Changes for version 1.2 include further corrections to chapter 4 and a correction to the numbering of the exercises in chapter 5.

Transcription

1 Versio 0507 klm Chages for versio icluded the isertio of a ew chapter 5 for the 007 specificatio ad some mior mathematical correctios i chapters ad 4 Please ote that these are ot side-barred Chages for versio iclude further correctios to chapter 4 ad a correctio to the umberig of the eercises i chapter 5 Chages for versio iclude mathematical correctios to eercises ad aswers GCE Mathematics (660) Further Pure uit (MFP) Tetbook The Assessmet ad Qualificatios Alliace (AQA) is a compay limited by guaratee registered i Eglad ad Wales 6447 ad a registered charity umber 074 Registered address AQA, Devas Street, Machester M5 6EX Dr Michael Cresswell Director Geeral Copyright 007 AQA ad its licesors All rights reserved

2 Further Pure : Cotets Further Pure (MFP) Tetbook Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5 The modulus ad argumet of a comple umber 6 4 The polar form of a comple umber 8 5 Additio, subtractio ad multiplicatio of comple umbers of the form + iy 9 6 The cojugate of a comple umber ad the divisio of comple umbers of the form +iy 0 7 Products ad quotiets of comple umbers i their polar form 8 Equatig real ad imagiary parts 9 Further cosideratio of z z ad arg(z z ) 4 0 Loci o Argad diagrams 5 Chapter : Roots of polyomial equatios Itroductio Quadratic equatios Cubic equatios 4 4 Relatioship betwee the roots of a cubic equatio ad its coefficiets 7 5 Cubic equatios with related roots 8 6 A importat result 7 Polyomial equatios of degree 8 Comple roots of polyomial equatios with real coefficiets Chapter : Summatio of fiite series 8 Itroductio 9 Summatio of series by the method of differeces 40 Summatio of series by the method of iductio 45 4 Proof by iductio eteded to other areas of mathematics 48 Chapter 4: De Moivre s theorem ad its applicatios 5 4 De Moivre s theorem 54 4 Usig de Moivre s theorem to evaluate powers of comple umbers 56 4 Applicatio of de Moivre s theorem i establishig trigoometric idetities Epoetial form of a comple umber The cube roots of uity 7 46 The th roots of uity The roots of z α, where α is a o-real umber 77 cotiued overleaf

3 Further Pure : Cotets (cotiued) Further Pure (MFP) Tetbook Chapter 5: Iverse trigoometrical fuctios 85 5 Itroductio ad revisio 86 5 The derivatives of stadard iverse trigoometrical fuctios 89 5 Applicatio to more comple differetiatio 9 54 Stadard itegrals itegratig to iverse trigoometrical fuctios Applicatios to more comple itegrals 96 Chapter 6: Hyperbolic fuctios 0 6 Defiitios of hyperbolic fuctios 0 6 Numerical values of hyperbolic fuctios 04 6 Graphs of hyperbolic fuctios Hyperbolic idetities Osbore s rule 0 66 Differetiatio of hyperbolic fuctios 67 Itegratio of hyperbolic fuctios 4 68 Iverse hyperbolic fuctios 5 69 Logarithmic form of iverse hyperbolic fuctios 6 60 Derivatives of iverse hyperbolic fuctios 9 6 Itegrals which itegrate to iverse hyperbolic fuctios 6 Solvig equatios 5 Chapter 7: Arc legth ad area of surface of revolutio 7 Itroductio 7 Arc legth 7 Area of surface of revolutio 7 Aswers to the eercises i Further Pure 4

4 Chapter : Comple Numbers Further Pure (MFP) Tetbook Itroductio The geeral comple umber The modulus ad argumet of a comple umber 4 The polar form of a comple umber 5 Additio, subtractio ad multiplicatio of comple umbers of the form + iy 6 The cojugate of a comple umber ad the divisio of comple umbers of the form + iy 7 Products ad quotiets of comple umbers i their polar form 8 Equatig real ad imagiary parts 9 Further cosideratio of z z ad arg( z z) 0 Loci o Argad diagrams This chapter itroduces the idea of a comple umber Whe you have completed it, you will: kow what is meat by a comple umber; kow what is meat by the modulus ad argumet of a comple umber; kow how to add, subtract, multiply ad divide comple umbers; kow how to solve equatios usig real ad imagiary parts; uderstad what a Argad diagram is; kow how to sketch loci o Argad diagrams 4

5 Itroductio Further Pure (MFP) Tetbook You will have discovered by ow that some problems caot be solved i terms of real umbers For eample, if you use a calculator to evaluate 64 you get a error message This is because squarig every real umber gives a positive value; both ( + 8) ad ( 8) are equal to 64 As caot be evaluated, a symbol is used to deote it the symbol used is i i i It follows that i The geeral comple umber The most geeral umber that ca be writte dow has the form + i y, where ad y are real umbers The term + iy is a comple umber with beig the real part ad iy the imagiary part So, both + i ad 4i are comple umbers The set of real umbers,! (with which you are familiar), is really a subset of the set of comple umbers, " This is because real umbers are actually umbers of the form + 0i 5

6 The modulus ad argumet of a comple umber Further Pure (MFP) Tetbook Just as real umbers ca be represeted by poits o a umber lie, comple umbers ca be represeted by poits i a plae The poit P(, y) i the plae of coordiates with aes O ad Oy represets the comple umber + iy ad the umber is uiquely represeted by that poit The diagram of poits i Cartesia coordiates represetig comple umbers is called a Argad diagram y r P(, y) O θ If the comple umber + iy is deoted by z, ad hece z + i, y z ( mod zed ) is defied as the distace from the origi O to the poit P represetig z Thus z OP r The modulus of a comple umber z is give by z + y The argumet of z, arg z, is defied as the agle betwee the lie OP ad the positive -ais usually i the rage (π, π) The argumet of a comple umber z is give y by arg z θ, where taθ You must be careful whe or y, or both, are egative 6

7 Further Pure (MFP) Tetbook Eample Fid the modulus ad argumet of the comple umber + i Solutio The poit P represetig this umber, z, is show o the diagram P (, ) y z ( ) + ad taθ θ Therefore, arg z π O Note that whe taθ, θ could equal + π or π However, the sketch clearly shows that θ lies i the secod quadrat This is why you eed to be careful whe evaluatig the argumet of a comple umber Eercise A Fid the modulus ad argumet of each of the followig comple umbers: (a) i, (b) i, (c) 4, (d) i Give your aswers for arg z i radias to two decimal places Fid the modulus ad argumet of each of the followig comple umbers: (a) + i, (b) + 4i, (c) 7i Give your aswers for arg z i radias to two decimal places 7

