B.A. (PROGRAMME) 1 YEAR

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1 Graduate Course B.A. (PROGRAMME) YEAR ALGEBRA AND CALCULUS (PART-A : ALGEBRA) CONTENTS Lesso Lesso Lesso Lesso Lesso Lesso : Complex Numbers : De Moivre s Theorem : Applicatios of De Moivre s Theorem 4 : Applicatios of De Moivre s Theorem to Summatio of Seric 5 : Theory of Equatios I 6 : Theory of Equatios II Editor: Dr. S.K. Verma SCHOOL OF OPEN LEARNING UNIVERSITY OF DELHI 5, Cavalry Lae, Delhi-0007

2 Sessio 0-0 (000 Copies) School of Ope Learig Published By: Executive Director, School of Ope Learig, 5, Cavalary Lae, Delhi-0007 Laser Composig by : M/s Computek System (0 )

3 LESSON COMPLEX NUMBERS. Complex Number Defiitio : A ordered pair (x, y) of real umbers x, y is called complex umber. Illustratio : (, 6) (π, ), (0, ), (, 0), (,), (/, ) are complex umbers. Two complex umbers : (x, y) ad (p, q) are said to be equal if x p, y q. I symbols, (x, y) (p, q) x p ad y q.. Additio of Complex Numbers If α (a, b) ad β (c, d) be to complex umbers, we defie their sum α + β to be the complex umber (a + c, b + d). Symbolically (a, b) + (c, d) (a + c, b + d) Illustratios : (, 7) + (4, ) (7, 6) (p, q) + (0, q) (p q). Multiplicatio of Complex Numbers If α (a, b) ad β (c, d) be two complex umbers, we defie the product αβ to be the complex umber (ac bd, ad + bd). I symbols (a, b) (c, d) [(ac bd), (ad + bc)] Illustratios : (i) (, 4) (, ) [( ) 4.,. + 4 ( )] ( 4, ) (ii) (, 0) (, ) [(.) 0.( ),.( ) + 0.)] (, ).4 Divisio If α, β, be ay two complex umber (β 0) the by defiitio α β αβ. We shall also use the symbols α/β, to deote α β..5 Usual Notatio for Complex Numbers We shall ow see at to how we ca express a complex umber (a, b) i the usual way as a + ib. The mappig a (a, 0) associates to each real umber a, a complex umber (a, 0) whose first umber is a ad whose secod umber is 0. This mappig is compatible with additio ad multiplicatio. That is, if a ad b be ay two umbers, the a + b (a + b, 0) (a, 0) + (b, 0) ab (ab, 0) (a, 0) (b, 0) The above relatios mea the if we have ay relatio ivolvig real umbers, the it remais true if each real umber a i the relatios is replaced by the complex umber (a, 0). Because of this importat property we ca write ay relatio ivolvig complex umbers. Also (0, ) (0, ) (, 0)

4 which suggest that we may write (0, ) i If we agree to write a for (a, 0) ad i for (0, ) the (a, b) (a, 0) + (0, b) (a, 0) + (0, ) (b, 0) a + ib Thus we fid that if we set up the covetio of writig a for (a, 0) ad i for (0, ), the (a, b) will be writte as a + ib (where i ) We shall give below, some examples to illustrate the Algebraic operatio o complex umbers, with usual Notatio. Example. Simplify (4 + 5i) + ( 7i) Solutio. (4 + 5i) + ( 7i) (4 + ) + (5i 7i) 6 (5 7) i 6 i We ote that to add two complex umbers, we add the real parts ad the pure imagiary parts separately. With the ew otatio, the same example ca be solved as follows : Example. Simplify ( + i) + (5 7i) (4 + 5i) + ( + 7i) (4, 5) + (, 7) (4 +, 5 7) (6, ) 6 i Solutio. ( + i) + (5 7i) (5 7i) + i (5 7i) With the ew otatio, we have 0 4i + 5i i 0 + i ( 4 + 5) ( ) i (5 4) + i ( + i) (5 7i) (, ) (5, 7) [.5 ( 7), ( 7) +.5] (0 +, 4 + 5) (, ) ( + i) Note : By the above two examples, it is evidet that ay oe of the two otatios ca be used..6 Real ad Imagiary Parts of a Complex Number Let z x + iy be ay complex umber. The x is called the real part of z ad y is called the imagiary part of z. We write R(z) x, I(z) y. We say that the complex umber z is purely real if I(z) 0, ad that z is purely imagiary if R(z) 0. Also, the complex umber x iy is called the complex- 4

5 cojugate of x + iy, the symbol z is usually used to deote the complex-cojugate of the complex umber z. That is, if z x + iy, the z x iy. If z is the cojugate of z, the z is the cojugate of z that the relatio of cojugacy betwee complex umber is a symmetric oe. We therefore, geerally say that z ad z are cojugate complex umbers istead of sayig that z the complex-cojugate of z. The followig results follow directly from the defiitio. (i) The sum of two cojugate complex umbers is purely real. (ii) The differece to two cojugate complex umbers is purely imagiary. (iii) The product of two cojugate complex umbers is a o-egative real umber. I order to see the truth of the above three statemets, we have z x + iy so that z x i y i.e. z (x, y) so that z (x, y), the z + z x which is purely real [ z + z ( x, y) + ( x, y) ( x, 0)] z z iy, which is purely imagiary [ z z ( x, y ) ( x, y) (0, y)] zz x + y, which is a o-egative real umber. [ zz ( x, y) ( x, y) ( x + y, 0] We give below some solved examples with the usual otatio which would help us at a later state. Example. Simplify a + bi c + di To divide two complex umbers, multiply both umerator ad deomiator of the fractio by the cojugate of the deomiator. a c + + bi di ( a + bi) ( c di) ( c + di) ( c di) ( ac + bd) + ( bc ad) i 5 c + d ac + bd bc ad + i c + d c + d Example 4. Fid the value of ( + i) ( i) (8 + i) Solutio. Here ( + i) ( i) 6 + i 4i i (6 + ) + (i 4i) [ i ] (8 + i) (8 i) (8 + i) 64 i Example 5. Fid the value of ( + i). Solutio. ( + i) + i + i + i Example 6. Divide ( + i) by ( + i). Solutio. + i + i + i i + i ( + i) ( i) ( + i) ( i)

6 Example 7. Express 5 i i the form x + iy 4i Solutio. Here 5 i 4i ( + ) + ( + 6) i i + i 5 (5 i) ( + 4 i) ( 4 i) ( + 4 i) Example 8. Fid the roots of the quadratic equatio x 6x (5 + 8) (0 6) i Solutio. This ca be writte as (x 6x + 9) + 0 or (x ) i or x ± i x ± i + 4i 4 i Thus + i ad i are the roots of the quadratic equatio x 6x It is see that the roots are cojugate. Verify that these values satisfy the equatio. The same examples ca also be solved as x 6 ± ± 4 6 ± i ± i.7 Graphical Represetatio of a Complex Number The complex umber z x + iy may be represetated graphically by a poit P whose rectagular coordiates are (x, y) i.e., we associate with each complex umber z x + iy the poit of the plae which has, with referece to a fixed rectagular system, the coordiates x ad y ad coversely with each poit havig the co-ordiates (x, y) we associate the complex umber z x + iy. From each theorem ivolvig complex umbers we ca deduce a defiite relatioship betwee the geometrical poits of the Cartesia plae ad coversely, the diagram showig poits which represet complex umbers geometrically is called Argad Diagram. 6

