powers of sin so this becomes cos ( + ) (( cos + i sin) + ( cos i sin) ) i i ( i cos i cos sin + i cos sin ) i cos cos sin + cos sin ( ) ( ( ) ( ) ) ( + + ( + )) ( + ) cos cos cos sin + sin cos cos cos cos + cos cos cos cos cos cos cos cos cos cos b Now consider the equation x x + x+ We make the substitution x cos so the equation becomes cos cos + cos i.e. cos for which the general solution is + kfor k Zwe take different values of 7 9 Let x cos so i i + x cos.9 x x x x cos.9 cos 7 cos cos.9 Now we only have distinct solutions so we still have to check this is all solutions, factorising out x+ gives x x + x+ x+ x x Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 9 b (Continued) And observe that + x x x x ( x x ) In other words the distinct roots that we have obtained completely factorise the equation and so these three solutions are the only solutions. Hence the three solutions are given by i a Let e cos isin + then sin i( ) +,.9,.9 so we have sin ( ) ( + + ) i i + + i ( isin isin isin) ( sin sin sin) b We wish to solve sin sin + 9sin Which we can rearrange to give sin sin + sin sin So sin sin So we either have sin in which case or sin hence on < the solutions correspond to, sin ± and a (cos + isin ) cos + isin cos + C cos (i sin ) + C cos (i sin ) + C cos (isin ) + C cos (isin ) + isin cos + icos sin + i cos sin + i cos sin + i cossin + i sin de Moivre s Theorem. Binomial expansion. Hence, cos + i sin cos + i cos sin cos sin i cos sin + cos sin + i sin Equating the real parts gives, cos cos cos sin + cos sin Hence, + cos (cos cos sin sin ) + cos (cos cos ( cos ) ( cos ) ) cos (cos cos cos ( cos cos )) cos (cos cos cos cos cos ) cos ( cos cos + ) cos cos (cos cos + ) (as required) Applying sin cos. Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

9 b cos α 7 9 So,,,, 7 9,,,, 7 9,,,, for< cos cos (cos cos + ) Five solutions must come from: Solution () cos α cos (cos cos + ) For <, (as found earlier) The final solutions come from: cos cos + ± ()() cos ± cos ± ± ± + Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 b (Continued) Due to symmetry and as cos > cos 9 7 cos cos > cos cos + cos c cos 7 cos cos 9 + cos cos Therefore, cos, cos, cos i a Let e cos + isin then we have ( ) sin ( cos + sin ) ( cos sin ) tan i i i cos ( + i ) ( + ) i ( cos + i sin) + ( cos i sin) i cos sin i sin cos sin sin i cos cossin cos cos sin Now divide top and bottom by tan tan tan tan cos to give Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

11 b We have tan cot tan tan tan Now divide top and bottom by cot cot cot cot tan to give We have C + kcos + k cos + k cos +... S ksin + k sin + k sin +... So we have i i i C+ is + ke + k e + k e +... Since k < this is a geometric series which we can sum giving C+ is k e i i ke kcos + kisin i i ke ke kcos + k sin kcos + kisin + k kcos Hence taking real and imaginary parts gives kcos C + k kcos ksin S + k kcos a i modulusr + ( ) + argument tan i cos + isin Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

12 b i for i, r, So, i e i + k e, k Z i + k Hence, e e e k, e k, e k, e + k i k i + i 7 i i k, e k, e 9 i 7 i de Moivre s Theorem. So, ( ) ( ) Therefore, i 7i i 9 7 i i e, e, e, e, e c Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

13 a We have i i e, now suppose ω then let ω r and ω ω ω e e e i 7i 9i re i then we have + k hence three distinct roots are given by r so b 7i i c We have w e + e but by geometrical considerations the argument of wis just 7 + and by considering the triangle with vertices at the origin, ABwe, have that the modulus is given by d We have w i i i i sin hence we have i e e e e w e i a Let + i now let ω e + i be a primitive rd root of unity then the coordinates of the other two vertices are given by + ω ( + i) + i + i + + ω ( + i) i i i Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.

14 b The length of one side is given by ω we have Challenge ω ω i We consider the solutions of, let ω then we have ω so ω e k Zhence we have ki + e So ki + e So ki e Therefore ki k k k k cos isin cos isin e ki ki ki e k k k e e cos + sin cos Assuming ki e i.e. k, when ki for k the equation becomes + which has no solutions, however, for k we get a different solution. So our six values of are cos isin i cos cos isin i cos cos isin cos cos isin i + cos cos isin i + cos Which all lie on the straight line Re( ) Pearson Education Ltd. Copying permitted for purchasing institution only. This material is not copyright free.