8 4 The polar form of a comple umber I the diagram alogside, y rsi θ rcosθ ad If P is the poit represetig the comple umber z + i, y it follows that z may be writte i the form rcosθ + irsi θ This is called the polar, or modulus argumet, form of a comple umber Further Pure (MFP) Tetbook y P(, y) r θ O A comple umber may be writte i the form z r(cosθ + isi θ ), where z r ad arg z θ For brevity, r(cosθ + isi θ ) ca be writte as (r, θ) Eercise B Write the comple umbers give i Eercise A i polar coordiate form Fid, i the form + i y, the comple umbers give i polar coordiate form by: (a) z cosπ π ( + isi ), (b) 4 4 ( π + π ) 4cos isi 8

9 Further Pure (MFP) Tetbook 5 Additio, subtractio ad multiplicatio of comple umbers of the form + iy Comple umbers ca be subjected to arithmetic operatios Cosider the eample below Eample 5 Give that z + 4i ad z i, fid (a) z + z, (b) z z ad (c) zz Solutio (a) z+ z ( + 4i) + ( i) 4+ i (b) z z ( + 4i) ( i) + 6i (c) zz ( + 4i)( i) + 4i 6i 8i + i 8 (sice i ) i I geeral, if z a+ i b ad z a + i b, z + z ( a + a ) + i( b + b ) z z ( a a ) + i( b b ) zz aa bb + i( ab+ ab) Eercise C Fid z + z ad z z whe: (a) z + i ad z i, (b) z + 6i ad z + i 9

10 Further Pure (MFP) Tetbook 6 The cojugate of a comple umber ad the divisio of comple umbers of the form + iy The cojugate of a comple umber z + iy (usually deoted by z * or z ) is the comple umber z* i y Thus, the cojugate of + i is i, ad that of i is + i O a Argad diagram, the poit represetig the comple umber * z is the reflectio of the poit represetig z i the -ais The most importat property of z * is that the product zz * is real sice zz* ( + i y)( i y) + iy iy i y + y * zz z Divisio of two comple umbers demads a little more care tha their additio or multiplicatio ad usually requires the use of the comple cojugate Eample 6 z Simplify z, where z + 4i ad z i Solutio + 4i (+ 4i)( + i) i ( i)(+ i) + 4i+ 6i+ 8i i + i 4i 5+ 0i 5 + i z multiply the umerator ad deomiator of z by z *, ie ( + i) so that the product of the deomiator becomes a real umber Eercise D z For the sets of comple umbers z ad z, fid z where (a) z 4 + i ad z i, (b) z + 6i ad z + i 0

11 Further Pure (MFP) Tetbook 7 Products ad quotiets of comple umbers i their polar form If two comple umbers are give i polar form they ca be multiplied ad divided without havig to rewrite them i the form + i y Eample 7 Fid z Solutio z if z π π cos + isi ad z π π ( 6 6) cos isi π π π π ( ) ( 6 6) π π π π π π π π ( ) π π π π π π π π 6 6 ( 6 6) zz cos + isi cos isi 6 cos cos + isi cos isi cos i si si 6 cos cos + si si + i si cos cos si π π π π ( 6) ( 6) π π ( 6 6) 6 cos + isi 6cos + isi Notig that arg z is π, it follows that the modulus of zz is the product of the modulus of 6 z ad the modulus of z, ad the argumet of zz is the sum of the argumets of z ad z Usig the idetities: cos( A B) cos Acos B+ si Asi B si( A B) si Acos B cos Asi B Eercise E z (a) Fid cos isi z z (b) What ca you say about the modulus ad argumet of z? if z cosπ π ( + isi ) ad z π π ( 6 6)

12 Further Pure (MFP) Tetbook Eample 7 If z ( r, θ ) ad z ( r, θ ), show that zz rr[ θ + θ + θ + θ ] Solutio zz r(cosθ + isi θ ) r(cosθ + isi θ ) cos isi rr cosθcosθ + i(siθcosθ + cosθsi θ) + i siθsiθ rr + + [(cosθ cosθ siθ si θ ) i(siθ cosθ cosθ si θ )] [ cos( θ θ ) isi( θ θ )] rr If z ( r, θ) ad z ( r, θ) the zz ( rr, θ+ θ) with the proviso that π may have to be added to, or subtracted from, θ+ θ if θ + θ is outside the permitted rage for θ There is a correspodig result for divisio you could try to prove it for yourself z r (, ) ad (, ) the z, r θ θ same proviso regardig the size of the agle θ θ If z r θ z r θ with the

13 8 Equatig real ad imagiary parts z Goig back to Eample 6, z ca be simplified by aother method Further Pure (MFP) Tetbook + 4i Suppose we let a + ib The, i ( i)( a+ i b) + 4i a+ b+ i( b a) + 4i Now, a ad b are real ad the comple umber o the left had side of the equatio is equal to the comple umber o the right had side, so the real parts ca be equated ad the imagiary parts ca also be equated: a+ b ad b a 4 Thus b ad a, givig + i as the aswer to a+ ib as i Eample 6 While this method is ot as straightforward as the method used earlier, it is still a valid method It also illustrates the cocept of equatig real ad imagiary parts If a+ ib c+ i d, where a, b, c ad d are real, the a c ad b d Eample 8 Fid the comple umber z satisfyig the equatio Solutio ( 4i) z ( + i) z* + i Let z ( a+ i b), the z* ( a i b) Thus, ( 4i)( a+ i b) ( + i)( a i b) + i Multiplyig out, a 4ia+ ib 4i b a ia+ ib+ i b + i Simplifyig, a+ b+ i( 5a+ 4 b) + i Equatig real ad imagiary parts, a+ b, 5a+ 4b So, a ad b Hece, z + i Eercise F If z ( π ) z π ( 6), ad,, fid, i polar form, the comple umbers (a) zz, (b) z z, (c) z, (d) z, (e) z z Fid the comple umber satisfyig each of these equatios: (a) ( + i) z + i, (b) ( z i)(+ i) 7i +, (c) z+ i z*

14 Further Pure (MFP) Tetbook 9 Further cosideratio of z z ad arg( z z) Sectio 5 cosidered simple cases of the sums ad differeces of comple umbers Cosider ow the comple umber z z z, where z + iy ad z + i y The poits A ad B represet z ad z, respectively, o a Argad diagram y A, ) B, ) ( y ( y O C The z z z ( ) + i( y y) ad is represeted by the poit C (, y y) This makes OABC a parallelogram From this it follows that z z OC ( ) + ( y y), that is to say z z is the legth AB i the Argad diagram Similarly arg( z z) is the agle betwee OC ad the positive directio of the -ais This i tur is the agle betwee AB ad the positive directio If the comple umber z is represeted by the poit A, ad the comple umber z is represeted by the poit B i a Argad diagram, the z z AB, ad arg( z z) is the agle betwee AB directio of the -ais ad the positive Eercise G Fid z z ad arg( z z) i (a) z + i, z 7+ 5i, (b) z i, z 4+ i, (c) z + i, z 4 5i 4