7 It may be oted that all poits o the x-axis are of the form (x, 0) ad are x + oi x ad so correspods to real umbers x. Similarly poits o the y-axis correspod to pure imagiary umbers yi. I additio, the complex umber may be repeseted by the directed lie segmet or vector OP..8 Trigoometric form of Complex Numbers The complex umber z x + iy is represeted by the vector OP. This vector ad hece the complex umber is described i terms of the legth r of the vector ad the agle θ which is vector makes with the positive directio of the x-axis (real axis) measured positively. real. Thus OP r x + y ad ta q y/x Thus a complex quatity ca always be put i the form r (cos θ + i si θ) where r ad θ are both Let x + iy r (cos θ + i si θ) Equatig real ad imagiary parts, we get By squarig ad addig, we get x r cos θ, ad y r si θ. x + y r r x + y 7

8 Here, we take oly positive sig of the square root. Also by dividig y x or θ r si θ ta θ r cos θ y ta ( x 0) x r x + y is called the modulus or absolute value of the complex umber x + iy ad θ is called amplitude (or argumet or phase). There are may values of θ satisfyig the equatio ta θ y/x, the value of θ, such that π < θ π is called the pricipal value of the amplitude. We shall geerally take the pricipal value of θ. I symbols the modulus of a complex umber z x + iy is also deoted by z or x + iy i.e., z x + iy x + y That amplitude is deoted by the symbols amp z or arg z. amp (x + iy) amp z θ here the value of θ is so choose that it is the pricipal value i.e., π < θ π The form r(cos θ + i sig θ) is called the stadard or the polar form of the complex umber x + iy, R, y R. Example. Express + i i the stadard form. Solutio. Let + i r (cos θ + i si θ) Equatig real ad imagiary parts, we get r cos θ; r si θ The r r Also ta θ i.e., θ Here cos θ is positive ad si θ is positive, ad hece θ lies i the first quadrat. θ π π Hece + i cos + i si 4 4 π The modulus of the complex umber is ad the amplitude is. 4 Example. Express i i the polar form. Solutio. Let i r (cos θ + i si θ) so that r cos q, π 4 ρ si θ r, ta θ π 4 ta y x 8

9 Here cos θ is +ve ad si θ is ve. θ lies i the fourth quadrat Also sice π < θ π θ π π π Hece i cos + i si Example. Express si α + i cos α i the stadard form. Solutio. Let si α + i cos α r (cos θ + i si θ) so that si α r cos θ cos α r si θ θ r ta θ cot α r, ta θ cot α π ta α π α π π Hece si α + cos α cos α + i si α.. Simplify the followig : (i) ( 5i) + ( + 4i) + (8 i) (ii) 5 + i i. Evaluate the followig : (i) ( 7 i) ( 5i) (ii) ( + 4 i) ( i). Solve the followig equatios : (i) [(x + y), (x y 6) (, ) (ii) (x y) + i (x + y) + i + i EXERCISE 4. Express the followig complex umber i the polar form (i) + i (ii) i 5. Fid the trigoometric represetatio of (i) si α i cos α (ii) + cos α + i si α 9

10 6. Show that cos θ + i si θ 7. Express the complex umber 5 + i + i ad amplitude. i the form (x + iy) where x ad y are real. Fid its modulus 0

11 LESSON DE MOIVRE S THEOREM I this lesso, we shall discuss a importat theorem which is used i fidig the roots of a complex umber, solutio of equatios, expasio of trigoometric fuctio etc. This theorem is kow as De Moivre s Theorem.. De Moivre s Theorem Statemet : (i) If is a iteger, positive, or egative, the (cos θ + i si θ) cos θ + i si θ. (ii) If is a ratioal umber, the oe of the values of (cos θ + i sig θ) is Proof : Part (i) case (a) : Let be a positive iteger By simple multiplicatio (cos α + i si α) (cos β + i si β) (cos α cos β si α si β) + i (si α cos β + cos α si β) cos (α + β) + i si (α + β) Agai multiplyig the above results by (cos γ + i si θ), we have (cos α + i si α) (cos β + i si β) (cos γ + i si θ) [cos (α + β) + i si (α + β)] (cos γ + i si γ) cos (α + β + γ) + i si (α + β + γ) This process ca be cotiued to ay umber of factors so that (coa α + i si α) (cos β + i si β) (cos γ + i si γ)... to factors I this expressio, put α β γ... θ. cos (α + β + γ +... to terms) + i si (α + β + γ +... to terms) So that we have (cos θ + i si θ) cos θ + i si θ. Case (b) : Let be a egative iteger. Suppose m where is a positive iteger. The (cos θ i si θ) (cos θ + i si θ) m (cos θ + i si θ) m (cos m θ + i si m θ) (cos m θ i si m θ) (cos m θ + i si m θ)(cos m θ si m θ) (cos m θ i si m θ ) cos m θ i si m θ (cos m θ + si m θ) cos( m) θ + i si ( m) θ co θ + i si θ (by the law of idices) (by case (a))

12 Thus (cos θ + i si θ) cos θ + i si θ for itegral values of, whether positive or egative p Proof of (ii) : If is (a rotatioal umber, positive or egative), let ( q 0) q We shall take q as a positive iteger ad p as a positive or a egative iteger. θ θ We haver cos + i si q q q θ θ cos q + i si q q q cos θ + i si θ θ θ i.e., qth power of cos + i si is cos θ + i si θ q q θ θ i.e., cos + i si is oe of the qth roots of (cos θ + i si θ) q q θ θ i.e., cos + i si is oe of the values of q q (cos θ + i si θ ) q Raisig each of these quatities of pth power, we have the results that oe of the values of p/ q θ θ (cos θ + i si θ ) is cos + i si q q p p i.e., cos θ i si θ q q p Hece cos θ + i si θ is oe of the values of (cos θ + i si θ) if is a ratioal umber positive or egative. Corollary. cos θ i si θ is oe of the values of (cos θ i si θ) for all ratioal values of. Illustratives : With the help of the De Moivre s Theorems, we see that. (cos θ + i si θ) 4 cos 4 θ + i si 4 θ. θ θ cos + i si 7 7θ 7θ cos + i si. (cos θ + i si θ) 7 cos ( 7θ) + i si ( 7θ) cos 7θ i si 7θ 4. Sice (cos θ + i si θ) (cos θ i si θ) cos θ i si θ cos θ + i si θ π π 5. si θ + i cos θ cos θ + i si θ (cos θ + si θ) + i (si θ cos θ si θ cos θ) + i0 6. (si θ + i cos θ) π π cos θ + i si θ

13 e.g., (si θ + i cos θ) π π cos θ + i si θ π π cos θ + i si 0 π π cos θ + i si θ si θ i cos θ Remarks : (i) As ay complex umber x + iy ca be put i the form r (cos θ + i si θ), therefore, De Moivre s Theorem may be writte as ( x + iy) r (cos θ + i si θ ) r (cos θ + i si θ ) where r is modulus of the complex umber ad θ is the amplitude. (ii) The product ad quotiet of ay two complex umbers ca be obtaied by usig this method. Let ay two complex umber be give by the their product is r (cos θ + i si θ ) ad r (cos θ + i si θ ), r r [cos (θ + θ ) + i si (θ + θ )] i.e., the product of two complex umbers is a complex umber whose modulus is the product of the modulli ad whose amplitude is the sum of the amplitudes two complex umbers. The quotiet of these two complex umbers, r (cos θ + i si θ ) ad r (cos θ + i si θ ) is Solved Examples Example. Simplify r (cos θ + i si θ ) r (cos θ + i si θ ) r (cos θ + i si θ) (cos θ i si θ ) r r [cos( θ θ ) + i si ( θ θ )] r 4 5 (cos θ + i si θ) (cos θ + i si θ) 7 (cos 6θ + i si 6 θ) (cos 4θ i si 4 θ) Solutio. By De Moivre s Theorem, we have Now, cos θ + i si θ (cos θ + i si θ) cos θ i si θ (cos θ + i si θ) cos 6θ + i si 6θ (cos θ + i si θ) cos 4θ i si 4θ (cos θ + i si θ) (cos θ + i si θ) (cos θ i si θ) 7 (cos 6θ + i si 6 θ) (cos 4θ i si 4 θ). 8 5 (cos θ + i si θ) (cos θ + si θ) 8 8 (cos θ + i si θ) (cos θ + i si θ) (cos θ + i si θ)

14 Example. Simplify (cos θ + i si θ) 6 cos θ + i si θ) cos θ + i si θ. 5 (cos θ + i si θ) (cos θ i si θ) 8 (cos θ i si θ) (cos 5θ + i si 5 θ) Solutio. By De Moiver s Theorem, we have θ + i θ θ i θ 8 θ i θ θ + i θ (cos θ + i si θ) (cos θ + si θ) (cos θ + i si θ) 4 (cos θ + i si θ) 0 (cos si ) (cos si ) (cos si ) (cos 5 si 5 ) (cos θ + i si θ) (cos θ + i si θ) 5 cos 5θ + i si 5θ + i Example. Obtai the quotiet of by usig De Moivre s Theorem. + i Solutio. Example 4. Obtai ( + i) 4. + i + i Solutio. Here ( + i) 4 Example 5. Prove that Solutio. Let + si φ + i cos φ + si φ i cos φ. i π π + cos si + i π π + i cos + i si 4 4 π π π π cos + i si 4 4 5π 5π cos + i si. 4 π π + cos + i si 6 6 4π 4π 6 cos + i si 6 6 π π 6 cos + i si i i π π cos φ + i si φ + si φ + i cos φ r (cos φ + i si θ). 4 4

15 The equatig real ad imagiary part, we have + si φ r cos θ, cos φ r si θ π si cos φ φ Hece, ta θ + si φ π + cos φ si α π, where α φ + cos α θ α α si cos cos α π φ 4 5 α ta α Also + si φ i cos φ r(cos θ si θ) The give expressio Example 6. Prove that : (a + bi) m/ + (a bi) m/ r (cos θ + si θ) (cos θ + i si θ) r (cos θ i si θ) (cos θ + i si θ) (cos θ + i si θ) (cos θ + i si θ) (cos θ + i si θ) (cos θ + i si θ) cos π φ i si π φ π π cos φ i si φ m/ m b ( a + b ) cos ta a Solutio. Here put a + bi r (cos θ + i si θ) The a r cos θ ad b r si θ a + b r (cos θ + si θ) r a + b ad ta θ Hece r b a a + b ad θ ta Also a bi r cos θ i si θ). Thus the give expressio ca be writte as (a + bi) m/ + (a bi) m/ b a m / m/ [ r(cos θ + i si θ )] + [ r(cos θ i si θ )] m / m m m/ m m r cos θ i si θ r cos θ i si θ + +

16 Example 7. If x cos θ + i si θ, the show that ad Solutio. Here ad Hece Also, x x x x m m m m m / mθ r cos m/ m b ( a + b ) cos ta a + m x cos mθ m x i si mθ. x m (cos θ + i si θ) cos mθ + i si mθ, m x (cos θ + i si θ) m cos mθ i si mθ. + m x cos mθ m x i si mθ Example 8. If x + cos, show that x cos. x θ + x θ Solutio. Here, it is give that x + cos θ x or x x cos θ + 0 or x x cos θ + (cos θ + si θ) 0 or (x x cos θ + cos θ) si θ or (x cos θ) ( i si θ) x cos θ ± i si θ. i.e., x cos θ ± i si θ Let x cos θ + i si θ (takig the + ve sig). The, x x cos θ i si θ cos θ + i si θ + x (cos θ + i si θ) + (cos θ si θ) (cos θ + i si θ) + (cos θ i si θ) cos θ If x cos θ i si θ, the cos θ + i si θ. The also the result ca be proved. x Example 9. If si α + si β + si γ 0 ad cos α + cos β + cos γ 0. Prove that cos α + cos β + cos γ cos (α + β + γ) 6

17 Solutio. Let si α + si β + si γ si (α + β + γ) x cos α + i si α y cos β + i si β z cos γ + i si γ The x + y + z (cos α + cos β + cos γ) + i (si α + si β + si γ) 0 + i0 0 x + y + z xyz (cos α + i si α) + (cos β + i si β) + (cos γ + i si γ) (cos α + i si α) + (cos β + i si β) + (cos γ + i si γ) (cos α + cos β + cos γ) + i (si α + si β + si γ) (cos α + i si α) (cos β + i si β) (cos γ + i si γ) [cos(α + β + ψ) + i si (α + β + ψ)] cos(α + β + γ) + i si (α + β + γ) Equatig real ad imagiary parts, we get cos α + cos β + cos γ cos(α + β + γ) ad si α + si β + si γ si (α + β + γ) Hece the results. Example 0. Show that for a iteger ( ) ( cos si ) ( cos si ) + cos π i i cos θ + θ + θ + θ θ Solutio. We have ( + cos θ + i si θ) + ( + cos θ i si θ) cos θ i si θ cos θ cos θ i si θ cos θ + + cos θ cos θ i si θ cos θ cos θ i si θ + + θ cos θ θ cos isi θ cos θ θ cos i si cos θ cos θ si θ cos θ i i si θ cos θ cos θ i si θ cos θ i si θ + + θ θ cos cos + θ θ cos cos Example. If si α + si β + si γ 0 cos α + cos β + cos γ. Prove that 7