15 0 Loci o Argad diagrams Further Pure (MFP) Tetbook A locus is a path traced out by a poit subjected to certai restrictios Paths ca be traced out by poits represetig variable comple umbers o a Argad diagram just as they ca i other coordiate systems Cosider the simplest case first, whe the poit P represets the comple umber z such that z k This meas that the distace of P from the origi O is costat ad so P will trace out a circle z k represets a circle with cetre O ad radius k If istead z z k, where z is a fied comple umber represeted by the poit A o a Argad diagram, the (from Sectio 9) z z represets the distace AP ad is costat It follows that P must lie o a circle with cetre A ad radius k z z k represets a circle with cetre z ad radius k Note that if z z k, the the poit P represetig z ca ot oly lie o the circumferece of the circle, but also aywhere iside the circle The locus of P is therefore the regio o ad withi the circle with cetre A ad radius k Now cosider the locus of a poit P represeted by the comple umber z subject to the coditios z z z z, where z ad z are fied comple umbers represeted by the poits A ad B o a Argad diagram Agai, usig the result of Sectio 9, it follows that AP BP because z z is the distace AP ad z z is the distace BP Hece, the locus of P is a straight lie z z z z represets a straight lie the perpedicular bisector of the lie joiig the poits z ad z Note also that if z z z z the locus of z is ot oly the perpedicular bisector of AB, but also the whole half plae, i which A lies, bouded by this bisector All the loci cosidered so far have bee related to distaces there are also simple loci i Argad diagrams ivolvig agles The simplest case is the locus of P subject to the coditio that arg z α, where α is a fied agle y O α P 5

16 Further Pure (MFP) Tetbook This coditio implies that the agle betwee OP ad O is fied ( α ) so that the locus of P is a straight lie arg z α represets the half lie through O iclied at a agle α to the positive directio of O Note that the locus of P is oly a half lie the other half lie, show dotted i the diagram above, would have the equatio arg z π + α, possibly ± π if π + α falls outside the specified rage for arg z I eactly the same way as before, the locus of a poit P satisfyig arg( z z) α, where z is a fied comple umber represeted by the poit A, is a lie through A arg( z z ) α iclied at a agle α to the positive directio of O represets the half lie through the poit z y P A α O Note agai that this locus is oly a half lie the other half lie would have the equatio arg( z z ) π + α, possibly ± π Fially, cosider the locus of ay poit P satisfyig α arg( z z) β This idicates that the agle betwee AP ad the positive -ais lies betwee α ad β, so that P ca lie o or withi the two half lies as show shaded i the diagram below y A β α O 6

17 Further Pure (MFP) Tetbook Eercise H Sketch o Argad diagrams the locus of poits satisfyig: (a) z, (b) arg( z ) π, (c) z i 5 4 Sketch o Argad diagrams the regios where: (a) z i, (b) π arg( z 4 i) 5π 6 Sketch o a Argad diagram the regio satisfyig both z i ad 0 arg z π 4 4 Sketch o a Argad diagram the locus of poits satisfyig both z i z+ + i ad z + i 4 7

18 Further Pure (MFP) Tetbook Miscellaeous eercises Fid the comple umber which satisfies the equatio z+ iz* 4 i, where z * deotes the comple cojugate of z [AQA Jue 00] The comple umber z satisfies the equatio ( z ) i( z+ ) (a) Fid z i the form a+ i b, where a ad b are real (b) Mark ad label o a Argad diagram the poits represetig z ad its cojugate, z * (c) Fid the values of z ad z z* [NEAB March 998] The comple umber z satisfies the equatio zz* z z* i, where z * deotes the comple cojugate of z Fid the two possible values of z, givig your aswers i the form a+ i b [AQA March 000] 4 By puttig z + i y, fid the comple umber z which satisfies the equatio z+ z* + i, + i where z * deotes the comple cojugate of z [AQA Specime] 5 (a) Sketch o a Argad diagram the circle C whose equatio is z i (b) Mark the poit P o C at which z is a miimum Fid this miimum value (c) Mark the poit Q o C at which arg z is a maimum Fid this maimum value [NEAB Jue 998] 8

19 Further Pure (MFP) Tetbook 6 (a) Sketch o a commo Argad diagram (i) the locus of poits for which z i, (ii) the locus of poits for which arg z π 4 (b) Idicate, by shadig, the regio for which z i < ad arg z < π 4 [AQA Jue 00] 7 The comple umber z is defied by z + i i (a) (i) Epress z i the form a+ i b (ii) Fid the modulus ad argumet of z, givig your aswer for the argumet i the form p π where < p (b) The comple umber z has modulus ad argumet 7π The comple umber z is defied by z zz (i) Show that z π 4 ad arg z 6 (ii) Mark o a Argad diagram the poits P ad P which represet z ad z, respectively (iii) Fid, i surd form, the distace betwee P ad P [AQA Jue 000] 9

20 Further Pure (MFP) Tetbook 8 (a) Idicate o a Argad diagram the regio of the comple plae i which 0 arg π ( z + ) (b) The comple umber z is such that 0 arg π ( z + ) ad π arg ( z + ) π 6 (i) Sketch aother Argad diagram showig the regio R i which z must lie (ii) Mark o this diagram the poit A belogig to R at which z has its least possible value (c) At the poit A defied i part (b)(ii), z z A (i) Calculate the value of z A (ii) Epress z A i the form a+ i b [AQA March 999] 9 (a) The comple umbers z ad w are such that z ( 4+ i)( i) ad w 4+ i i Epress each of z ad w i the form a+ i b, where a ad b are real (b) (i) Write dow the modulus ad argumet of each of the comple umbers 4 + i ad i Give each modulus i a eact surd form ad each argumet i radias betwee π ad π (ii) The poits O, P ad Q i the comple plae represet the comple umbers 0 + 0i, 4 + i ad i, respectively Fid the eact legth of PQ ad hece, or otherwise, show that the triagle OPQ is right-agled [AEB Jue 997] 0

21 Further Pure (MFP) Tetbook Chapter : Roots of Polyomial Equatios Itroductio Quadratic equatios Cubic equatios 4 Relatioship betwee the roots of a cubic equatio ad its coefficiets 5 Cubic equatios with related roots 6 A importat result 7 Polyomial equatios of degree 8 Comple roots of polyomial equatios with real coefficiets This chapter revises work already covered o roots of equatios ad eteds those ideas Whe you have completed it, you will: kow how to solve ay quadratic equatio; kow that there is a relatioship betwee the umber of real roots ad form of a polyomial equatio, ad be able to sketch graphs; kow the relatioship betwee the roots of a cubic equatio ad its coefficiets; be able to form cubic equatios with related roots; kow how to eted these results to polyomials of higher degree; kow that comple cojugates are roots of polyomials with real coefficiets