18 Solutio. Let (i) cos(β + γ) + cos(γ + α) + cos(α + β) 0 (ii) si(β + γ) + si(γ + α) + si(α + β) 0 x cos α + i si α, y cos β + i si β, z cos γ + i si γ so that x + y + z (cos α + cos β + cos γ) + i(si α + si β + si γ) Also 0 + i0 0 x cos α si, cos si, i α i cos i y β β z γ si γ + + (cos α + cos β + cos γ) i (si α + si β + si γ ) x y z yz + zx + xy 0 xy + yz + zx 0 0 i0 0 (cos α + i si α) (cos β + i si β) + (cos β + i si β) (cos γ + i si γ) [cos (α + β) + i si (α + β)] + [cos (β + γ) + i si (β + γ)] [cos (α + β) + cos (β + γ)] + cos (γ + α)] + i [si (α + β) + si (β + γ)] Equatig real ad imagiary parts o both sides, we get cos (α + β) + cos (β + γ) + cos (γ + α) 0 si (α + β) + si (β + γ) + si (γ + α) 0 Hece the result. Example. If Solutio. Z Z Z Z Z cos θ + i si θ, show that i ta θ, beig a iteger. + + Z Z Z + Z 8 + (cos γ + i si γ) (cos α + i si α) 0 + [cos (γ + α) + i si (γ + α)] 0 (Dividig umerator ad deomiator by Z ) (cos θ + i si θ) (cos θ + i si θ) (cos θ + i si θ ) + (cos θ + i si θ) (cos θ + i si θ) (cos θ i si θ) (cos θ + i si θ ) + (cos θ i si θ) i si θ i ta θ proved cos θ Example. If (cos θ + i si θ) (cos θ + i si θ)... (cos θ + i si θ) prove that θ Solutio. We have cos 0 + i si 0 4kπ, where k is a iteger. ( + ) + si (γ + α)] 0

19 cos kπ + i si kπ..., where k is a iteger....() Also (cos θ + i si θ) (cos θ + i si θ)... (cos θ + i si θ) Thus L.H.S. cos (θ + θ θ) + i si (θ + θ θ) ( + ) ( + ) cos θ + i si θ Q ( + ) ( + ) cos θ + i si θ cos kπ + i si kπ by () This gives us Example 4. If ( + x) (i) (ii) ( + ) 4kπ θ kπ θ ( + ) x, ( + ) Hece the result. C C x C x KK C x is a positive iteger. Prove that C0 + C4 + C8 + KK + cos π 4 π C0 C + C4 C 6 + KK cos 4 / (iii) π C C + C5 C 7 KK si 4 / Solutio. We are give that ( + x) 4 C0 Cx Cx C x C4x KK Puttig x,, i, i i successio, we get Addig (), (), (4) ad (5) we get C0 + C + C + C + C4 +KK C...() 0 C0 C + C C + C4 KK...() ( + i) C0 i C C C i + C4 +KK...() C0 C C4 i C C C5 ( + ) + ( + KK )...(4) ( i) C0 i C C + i C + C4KK...(5) + ( + i) + ( i) 4( C0 + C4 + C8 +...)...(6) Now ( + i) π π cos + i si 4 4 Similarly, ( i) Addig (7) ad (8) ad substitutig i (6), we get C0 C4 C8 4( + + +KK ) / π π cos + i si (7) / π π cos i si (8) / +. cos π 4 9

20 + π + cos 4 C0 C4 C8 (ii) From (4) ad (7) + + +KK π cos 4 C0 + C4 + C8 + KK + i C C + C5 + Proved ( ) (...) / π π cos + i si 4 4 Equatig real ad imagiary parts, we get C0 C C4 + KK / cos π/4 C C C5 + KK / si π / 4. Simplify. Evaluate: 8 EXERCISE. (cos θ + i si θ) (cos θ i si θ) 9 8 (cos θ i si θ) (cos θ i si θ) 0 0 π π π π cos + i si + cos i si π π cos + i si. Use De Moivre s Theorem to simplify 6 9 (cos θ + i si θ) (si θ + i cos θ) 4. Prove that (si x i cos x) π π cos x i si x 5. Obtai the value of ( + 4i) with the help of De Moivre s Theorem. 6. If is a positive iteger, show that : ( + i) + ( i) π cos Prove that : [cos θ + cos φ) + i (si θ + si φ)] + [cos θ + cos φ) i si θ + si φ)] + θ φ ( θ + φ) cos cos Hit : (cos θ + cos φ + i(si θ + si φ) θ + φ θ φ θ + φ θ φ cos cos + i si cos θ φ θ + φ θ + φ cos cos + i si 0 4

21 8. Show that [(cos θ cos φ) + i (si θ si φ)] + [(cos θ cos φ) i(si θ si φ)] + θ φ π + θ + φ si cos Hit : (cos θ cos φ + i (si θ si φ) θ + φ θ φ θ φ θ + φ si si + i si cos θ φ θ + φ θ + φ si si + i cos θ φ π θ + φ π θ + φ si cos + + i si +. If cos θ Show that x +, cos φ y + x y m x y EXERCISE. + m x y cos (mθ + θ) x y m y + cos (mθ θ) m x where m ad are itegers. What happes if m ad are ratioal umbers?. If x cos α + i si α, y cos β + i si β ad z cos γ + i si γ such athat x + y + z 0 the prove that : ad x y z xyz + cos (α + β + γ) xyz. Let the complex umbers x, y, z be give respectively by cos a + i si a, cos b + i si b ad cos c + i si c, the prove that for ay itegers p, q, r. p q r x y z + p q r x y z cos (pa + qb + rc) 4. Let x + iy, x + iy..., x + iy be ay complex umber ad A + ib be some other complex umber, such that (x + iy ) (x + iy )... ( x + iy ) A + ib The show thats y y y B (i) ta ta ta ta x x x A (ii) + + KK + + ( x y ) ( x y ) ( x y ) A B

22 5. Expad ( + i) i two differet ways ad fid the sum of the two series, C C4 C6 KK C C C5 ( + + ) ad + KK ) [Hit : First expad ( + i) by Boiomial theorem ad the expad it by help of De Moivre s Theorem. 6. If si α + si β + si γ cos α + cos β + cos γ 0. Prove that (i) si α + si β + si γ 0 (ii) cos α + cos β + cos γ 0 (iii) cos α + cos β + cos γ /

23 LESSON APPLICATIONS OF DE MOIVRE S THEOREM I this lesso, we shall discuss certai applicatios ad uses of De Moivre s Theorem. As metioed earlier, it helps i fidig the roots of a umber, solutio of some equatios ad expressios of trigoometric fuctios, etc. We have see that De Moivre s Theorem ca be stated as θ θ θ θ cos q + i si q cos + i si q q q q θ θ i.e., cos θ + i si θ cos + i si q q θ θ cos + i si is a q th root of cos θ + i si θ q q θ θ i.e., cos + i si is oe of the values of (cos θ + i si θ) /q q q Now the questios arises : q q, where q is a positive iteger. What about the other values of (cos θ + i si θ) for ratioal values of? We shall ow determie all the values of (cos θ + i si θ) /q, where θ is a positive iteger. We kow that (cos θ + i si θ) cos (πr + θ) + i si (πr + θ) (cos θ + i si θ) /q [cos (πr + θ) + i si (πr + θ)] /q where q is ay iteger. Now by De Moiver s Theorem oe of the values of (cos θ + i si θ) /q is π r + θ π r + θ cos + i si q q By givig r various values, we get differet values of (cos θ + i si θ) /q. Let us give r i successio the values 0,,,... q ad we see that each of the followig quatities θ θ cos + i si q q π + θ π + θ cos + i si q q 4π + θ 4π + θ cos + i si q q ( q ) π + θ ( q ) π + θ cos + i si q q

24 is equal to oe of the values of (cos θ + i si θ) /q. We ote here that if we give to r itegral values greater tha (q ) viz., the values q, (q + ), (q + ),... etc. the we do ot obtai ay ew values of (cos θ + i si θ) /q. For example, for r q we obtai the value π q + θ π q + θ cos + i si q q i.e., θ θ cos π + + i si π + q q i.e., θ θ cos + i si. q q which is the same values as obtaied by puttig r 0. Thus the greatest values we eed to assig to r is q for the values q, q +, q +... will be foud to give the same result as the values, 0,,,... q, for r. Also o two of the quatities obtai by givig r, the values 0,,,... q will be the same for all the agles ivolved there i differ from oe aother by less tha π ad o two agles differig by less tha π have their cosies the same ad also sies the same. Thus the expressio cos ( π r + θ) ( π r +θ) + i si, where r 0,,,... (q ) gives q ad oly q q q differet values of (cos θ + i si θ) /q. We may exted the above result to (cos θ + i si θ) /q where p ad q are itegers, q beig take as positive. Thus (cos θ + i si θ) /q has q ad oly q differet values ad the these are give by ( π r + θ) p ( π r + θ) p cos + i ai, where r 0,,,... ( q ). q q The, the θ values are pθ pθ cos + i si, whe r 0 q q ( π + θ) p ( π + θ) p cos + i si, where r q q (4 π + θ) p (4 π + θ) p cos + i si, where r q q... [ π( q ) + θ] p [ π( q ) + θ] p cos + i si, whe r q. q q Note. We may ote here that the q distict values of (cos θ + i si θ) p/q ca be arraged i a geometrical progressio, for if µ cos π p p p p + i si π ad a cos θ + i si θ. q q q q the the q values of (cos θ + i si θ) p/q ca be arraged as a, aµ, aµ,... aµ q which is a G.P. The studet should ote that 4

25 p( π r + θ) p( π r + θ) πp πp pθ pθ cos + i si cos + i si cos + isi q q q q q q. Extractio of Roots of Complex Quatity We kow that where r x + iy r (cos θ + i si θ), x y + y ad θ ta. x Hece (x + iy) /q r /q [cos θ + i si θ] /q / q / q r [cos( π + θ ) + i si ( π + θ )] / q ( π + θ) ( π + θ) r cos + i si q q whe is give i successio the values, 0,,,..., q. This gives the required q th roots of x + iy. It may be oted that eve if we write x + iy r [cos ( kπ + θ ) + i si ( kπ + θ )], k beig a iteger, the same values are obtaied. Example. Fid the values of ( + i) /6. π π Solutio. We kow that cos si. 4 4 We ca write ( + i) ( + i) /6 whe is give the six values 0,,,, 4, ad 5. The six values of ( + i) /6, are therefore + i π π cos + i si r π π cos π + + i si π π + π/4 π + π/4 cos + i si cos π i si π, cos π i si π +, cos π i si π, cos π i si π +, cos π i si π, cos π i si π It may be observed that the 4 th value, viz.,

26 5π 5π cos + i si 4 4 π π cos π + + i si π π π cos + i si, 4 4 sice cos (π + θ) cos θ ad si (π + θ) si θ. Similarly the 5th ad 6th values ca simplified. The six value of ( + i) /6 ca the be writte as or simply as where r, 9 ad 7. π π ± cos + i si, 4 4 9π 9π ± cos + i si, 4 4 7π 7π ± cos + i si, 4 4 rπ rπ ± cos + i si, 4 4 Example. Fid all the values of values is. Solutio. Let the ad so that r + i /4 + i r (cos θ + i si θ ), r cos θ r si θ ad ta θ ta π ad show that the cotiued product of all the or Now i + /4 r ad π θ 4 π π cos + i si (cos π + i si π) /4 6

27 [cos (kπ + π) + i si (kπ + π)] /4 π π cos (k + ) + i si (k + ), k 0,,, 4 4 Hece the required four values are π π whe k 0 cos + i si, 4 4 k k π π cos + i si, 4 4 5π 5π cos + i si, 4 4 7π 7π k cos + i si, 4 4 Ad the cotiued product of four values cos π i si π cos π i si π cos π i si 5π cos π i si π cos π π π π i si π π π π cos 4π + i si 4π + i.0 [ cos 4π, si 4π 0]. Example. Fid the seve 7th roots of uity ad prove that the sum of their th powers always vaishes uless be a multiple of 7, beig iteger ad that the sum is 7 whe is multiple of 7. Solutio. We have to fid the 7 values of () /7, We kow that cos 0 ad si 0 0 We ca write cos 0 + i si 0 cos π + i si π () /7 π π cos + i si 7 7 where 0,,,, 4, 5, 6. The 7 roots are, therefore (cos 0 + i si 0) i.e.,, π π cos + i si ω (say) 7 7 4π 4π π π cos + i si i. e., cos + i si ω ad similarly for the other roots. Thus the seve 7th roots of uity are, ω, ω, ω, ω 4, ω 5, ω 6 7

28 where ω The sum of the th power of the roots Sice ω 7 Also ω, sice is ot a multiple of 7. π π cos + i si. 7 7 () +s (ω) + (ω ) + (ω ) + (ω 4 ) + (ω 5 ) + (ω 6 ) + ω + ω + ω + ω 4 + ω 5 +ω ( ω ) ( ω ) ( ) 0 ω ω ω ω If is a multiple of 7, i.e., if 7 m say where m is a iteger, the ω cos π i si π cos π i si π cos πm + i si πm. 8 (which is geometric progressio) Hece whe is a multiple of 7, the sum of the th power of the 7th roots of uity is Example 4. Prove that x + iy + x iy has real values. Solutio. Let x r cos θ ad y r si θ. The where k is a iteger. (x + iy) / + (x + iy) / x + iy r[cos (kπ + θ) + i si (kπ + θ)], / / / r {[cos( kπ + θ ) + i si( kπ + θ )] + [cos( kπ + θ) i si( kπ + θ )] } / kπ + θ r cos, where k 0,,,... ( ) Here r y x + y, ad ta θ. x Hece (x + iy) / + (x iy) / kπ + ta / ( x + y ) cos where k 0,,,... ( ). This is real ad has distict values.. Solutio of Equatios De Moivre s Theorem ca be used to obtai solutios of certai types of equatios. This will be illustrated by a few solved examples. Coversely De Moivre s Theorem ca be used to fid the equatio whose roots are give as trigoometical fuctios. Example 5. Solve the equatio z, where is a positive iteger. Solutio. Here we write y x 7m