22 Itroductio Further Pure (MFP) Tetbook You should have already met the idea of a polyomial equatio A polyomial equatio of degree, oe with as the highest power of, is called a quadratic equatio Similarly, a polyomial equatio of degree has as the highest power of ad is called a cubic equatio; oe with 4 as the highest power of is called a quartic equatio I this chapter you are goig to study the properties of the roots of these equatios ad ivestigate methods of solvig them Quadratic equatios You should be familiar with quadratic equatios ad their properties from your earlier studies of pure mathematics However, eve if this sectio is familiar to you it provides a suitable base from which to move o to equatios of higher degree You will kow, for eample, that quadratic equatios of the type you have met have two roots (which may be coicidet) There are ormally two ways of solvig a quadratic equatio by factorizig ad, i cases where this is impossible, by the quadratic formula Graphically, the roots of the equatio a + b + c 0 are the poits of itersectio of the curve y a + b+ c ad the lie y 0 (ie the -ais) For eample, a sketch of part of y + 8 is show below y ( 4, 0) (, 0) (0, 8) The roots of this quadratic equatio are those of ( )( + 4) 0, which are ad 4

23 Further Pure (MFP) Tetbook A sketch of part of the curve y is show below y (0, 4) O (, 0) I this case, the curve touches the -ais The equatio ( ) 0 ad, a repeated root may be writte as Not all quadratic equatios are as straightforward as the oes cosidered so far A sketch of part of the curve y 4+ 5 is show below y (0, 5) O (, ) This curve does ot touch the -ais so the equatio caot have real roots Certaily, will ot factorize so the quadratic formula b± b 4ac a has to be used to solve this equatio This leads to 4± 6 0 ad, usig ideas from Chapter, this becomes 4 ± i or ± i It follows that the equatio does have two roots, but they are both comple umbers I fact the two roots are comple cojugates You may also have observed that whether a quadratic equatio has real or comple roots depeds o the value of the discrimiat b 4 ac The quadratic equatio a + b + c 0, where a, b ad c are real, has comple roots if b 4ac< 0

24 Further Pure (MFP) Tetbook Eercise A Solve the equatios (a) , (b) Cubic equatios As metioed i the itroductio to this chapter, equatios of the form a + b + c + d 0 are called cubic equatios All cubic equatios have at least oe real root ad this real root is ot always easy to locate The reaso for this is that cubic curves are cotiuous they do ot have asymptotes or ay other form of discotiuity Also, as, the term a becomes the domiat part of the epressio ad a (if a > 0), whilst a whe Hece the curve must cross the lie y 0 at least oce If a < 0, the as, ad a as ad this does ot affect the result a 4

25 Further Pure (MFP) Tetbook A typical cubic equatio, below y y a + b + c+ d with a > 0, ca look like ay of the sketches The equatio of this curve has three real roots because the curve crosses the lie y 0 at three poits O y y O O I each of the two sketch graphs above, the curve crosses the lie y 0 just oce, idicatig just oe real root I both cases, the cubic equatio will have two comple roots as well as the sigle real root Eample (a) Fid the roots of the cubic equatio + 0 (b) Sketch a graph of y + Solutio y 4 (a) If f +, the f() + 0 Therefore is a factor of f() f ( )( + 4+ ) ( )( + )( + ) Hece the roots of f() 0 are, ad (b)

26 Further Pure (MFP) Tetbook Eample Fid the roots of the cubic equatio Solutio Let f The f () Therefore is a factor of f(), ad f ( )( + 6+ ) The quadratic i this epressio has o simple roots, so usig the quadratic formula o , ± 4 a 6± 6 5 6± 4i ± i b b ac Hece the roots of f ( ) 0 are ad ± i Eercise B Solve the equatios (a) 5 0, (b) (c) + 4 0,

27 Further Pure (MFP) Tetbook 4 Relatioship betwee the roots of a cubic equatio ad its coefficiets As a cubic equatio has three roots, which may be real or comple, it follows that if the geeral cubic equatio a + b + c + d 0 has roots α, β ad γ, it may be writte as a ( α)( β)( γ) 0 Note that the factor a is required to esure that the coefficiets of are the same, so makig the equatios idetical Thus, o epadig the right had side of the idetity, a + b + c + d a( α)( β)( γ) a a( α + β + γ ) + a( αβ + βγ + γα) aαβγ The two sides are idetical so the coefficiets of ad ca be compared, ad also the umber terms, b a( α + β + γ) c a( αβ + βγ + γα) d aαβγ If the cubic equatio a + b + c + d 0 has roots α, β ad γ, the b α, a c αβ, a αβγ d a Note that α meas the sum of all the roots, ad that αβ meas the sum of all the possible products of roots take two at a time Eercise C Fid α, αβ ad αβγ for the followig cubic equatios: (a) , (b) The roots of a cubic equatio are α, β ad γ If α, the cubic equatio 7 αβ ad αβγ 5, state 7

28 5 Cubic equatios with related roots Further Pure (MFP) Tetbook The eample below shows how you ca fid equatios whose roots are related to the roots of a give equatio without havig to fid the actual roots Two methods are give Eample 5 The cubic equatio has roots α, β ad γ Fid the cubic equatios with: (a) roots α, β ad γ, (b) roots α, β ad γ, (c) roots, ad α β γ Solutio: method From the give equatio, α αβ 0 αβγ 4 (a) Hece α α 6 α β 4 αβ 0 α β γ 8αβγ From which the equatio of the cubic must be or (b) α ( α) αβ 4 α + ( ) 6 6 ( α )( β ) αβ α β + (4 ) ( α )( β )( γ ) αβγ αβ + 4 α Hece the equatio of the cubic must be + 0 8

29 Further Pure (MFP) Tetbook (c) + + α α β γ αβ αβγ α β αβ βγ γα α αβγ 4 4 α β γ 4 4 So that the cubic equatio with roots, ad α β γ is or

30 Further Pure (MFP) Tetbook The secod method of fidig the cubic equatios i Eample 5 is show below It is ot always possible to use this secod method, but whe you ca it is much quicker tha the first Solutio: method (a) As the roots are to be α, β ad γ, it follows that, if X, the a cubic equatio i X must have roots which are twice the roots of the cubic equatio i As the equatio i is + 4 0, if you substitute X the equatio i X becomes X X + 4 0, or X 6X + 0 as before (b) I this case, if you put X i + 4 0, the ay root of a equatio i X must be less tha the correspodig root of the cubic i Now, X gives X + ad substitutig ito gives ( X + ) ( X + ) which reduces to X + X 0 (c) I this case you use the substitutio X or X For X X O multiplyig by X, this gives X + 4X 0 or 4X X + 0 as before this gives Eercise D The cubic equatio has roots α, β ad γ Usig the first method described above, fid the cubic equatios whose roots are (a) α, β ad γ, (b) α +, β + ad γ +, (c), ad α β γ Repeat Questio above usig the secod method described above Repeat Questios ad above for the cubic equatio