29 z z cos (kπ + 0) + i si (kθ + 0) / kπ kπ () cos + i si where k 0,,,... ( ) Hece the th roots of uity are π π (cos 0 + i si 0), i.e.,, cos + i si 4π 4π cos + i si It may be oted that if we write ω π π cos + i si ( ) π ( ) π π π cos + i si cos + i si π π cos + i si, the the th root of ca be writte as, ω, ω,... ω. Thus we get a iterestig result that the th roots of uity form a G.P. Also it is see that each of these roots ca be expressed as a power of aother. Exmple 6. Solve the equatio z 9 z 5 + z 4 0. Solutio. Here (z 9 z 5 + z 4 ) 0 z 5 (z 4 ) + (z 4 ) 0 i.e., (z 5 ) (z 4 ) 0 Either (i) z 4 or (ii) z 5 sice ad From (i), we have ad form (ii), we have (cos π + i si π) /4 cos π + i si π cos π + i si π cos (π + π) + i si (π + π) cos ( + ) π + i si ( + ) π. π π cos + i si π π z cos + i si, where 0,,,, z ( ) /5 [cos ( + ) p + i si ( + ) π] /5 ( + ) π ( + ) π cos + i si, whe 0,,,, These (4 + 5) 9 values give the complete solutio of the give equatios. Example 7. Solve the expressio z Solutio. z 5 [cos (r + ) π + i si (r + )π]

30 where r is a iteger. z π π Whe r 0, z cos + i si. 5 5 Whe r, (r + ) π (r + ) π cos + i si, where r 0,,,, π π z cos + i si 5 5 Ad for r, z cos π + i si π. For r, the value of z is For 7π 7π cos + i si 5 5 π π cos π + i si π 5 5 π π cos + i si 5 5 π π cos i si 5 5 r 4, the value of z is 9π 9π cos + i si 5 5 π π cos π + i si π 5 5 π π cos + i si 5 5 π π cos i si 5 5 Thus the 5 roots of the equatio z are, cos π ± i si π, cos π ± i si π Example 8. Fid the roots of the equatio z 6 + z + 0 Solutio. O solvig the equatio a quadratic i z, we fid z ± ± i π π cos ± i si π π cos rπ + ± i si rπ + π π cos (r + ) ± i si (r + ) 0

31 0. z ( ) ( ) cos π r + r i si π + ± π (r + ) π (r + ) cos ± i si, r 0,,. 9 9 O givig r the values, 0,,, the six roots of the give equatio ca be writte as Now π π 8π 8π 4π 4π cos ± i si, cos + i si, cos ± i si cos π i si π cos π i si π ± π ± π cos π i si π + cos π ± i si π 9 9 p 9 π π 4π 4π 8π 8π The roots are cos ± isi, cos ± isi, cos ± isi Example 9. Solve the equatio z 5 + z 4 + z + z + z + 0 multiply through by z we get z 6 Solutio. z 5 + z 4 + z + z + z + 0 Now z 6 0 z z 6 cos kπ + i si kπ, where k is a iteger cos kπ kπ + i si 6 6 kπ kπ cos + i si, where k 0,,,, 4, 5. π π π π π z cos 0 + i si 0, cos + i si ; cos + i si + i si ; cos π + i si π ±, ± + i The root correspods to the factor (z ) The remaiig roots cos π + i si π ; cos π + i si π ;, + i, + i,, i, i, ± ± i are the roots of the give equatio Exercise. Fid all the values of (i) ( + i) / (ii) ( + i ) (iii) ( ) /6 4

32 (iv) ( + i) /. Solve the equatio: (i) z 7 (ii) z 7 + z 0. Fid the th roots of uity ad show that from a series i G.P. whose sum is zero. Also prove that the sum of their pth power always vaishes uless p be a multiple of, p beig a iteger ad that the the sum is. 4. Solve the equatio : z 7 + z 4 + z + 0. [Hit : z 7 + z 4 + z + (z 4 + ) (z + ) 0] 5. Use De Moivre s Theorem to solve : z 0 + z [Hit : Put z 5 y] 6. Solve the equatio : z 4 z + z z + 0. [Hit : Multiply the equatio by (z + )]. Expasio of si θ ad cos θ De Moivre s Theorem may be applied to express trigoometric fuctios of multiple agles i terms of the trigoometric fuctio of the agles. The theorem ca be employed to express powers of sies ad cosies of agle i terms of trigoometric fuctios of multiple agles. First we shall lear to express cos θ, si θ, ta θ i terms of cos θ, si θ, ta θ, etc. For example, by De Moivre s Theorem cos θ + i si θ (cos θ + i si θ) cos θ + i si θ cos θ + i si θ (cos θ si θ) + i cos θ si θ) Equatig the real ad imagiary parts, we get Geerally cos θ cos θ si θ, si θ cos θ si θ (cos θ + i si θ) (cos θ + i si θ), beig a positive iteger. By the Biomial Theorem for a positive itegral idex, we have (cos θ + i si θ) But Hece we ca write cos θ + i si θ ( ) cos θ + cos θ ( isi θ ) + cos θ ( isi θ ) ( + ) ( ) cos ( si )... ( si ) + θ i θ + i θ i, i 4, i etc. ( ) cos θ cos θ si θ ( ) ( ) ( ) cos 4 si 4 + θ θ ( )( ) + i cos θ si θ cos θ si θ + KK