31 6 A importat result If you square α + β + γ you get ( α + β + γ) ( α + β + γ)( α + β + γ) α + αβ + αγ + βα + β + βγ + γα + γβ + γ α + β + γ + αβ + βγ + γα So, ( α) α + αβ, or α ( α) αβ for three umbers α, β ad Further Pure (MFP) Tetbook γ This result is well worth rememberig it is frequetly eeded i questios ivolvig the symmetric properties of roots of a cubic equatio Eample 6 The cubic equatio has roots α, β ad γ Fid the cubic equatios with (a) roots βγ, γα ad αβ, (b) roots α, β ad γ [Note that the direct approach illustrated below is the most straightforward way of solvig this type of problem] Solutio (a) α 5 αβ 6 αβγ αβ βγ αβ γ αβγ β αβγ α 5 5 αβ βγ γα α β γ ( ) + Hece the cubic equatio is (b) α ( α) αβ 5 ( 6) α β ( αβ) αβ βγ usig the same result but replacig with β with βγ, ad γ with γα Thus ( ) ( ) α β αβ αβ γ αβ αβγ α 6 ( 5) 46 α αβ, α β γ ( ) Hece the cubic equatio is

32 7 Polyomial equatios of degree Further Pure (MFP) Tetbook The ideas covered so far o quadratic ad cubic equatios ca be eteded to equatios of ay degree A equatio of degree has two roots, oe of degree has three roots so a equatio of degree has roots Suppose the equatio has roots α, β, γ, the b α, a c αβ, a αβγ d, a a b c d k ( ) k util, fially, the product of the roots αβγ a Remember that αβ is the sum of the products of all possible pairs of roots, αβγ is the sum of the products of all possible combiatios of roots take three at a time, ad so o I practice, you are ulikely to meet equatios of degree higher tha 4 so this sectio cocludes with a eample usig a quartic equatio Eample 7 4 The quartic equatio has roots α, β, γ ad δ Write dow (a) α, (b) αβ (c) Hece fid α Solutio (a) (b) (c) α α + β + γ + δ 4 αβ αβ + βγ + γδ + δα + αγ + βδ 6 α ( α + β + γ + δ) Now α β γ δ αβ βγ γδ δα αγ βδ ( ) α + αβ This shows that the importat result i Sectio 6 ca be eteded to ay umber of letters Hece ( ) α α αβ 0

33 Further Pure (MFP) Tetbook Eercise E 4 The quartic equatio has roots α, β, γ ad δ (a) Fid the equatio with roots α β γ,, ad δ (b) Fid α 8 Comple roots of polyomial equatios with real coefficiets Cosider the polyomial equatio f a b + + c + k Usig the ideas from Chapter, if p ad q are real, f( p+ i q) a( p+ i q) + b( p+ i q) + + k u+ i v, where u ad v are real Now, f( i ) ( i ) ( i ) p q a p q + b p q + + k u iv sice i raised to a eve power is real ad is the same as + i raised to a eve power, makig the real part of f ( p i q) the same as the real part of f ( p + i q) But i raised to a odd power is the same as + i raised to a odd power multiplied by, ad odd powers of i comprise the imagiary part of f ( p i q) Thus, the imagiary part of f ( p i q) is times the imagiary part of f ( p + i q) Now if p + iq is a root of f ( ) 0, it follows that u+ iv 0 ad so u 0 ad v 0 Hece, u iv 0 makig f ( p i q) 0 ad p iq a root of f ( ) 0 If a polyomial equatio has real coefficiets ad if p + i q, where p ad q are real, is a root of the polyomial, the its comple cojugate, p i q, is also a root of the equatio It is very importat to ote that the coefficiets i f ( ) 0 must be real If f ( ) 0 has comple coefficiets, this result does ot apply

34 Further Pure (MFP) Tetbook Eample 8 The cubic equatio + + k 0, where k is real, has oe root equal to i Fid the other two roots ad the value of k Solutio As the coefficiets of the cubic equatio are real, it follows that + i is also a root Cosiderig the sum of the roots of the equatio, if γ is the third root, ( i) + ( + i) + γ, γ To fid k, k αβγ ( i)( + i)( ) 5, k 5 Eample 8 The quartic equatio roots has oe root equal to + i Fid the other three Solutio As the coefficiets of the quartic are real, it follows that i is also a root (+ i) ( i) is a quadratic factor of the quartic Now, Hece [ ][ ] [ ][ ] (+ i) ( i) (+ i) ( i) + (+ i)( i) Hece + 5 is a factor of Therefore ( + 5)( + a+ b) Comparig the coefficiets of, a a 4 Cosiderig the umber terms, 5 b b 5 Hece the quartic equatio may be writte as ( + 5)( + 4+ ) 0 ( + 5)( + )( + ) 0, ad the four roots are + i, i, ad 4

35 Further Pure (MFP) Tetbook Eercise F A cubic equatio has real coefficiets Oe root is ad aother is + i Fid the cubic equatio i the form + a + b+ c 0 The cubic equatio roots has oe root equal to i Fid the other two The quartic equatio three roots has oe root equal to i Fid the other 5

36 Further Pure (MFP) Tetbook Miscellaeous eercises The equatio , p where p is a costat, has roots α β, α ad α + β, where β > 0 (a) Fid the values of α ad β (b) Fid the value of p [NEAB Jue 998] The umbers α, β ad γ satisfy the equatios α + β + γ ad αβ + βγ + γα (a) Show that α + β + γ 0 (b) The umbers α, β ad γ are also the roots of the equatio where p, q ad r are real , p q r (i) Give that α + 4i ad that γ is real, obtai β ad γ (ii) Calculate the product of the three roots (iii) Write dow, or determie, the values of p, q ad r [AQA Jue 00] The roots of the cubic equatio are α, β ad γ (a) Write dow the values of α + β + γ, αβ + βγ + γα ad αβγ (b) Fid the cubic equatio, with iteger coefficiets, havig roots αβ, βγ ad γα [AQA March 000] 4 The roots of the equatio are α, β ad γ (a) Write dow the value of α + β + γ (b) Give that + i is a root of the equatio, fid the other two roots [AQA Specime] 6

37 Further Pure (MFP) Tetbook 5 The roots of the cubic equatio , p q r where p, q ad r are real, are α, β ad γ (a) Give that α + β + γ, write dow the value of p (b) Give also that α + β + γ 5, (i) fid the value of q, (ii) eplai why the equatio must have two o-real roots ad oe real root (c) Oe of the two o-real roots of the cubic equatio is 4i (i) Fid the real root (ii) Fid the value of r [AQA March 999] 6 (a) Prove that whe a polyomial (b) The polyomial g ( ) is defied by 5 f is divided by a, g 6 + p + q, the remaider is ( a ) f where p ad q are real costats Whe g ( ) is divided by i, where i, the remaider is (i) Fid the values of p ad q [AQA Jue 999] (ii) Show that whe g ( ) is divided by i, the remaider is 6i 7

38 Further Pure (MFP) Tetbook Chapter : Summatio of Fiite Series Itroductio Summatio of series by the method of differeces Summatio of series by the method of iductio 4 Proof by iductio eteded to other areas of mathematics This chapter eteds the idea of summatio of simple series, with which you are familiar from earlier studies, to other kids of series Whe you have completed it, you will: kow ew methods of summig series; kow which method is appropriate for the summatio of a particular series; uderstad a importat method kow as the method of iductio; be able to apply the method of iductio i circumstaces other tha i the summatio of series 8

39 Itroductio Further Pure (MFP) Tetbook You should already be familiar with the idea of a series a series is the sum of the terms of a sequece That is, the sum of a umber of terms where the terms follow a defiite patter For istace, the sum of a arithmetic progressio is a series I this case each term is bigger tha the preceedig term by a costat umber this costat umber is usually called the commo differece Thus, is a series of 5 terms, i arithmetic progressio, with commo differece The sum of a geometric progressio is also a series Istead of addig a fied umber to fid the et cosecutive umber i the series, you multiply by a fied umber (called the commo ratio) Thus, is a series of 6 terms, i geometric progressio, with commo ratio A fiite series is a series with a fiite umber of terms The two series above are eamples of fiite series 9

40 Summatio of series by the method of differeces Some problems require you to fid the sum of a give series, for eample sum the series Further Pure (MFP) Tetbook I others you have to show that the sum of a series is a specific umber or a give epressio A eample of this kid of problem is show that The method of differeces is usually used whe the sum of the series is ot give Suppose you wat to fid the sum, ur, of a series r u + u + u + + u where the terms follow a certai patter The aim i the method of differeces is to epress th the r term, which will be a fuctio of r (just as th is the r term of the first series r( r+) above), as the differece of two epressios i r of the same form I other words, u r is epressed as f ( r) f( r+ ), or possibly f ( r+ ) f ( r), where f ( r ) is some fuctio of r If you ca epress u r i this way, it ca be see that settig r ad the r gives () () u+ u f f + f f f f If this idea is eteded to the whole series, the r u f f r u f f r u f f 4 $ $ r u f f r u f f + Now, addig these terms gives u+ u + u+ + u + u f f +f f +f f 4 +f 4 + f f +f f + The left had side of this epressio is the required sum of the series, had side, early all the terms cacel out: f ( ),f ( ),f ( 4 ),,f( ) f () f ( + ) as the sum of the series ur O the right r all cacel leavig just 40

41 Further Pure (MFP) Tetbook Eample Fid the sum of the series Solutio Clearly, this is ot a familiar stadard series, such as a arithmetic or geometric series Nor is the aswer give So it seems that the method of differeces ca be applied th As above, the r term, u r, is give by We eed to try to split up u r The oly r( r+) sesible way to do this is to epress i partial fractios Suppose r( r+) A + B r( r+ ) r r + The, A( r+ ) + Br Comparig the coefficiets of r, A+ B 0 Comparig the costat terms, A Hece B ad u r r( r+ ) r r + Hece, i this case the f ( r ) metioed previously would be, with f ( r ), r + r + ad so o Now, writig dow the series term by term, r + r + r $ $ $ $ r ( ) ( ) + r ( + ) + Addig the colums, the left had side becomes r rr+ Because the,, etc terms cacel, the right had side becomes, r + amely the first left had side term ad the last right had side term Hece, r rr ( + ) + ( + ) + + 4

42 Further Pure (MFP) Tetbook Eample Show that Solutio r r+ r r 4 r Hece fid r r The left had side of the idetity has a commo factor, r r ( r+ ) ( r ) r r ( r+ ) ( r ) r ( r + r+ ) ( r r+ ) r r r r r r 4r Now, if r r r so that f, the 4r is of the form 4 r r ( r ), f( r+ ) r+ r+ + f r+ f( r) Listig the terms i colums, as i Eample, r 4 0 r 4 r 4 4 $ $ $ 4 4 ( ) r r + Addig the colums, it ca be see that the left had side is % Summig the right had side, all the terms cacel out ecept those shaded i the scheme above, so the sum is Hece, r ( + ) r Hece, r 4 r + 0, as required ( ) + r 4

43 Further Pure (MFP) Tetbook I both Eamples ad, oe term o each lie cacelled out with a term o the et lie whe the additio was doe Some series may be such that a term i oe lie cacels with a term o a lie two rows below it Eample Sum the series ( )( ) Solutio As i Eample, the way forward is to epress ( )( + ) Let A + B +, ( r )( r+ ) r r Multiplyig both sides by ( r )( r+ ) A r+ + B r 4 i partial fractios Comparig the coefficiets of r, A+ B 0, so A B Comparig the costat terms, A B Hece A ad B Thus, 4 4 r r+ 4 r 4 r+ Now substitute r,,, % r r [ote that othig will cacel at this stage] r 5 9 4( 5) 4( 9) [ote that will cacel o the first row ad the third row] 4 5 r 4 7 4( 7) 4 $ $ $ r ( ) ( 5)( ) 4 ( ) ( 5 4 ) ( ) ( )( + ) 4( ) 4( + ) ( )( + ) 4( ) 4( + ) r r

44 Further Pure (MFP) Tetbook There will be two terms left at the begiig of the series whe the colums are added, ad Likewise, there will be two terms left at the ed of the series the right 4 4 had part of the r ad r rows Therefore, additio gives ( )( + ) ( ) ( ) + ( + ) ( + )( + ) 4( + ) ( + )( + ) ( + )( + ) Eercise A (a) Simplify ( + ) ( ) r r r r (b) Use your result to obtai r r (a) Show that rr ( + )( r+ ) ( r+ )( r+ )( r+ ) rr ( + )( r+ )( r+ ) (b) Hece sum the series ( + )( + )( + ) r rr r r (a) Show that (b) Deduce r r+ r 6r + r 44

45 Summatio of a series by the method of iductio Further Pure (MFP) Tetbook The method of iductio is a method of summig a series of, say, terms whe the sum is give i terms of Suppose you have to show that the sum of terms of a series is S() If you assume that the summatio is true for oe particular iteger, say k, where k <, the you are assumig that the sum of the first k terms is S(k) You may thik that this rather begs the questio but it must be uderstood that the result is assumed to be true for oly oe value of, amely k You the use this assumptio to prove that the sum of the series to k + S k+ that is to say that by addig oe etra term, the et term i the series, the terms is sum has eactly the same form as S() but with replaced by k + Fially, it is demostrated that the result is true for To summarise: Assume that the result of the summatio is true for k ad prove that it is true for k+ Prove that the result is true for Statemet shows that, by puttig k (which is kow to be true from Statemet ), the result must be true for ; ad Statemet shows that by puttig k the result must be true for ; ad so o By buildig up the result, it ca be said that the summatio result is true for all positive itegers There is a formal way of writig out the method of iductio which is show i the eamples below For coveiece, ad compariso, the eamples worked i Sectio are used agai here Eample Show that r r r ( + ) + Solutio Assume that the result is true for k; that is to say k + k( k+ ) k + Addig the et term to both sides, k + k( k+ ) + ( k+ )( k+ ) k + + ( k+ )( k+ ) The k+ k + r r+ k + k+ k+ r k( k+ ) + ( k+ )( k+ ) k + k+ ( k+ )( k+ ) 45