33 By equatig the real ad imagiary parts, we get cos θ si θ ( ) cos θ cos θ si θ ( ) ( ) ( ) cos 4 si 4 + θ θ +KK 4 ( )( ) cos θ si θ cos θ si θ +KK By usig the biomial coefficiets, we ca also express these results as follows cos θ si θ Dividig (ii) by (i), we have ta θ 4 4 C C4 cos θ cos θ si θ + cos θ si θ C C4 cos θ[ ta θ + ta θ +...]...(i) C C cos θ si θ + cos θ si θ +... cos θ[ C ta θ C ta θ +...]...(ii) 5 C ta C ta C5 ta 4 C ta θ + C4 ta q θ θ + θ KK KK We have thus obtaied the values of cos θ, si θ, ta θ i terms of cos θ, si θ, ta θ. Example 0. Obtai the values of cos 5θ, si 5θ, ad ta 5θ. Solutio. We ca either apply the formulae or proceed as below: cos 5θ + i si 5θ (cos θ + i si θ) 5 cos 5 θ + 5 cos 4 θ (i si θ) + 0 cos θ(i si θ) + 0 cos θ (i si θ) + 5 cos θ (i 4 si 4 θ) + i 5 si 5 θ Thus equatig real ad imagiary parts, cos 5θ cos 5 θ 0 cos θ si θ + 5 cos θ si 4 θ cos 5 θ 0 cos θ ( cos θ) + 5 cos θ ( cos θ) 6 cos 5 θ 0 cos θ + 5 cos θ (Note that cos 5θ has bee expressed i powers of cos θ oly). Also ta 5θ 4 5 si 5θ 5 cos θ si θ 0 cos θ si θ + si θ cos 5 cos 5 0 cos si 5 cos si 4 θ θ θ θ + θ + θ 5 5 ta θ 0 ta θ + ta θ 5 0 ta θ + 5 ta θ (Note that ta 5θ has bee expressed i powers of cos θ oly. Note. We ca apply a similar method to obtai si (α + β + γ +...), cos (α + β + γ) +...), ta (α + β + γ +...) ad thus obtai expressio for the sie, cosie ad taget of the sum of ay umbers of agles i terms of trigoometric ratio of idividual agles. We have

34 Sice We have cos (α + β + γ +...) + i si (α + β + γ +...) (cos θ + i si α) (cos β + i si β) (cos γ + i si γ)... cos α + i si α cos α( + ta α) cos β + i si β cos β( + i ta β) cos ψ + i si γ cos γ( + i ta γσ) etc. cos (α + β + γ +...) + i si (α + β + γ +...) cos α cos β cos cos γ... [( + i ta α) ( + i ta β) ( + i ta γ)... cos α cos β cos γ... [ + i(ta α + ta β +...) + i (ta α ta β + ta β ta γ +...] + i (ta α ta β ta γ +... ) +...] cos α cos β cos γ... [ + is S is +...] where S ta α + ta β cos γ... Σ ta α, S ta α ta β + ta β ta γ +... Σ ta α ta β S ta α ta β ta γ +... Σ ta α ta β ta γ, i.e., S idicates the sum of the tagets of the agles, S idicates the sum of the products of the tagets of the agles take two at a time ad S idicates the sum of the products take three at a time ad so o. Equatig real ad imagiary parts, we have cos (α + β + γ...) cos α cos β cos γ... [ S + S 4 S ] si (α + β + γ...) cos α cos β cos γ... [S S + S 5...] Also, by divisio, we get S S + S5... ta (α + β + γ) S + S S Expasio of si θ ad cos θ i Powers of θ We kow that for a positive iteger ad a agle α, ( ) cos α cos α cos α si α + α α α 4 ( ) ( ) ( ) cos 4 si 4 si 4... Now put α θ, the θ α ad 4

35 cos θ cos θ( θ α) si α α cos α α θ ( θ α ) ( θ α ) ( θ α ) si cos α + α 4 α KK Now if α is mode idefiitely small, θ remaiig costat ad cosequetly becomig idefiitely large, cos α ad si α α both ted to uity, ad hece i the limit whe a 0 we have cos θ 4 7 θ θ θ Similarly by writig the value of si α ad proceedig as before, we ca obtai. si θ 4 7 θ θ θ θ The method we have applied here is oly a idicatio to get the values of cos θ ad si θ ad does ot give a rigorous proof. We shall ow obtai a series for ta θ i terms of θ by usig the series for cos θ ad si θ. si θ Thus ta θ cos θ 5 θ θ θ + KK 5 4 θ θ + KK θ θ θ θ θ + KK 6 0 KK 4 Expadig the secod bracket by the Biomial theorem, we have ta θ Thus ta θ 5 4 θ θ θ θ θ θ θ + KK + K + K + K θ θ θ θ q θ + K K θ θ θ 5 4 θ + K q K θ θ θ θ θ + + θ + K θ 5 θ + + θ +K 5 θ 5 θ + + θ +K 5 5 4

36 If powers of θ more tha five are eglected, i.e., for small values of θ, we have si θ cos θ ta θ θ θ, 6 4 θ θ +, 4 θ θ +, ad if power more tha are to be eglected the θ si θ θ, cos θ, ta θ θ It is uderstood that the agle θ is expressed i radias. Example. Fid the value of θ whe cos θ Solutio. Here cos θ is early ad θ is small, Hece we ca take i.e., θ cos θ θ or θ θ 99 cos θ radias degrees π 0.8 degrees approx. Example. Fid the value of si º correct to theree places of decimals. Solutio. We have º Now π radias 80 π π π si si º π π early π approx Expressio of the Products of the Form cos m θ si θ Terms of sies or cosies of Multiples of θ. De Moivre s Theorem ca be employed to express a product of the form cos m θ si m θ where m ad are positive itegers i terms of sies or cosies of multiples or θ. 6

37 Let z cos θ + i si θ the z cos θ i si θ cos θ + i si θ Hece z + cos θ z z i si θ z Also from De Moivre s Theorem, if r is a iteger. (i) z r cos r θ i si r θ (ii) z r cos rθ i si rθ ad, therefore, by addig ad subtractio (i) ad (ii), we get cos rθ ad i si rθ We kow that ( cos θ) m (i si θ) z z r r + r z r z m z + z z z m The right had side of this equatio is expaded of z ad terms with idices equal but opposite i r sig are grouped together. The resultig sum is made up, if is eve, of term of the form z + r z ad if r is odd, of term of the form z r. z r r Also we ca express these terms z + ad z r r as cos rθ ad i si rθ respectively. z z The followig examples will illustrate the method. Example. Express si 6 θ i terms of cosies of multiples of θ. Solutio. Let z cos θ + i si θ, so that z + cos θ, z i si θ, z z z r + r z r cos rθ, z i si rθ r z r beig ay positive iteger. 6 Now (i si θ) 6 z z Cz + Cz + Cz + C4z C5z + C6z z 64 i 6 si 6 θ z 6z + 5z 0 + 5z 6z + z 7

38 si 6 θ 6 4 z 6 z 5 z z z z [ cos 6 θ 6. cos 4 θ + 5. cos θ θ ] 64 [cos 6 θ 6 cos 4 θ + 5 cos θ 0] Example 4. Express si 4 θ cos θ i cosies of multiples of θ. Solutio. Let z so that Also z r cos θ + i si θ the cos θ i si θ. z z + cos θ ad z + r z (i si θ) 4 ( cos θ) z i si θ z r cos rθ ad z i si rθ r z 4 z z + z z z z z + z z z z z z z 4 z z z z z + z + + z z z z i.e., 64 si 4 θ cos θ cos 6θ 4 cos 4θ + cos θ + 4. Hece si 4 θ cos θ (cos 6θ cos 4θ + cos θ + ). Express si 5θ i terms of si θ oly.. Express cos 6θ i terms of cos θ oly.. Express ta 7θ i terms of ta θ oly. 4. Fid the value of si º approximately. 5. Fid the value of θ, where si θ 0. θ 04 EXERCISE 6. Show that 6 si 5 θ si 5θ 5 si θ + 0 si θ. 7. Express si 7 θ cos θ i terms of sies of multiples of θ. 8. Express cos 6 θ i terms of cosies of multiples of θ. 8