46 Further Pure (MFP) Tetbook ( k+ )( k+ ) ( k+ )( k+ ) k + k + k +, + + ( k ) which is of the same form but with k + replacig k Hece, if the result is true for k, it is true for k+ But it is true for because the left had side is, ad the right had side is Therefore the result is true for all positive itegers by iductio + Eample Show that r ( + ) r 4 Solutio Assume that the result is true for k, that is to say k r The, addig the et term to both sides r k ( k+ ) 4 k+ r r k k+ + k+ 4 ( k+ ) k 4( k ) ( k+ ) k 4k ( k+ ) ( k+ ) 4 ( k+ ) ( k+ ) +, 4 which is of the same form but with k + replacig k Hece, if the result is true for k, it is true for k+ But it is true for k because the left had side is, ad the right had side is Therefore the result is true for all positive itegers by iductio 4 46

47 Further Pure (MFP) Tetbook Eercise B Prove the followig results by the method of iductio: 4 (a) ( ) + ( ) + ( ) + + ( + ) ( + )( + ) (b) ( + )( + ) 6 (c) r( r+ ) ( + )( + ) r (d) r r ( ) r 7 6! +! 47

48 Further Pure (MFP) Tetbook 4 Proof by iductio eteded to other areas of mathematics The method of iductio is certaily useful i the summatio of series but it is ot cofied to this area of mathematics This chapter cocludes with a look at its use i three other coectios sequeces, divisibility ad de Moivre s theorem for positive itegers Eample 4 applicatio to sequeces A sequece,,, u u u u is defied by u u ( ) Prove by iductio that for all Solutio + +, u Assume that the result is true for k, that is to say k u k + k The, usig the relatioship give, u k+ u k k+ k k ( ) k+ k+ k ( ) ( ) k+ k+ k+ ( ) + k+ k+ ( ) k+ k+ k+ ( k+ ) + k+ which is of the same form as uk but with k + replacig k Hece, if the result is true for k, it is true for k+ But whe k, u + as give Therefore the result is true for all positive itegers by iductio 48

49 Further Pure (MFP) Tetbook Eample 4 applicatio to divisibility Prove by iductio that if is a positive iteger, Solutio The best approach is a little differet to that used so far + 7 is divisible by 8 Assume that the result is true for k, i other words that + 7 is divisible by 8 ( k+ ) k+ Whe k+ the epressio is + 7 Cosider + 7 k + 7, the differece betwee the values whe k ad k+ This epressio is equal to k+ k k k ( ) k k 8 ( k+ ) k Thus, if k + 7 is divisible by 8, ad clearly 8 k is divisible by 8, it follows that ( k+ ) + 7 is also divisible by 8 I other words, if the result is true for k, it is true for k+ But for, ad is divisible by 8 Hece, + 7 is divisible by 8 for all positive itegers by iductio Eample 4 applicatio to de Moivre s theorem for positive itegers Prove by iductio that for itegers, Solutio Assume that the result is true for k, that is to say k cosθ + isiθ cos kθ + isi kθ Multiplyig both sides by cosθ + isi θ, cosθ + isiθ cos θ + isi θ k ( cosθ + isiθ) ( cosθ + isiθ) ( cos kθ + isi kθ)( cosθ + isiθ) k+ ( cosθ + isiθ) cos kθ cosθ + isi kθ cosθ + isiθ cos kθ + i si kθsiθ ( kθ θ kθ θ) ( kθ θ kθ θ) cos( k+ ) θ + isi ( k+ ) θ, cos cos si si + i si cos + cos si i which is of the same form but with k + replacig k Hece, if the result is true for k it is true for k+ But whe k, ( cosθ + i siθ) cosθ + i si θ Therefore the result is true for all positive itegers by iductio 49

50 Further Pure (MFP) Tetbook Eercise C Prove the followig results by the method of iductio i all eamples is a positive iteger: (a) (b) is divisible by 6 + is divisible by 7 [Hit: cosider f ( ) 5f ( ) 5 ( ) + f 5 ] d d (c) ( ) + where [Hit: use the formula for differetiatig a product] (d) is divisible by 50

51 Further Pure (MFP) Tetbook Miscellaeous eercises Use the idetity to show that [AQA Jue 999] r r r r + ( + ) r r r ( + ) + (a) Use the idetity 4 ( + ) ( ) r r r r r r to show that r 4r + (b) Hece fid r( r + ), givig your aswer as a product of three factors i terms of [AQA Jue 000] Prove by iductio that [AQA March 999] r r r + ( ) Prove by iductio, or otherwise, that [NEAB Jue 998] r ( r r ) ( )! +! 5 Prove by iductio that for all itegers 0, 7 + is divisible by [AQA Specime] 6 Use mathematical iductio to prove that for all positive itegers [AEB Jue 997] r ( r )( r ) ( ) 5

52 Further Pure (MFP) Tetbook 7 A sequece u, u, u,% is defied by u, u +, u Prove by iductio that for all, u + [AQA Jue 999] 8 Verify the idetity r r+ r r r r r r ( ) ( + ) ( )( + ) Hece, usig the method of differeces, prove that [AEB Jauary 998] r + ( r )( r+ ) ( + ) 9 The fuctio f is defied for all o-egative itegers r by f r r + r (a) Verify that f f r r Ar for some iteger A, statig the value of A (b) Hece, usig the method of differeces, prove that [AEB Jauary 000] r r + 0 For some value of the costat A, r r A + r r r ( + )( + ) ( + )( + ) (a) By settig, or otherwise, determie the value of A (b) Use mathematical iductio to prove the result for all positive itegers (c) Deduce the sum of the ifiite series % + + % [AEB Jue 000] 5

53 Further Pure (MFP) Tetbook Chapter 4: De Moivre s Theorem ad its Applicatios 4 De Moivre s theorem 4 Usig de Moivre s theorem to evaluate powers of comple umbers 4 Applicatio of de Moivre s theorem i establishig trigoometric idetities 44 Epoetial form of a comple umber 45 The cube roots of uity 46 The th roots of uity 47 The roots of z α, where α is a o-real umber This chapter itroduces de Moivre s theorem ad may of its applicatios Whe you have completed it, you will: kow the basic theorem; be able to fid shorter ways of workig out powers of comple umbers; discover alterative methods for establishig some trigoometric idetities; kow a ew way of epressig comple umbers; kow how to work out the th roots of uity ad, i particular, the cube roots; be able to solve certai types of polyomial equatios 5

54 4 De Moivre s theorem Further Pure (MFP) Tetbook I Chapter (sectio 4), you saw a very importat result kow as de Moivre s theorem It was proved by iductio that, if is a positive iteger, the cosθ + isiθ cos θ + isi θ De Moivre s theorem holds ot oly whe is a positive iteger, but also whe it is egative ad eve whe it is fractioal Let be a egative iteger ad suppose k The k is a positive iteger ad k ( cosθ + isiθ) ( cosθ + isiθ) k ( cosθ + isiθ ) cos kθ + isi kθ Some of the results obtaied i Chapter ca ow be put to use I order to remove i from the deomiator of the epressio above, the umerator ad deomiator are multiplied by the comple cojugate of the deomiator, i this case cos kθ isi kθ Thus, coskθ isikθ cos kθ + i si kθ cos kθ + i si kθ cos kθ i si kθ cos kθ isi kθ cos kθ + i si kθ cos kθ i si kθ cos kθ i si kθ cos kθ isi kθ cos kθ + si kθ cos kθ isi kθ cos( kθ) + isi ( kθ) cos θ + isi θ, as required If is a fractio, say p q where p ad q are itegers, the q p p pθ pθ cos θ+ isi θ cos q+ isi q q q q q cos pθ + isi pθ But p is also a iteger ad so Takig the th q root of both sides, cos pθ + isi pθ cosθ + isi θ p p q cos θ + isi θ ( cosθ + isi θ) q q p p [ q is a iteger] 54

55 Further Pure (MFP) Tetbook p p It is importat to poit out at this stage that cos θ + isi θ is just oe value of q q ( θ θ) p q cos + isi A simple eample will illustrate this If θ π, p ad q, the + + cos π isi π cos π isi π i But ( cos π isiπ) ( cosπ ad si π 0) + ad ± i So i is oly oe value of cos π + isiπ There are, i fact, q differet values of ( cos π ) show i sectio 46 cosθ + isiθ cos θ + isi θ for positive ad egative itegers, ad fractioal values of + isiπ p q ad this will be 55

56 Further Pure (MFP) Tetbook 4 Usig de Moivre s theorem to evaluate powers of comple umbers Oe very importat applicatio of de Moivre s theorem is i the additio of comple umbers of the form ( a b) Eample 4 Simplify π π + i The method for doig this will be illustrated through eamples cos + isi 6 6 Solutio It would, of course, be possible to multiply cos π + isi π by itself three times, but this would 6 6 be laborious ad time cosumig eve more so had the power bee greater tha Istead, cos π + isi π cos π + isi π cos π + isi π 0+ i i 56

57 Further Pure (MFP) Tetbook Eample 4 Fid ( i) 0 + i the form a+ i b Solutio Clearly it would ot be practical to multiply ( + i) by itself te times De Moivre s theorem could provide a alterative method but it ca be used oly for comple umbers i the form cosθ + isi θ, ad + i is ot i this form A techique itroduced i Chapter (sectio 4) ca be used to epress it i polar form# y O a Argad diagram, + i is represeted by the poit whose Cartesia coordiates are (, ) taθ so that + i cos π + isi π i cos π + isi π π 0π ( cos + isi 6 6 ) 04 i 5 i Now, r + ad Thus, θ π 6 ad [ ote that is raised to the power 0 as well] O θ r + i 57

58 Further Pure (MFP) Tetbook Eample 4 Simplify π π Solutio cos isi 6 6 De Moivre s theorem applies oly to epressios i the form cosθ + isiθ ad ot cosθ isi θ, so the epressio to be simplified must be writte i the form cos( π π ) + isi ( ) 6 6 ( cos π isi π ) cos( π ) isi ( π ) cos π π ( ) + isi 6 ( 6 ) cos π π ( ) + isi ( ) cos π isi π i Note that it is apparet from this eample that cosθ isiθ cos θ isi θ It is very importat to realise that this is a deductio from de Moivre s theorem ad it must ot be quoted as the theorem 58

59 Further Pure (MFP) Tetbook Eample 44 Fid ( + i ) Solutio i the form a+ i b The comple umber + i is represeted by the poit whose Cartesia coordiates are (, ) o the Argad diagram show here (, ) r α y O θ r + 6 4, ad taθ taα so that θ π Thus ( + i) + i Hece, π ( isi π ) π π 4cos + 4 cos + isi cos ( π ) isi ( π ) 64 + ( + 0 ) Eercise 4A Prove that cosθ isiθ cos θ isi θ Epress each of the followig i the form a+ i b: (a) ( cosθ isi θ ) 5 cos π isi π (b) 0 + (c) π π cos + isi (d) ( + i) 6 (e) ( i) 4 (g) ( + i) (f) 5 + i 59

60 Further Pure (MFP) Tetbook 4 Applicatio of de Moivre s theorem i establishig trigoometric idetities Oe way of showig how these idetities ca be derived is to use eamples The same priciples are used whichever idetity is required Eample 4 Show that cos θ 4 cos θ cos θ Solutio There are several ways of establishig this result The epasio of cos( A B) + ca be used to epress cos θ i terms of cosθ settig A θ ad B θ Similarly, the epasio of cos( θ + θ ) ca be used to give cosθ i terms of cos θ Usig de Moivre s theorem gives a straightforward alterative method cosθ + isi θ cosθ + isiθ usig the biomial epasio of ( p+ q) usig i cos θ + cos θ i siθ + cosθ i siθ + i siθ cos θ + i cos θ siθ cosθ si θ i si θ Now cosθ is the real part of the left-had side of the equatio, ad the real parts of both sides ca be equated, cos θ cos θ cosθ si θ cos θ cosθ( cos θ) sice cos θ + si θ 4cos θ cos θ Note that this equatio will also give si θ by equatig the imagiary parts of both sides of the equatio 60

61 Further Pure (MFP) Tetbook Eample 4 Epress ta 4θ i terms of ta θ Solutio ta 4θ si 4θ so epressios for si 4 θ ad cos 4θ i terms of si θ ad cosθ must be cos 4θ established to start with Usig de Moivre s theorem, cos4 θ + isi4θ cosθ + isiθ [ usig the biomial epasio] cos θ + 4cos θ isiθ + 6cos θ isiθ + 4cosθ isiθ + isiθ 4 4 cos θ + 4i cos θ siθ 6 cos θ si θ 4i cosθ si θ + si θ usig i Equatig the real parts o both sides of the equatio, 4 4 cos 4θ cos θ 6 cos θ si θ + si θ, ad equatig the imagiary parts, si 4θ 4cos θsiθ 4cosθsi θ Now, ta 4θ si 4θ cos 4θ 4cos θ siθ 4cosθsi θ 4 4 cos θ 6 cos θ si θ + si θ 4 Dividig every term by cos θ gives 4 siθ 4 si θ cosθ ta 4 θ cos θ 4 6 si θ + si θ cos cos 4 θ θ But taθ siθ so cosθ ta 4 θ 4taθ 4ta θ 4 6ta θ + ta θ 6