39 LESSON 4 APPLICATIONS OF DE MOIVRE S THEOREM I this lesso, we shall study the applicatio of De Moivre s theorem to fid the sum of certai types of trigoometric series. The series may be fiite or ifiite. The geeral method is kow as the C + is method. If we are give as series of cosies such as C cos θ + cos θ +... cos θ, we cosider a similar series of sies i.e. S si θ + si θ +... si θ. The ew series C + is so formed ca be summed up by differet methods. Equatig real ad imagiary parts we obtai the values of C ad S. The followig formulae are maily used i the summatio of series. x (i) + x + x + K x, x x (ii) ( + x) + C x + C x +K C x (iii) Euler s Formula: If z is complex variable, the the complex expoetial fuctio e z is defied by e z + z + z + z + K!! Replacig z by i θ we have Similarly ad Also e iθ 4 5 ( iθ) ( iθ) ( iθ) ( iθ) + iθ K θ θ θ θ + iθ i + + i θ θ θ + K + i θ cos θ + i si θ e iθ cos θ i si θ e iθ + e iθ cos θ e iθ e iθ i si θ (e iθ ) e iθ cos θ + i si θ where is a iteger. We shall illustrate the method C + is i the followig solved examples. Example. Fid the sum to terms of the series. + x cos θ + x cos θ x cos ( )θ Let C + x cos θ + x cos θ x si ( ) θ S x si θ + x si θ x si ( )θ C + is + x(cos θ + i si θ) + x (cos θ + i si θ) x [cos( )θ + i si ( )θ] + i si ( ) θ)] + xe iθ + x e iθ x e ( )iθ 9

40 iθ x e, iθ xe summig the GP to terms iθ iθ ( x e ) ( xe ) iθ iθ ( xe ) ( i xe ) [Multiply the umerator ad deomiator by xe iθ ] iθ iθ + ( ) iθ x e xe + x e iθ iθ x( e + e ) + x + x [cos θ + i si θ] x (cos θ i si θ ) + x [cos ( ) θ + i si ( ) θ] x cos θ + x (Usig Euler s Formulae) Equatig the real part we get C + x cos θ x cos θ + x cos( ) θ x cos θ + x Example. Fid the sum to term of the series Let si θ + si 4θ + si 6θ +... C cos θ + cos 4θ + cos 6θ cos θ S si θ + si 4θ + si 6θ si θ C + is e iθ + e 4iθ +... e iθ iθ iθ ( ) iθ C + is e [ + e +K e ] x [ + x + x +K x ] where x e iθ x ( x ) x e iθ iθ [ e ] e iθ iθ iθ iθ e ( e ) ( e ) iθ iθ ( e ) ( e ) (Multiplyig umerator ad deomiator by the cojugate of the deomiator) iθ iθ iθ ( ) iθ e [ e e + e iθ iθ ( e + e ) + iθ ( + ) iθ iθ e e + e cos θ + (cos θ + i si θ) [cos ( + ) θ + i si ( + ) θ] + [cos θ θ + i si θ] ( cos θ) Equatig the imagiary parts o both sides, we get 40

41 S si θ si ( + ) θ + si θ ( cos θ) si θ cos θ [si ( + ) θ si θ)] 4 si si θ cos θ cos ( + ) θ si θ 4 si si θ[cos θ cos ( + ) θ si θ] ( ) 4 si si + θ si θ si θ si ( + ) θ si π si θ Example. Fid the sum of the series θ θ θ cos θ C cos ( θ + φ ) + C cos ( θ + φ ) C cos ( θ + φ ) 4 where si θ 0 i.e., θ π Let C cos θ C cos ( θ + φ ) + C cos ( θ + φ ) C cos ( θ + θ ) ad S si θ + C si ( θ + φ ) C si ( θ + θ ) Here C + is iθ i( θ + φ) i( θ + φ) i( θ + φ) e C e C e C e iθ iφ iφ iφ [ e C e C e C e i θ i φ e [ + e ] iθ e [ + cos φ + i si φ ] iθ φ φ φ e cos + i si cos i e θ cos φ cos φ i φ + Equatig the real part o both sides, we get cos φ [cos i si ] cos φ i si φ θ + θ + φ φ φ cos cos θ + + i si q + φ φ C cos cos θ + Sometimes the sum to terms of a trigoometric series ca be obtaied by usig simple trigoometric formulae viz. si A si B cos (A B) cos (A + B) cos A si B si (A B) si (A + B)

42 These formulae ca be useful whe the th term of a series ca be expressed at the differece of two cosies of sies. The followig example will illustrate this method. Example 4. Fid the sum to terms of the series. si θ + si θ + si 5θ +... Let S si θ + si θ + si 5θ si ( )θ Multiplyig both sides by si θ ( 0) we get, si θ S si θ + si θ si θ + si θ si 5θ si θ si ( )θ S [ cos θ] + [cos θ cos 4θ] + [cos 4θ cos 6θ]... cos θ si θ si si θ θ si θ cosec θ provided siθ 0 i.e., θ π Example 5. Fid the sum of the ifiite series. Let S ad we have C C + is si θ si 5θ si θ + +K 5 si θ si 5θ si θ + +K 5 cos θ cos 5θ cos θ + +K 5 iθ 5iθ i e e e θ + K [cos ( ) θ cos θ) x x x + K where x e i θ 5 si x si [cos θ + i si θ] si (cos θ) cos (i si θ) + cos(cos θ) si (i si θ) si (cos θ) cos h (si θ) + i cos(cos θ) si h (si θ) Equatig imagiary parts o both sides, we get Example 6. Sum of the series Let C S cos (cos θ) si h (si θ)...5 cos θ + cos θ cos θ + K cos θ + cos θ +K.4 [Q cos iθ cos hθ, si iq i si hθ] from the defiitio of Hyperbolic fuctios) 4

43 S Here C + is. si θ + si θ +K.4 iθ. iθ..5 iθ e + e e iθ ( + e ) [ + cos θ + i si θ ] θ θ θ cos + i si cos θ θ θ cos cos + i si Equatig real parts o both sides we get, C θ θ θ cos cos i si 4 θ cos 4 θ cos Example 7. Fid the sum to ifiitely of the series Let C S C + is si α + si α + si α +... cos α + cos α + cos α +... si α + si α + si α +... iα iα iα e + e + e +... iα iα iα e + e + e +... e iα e iα [Summig the G.P. to ] iα iα e e iα iα e e 4

44 iα e iα iα ( e + e ) + 4 cos α + isi α 5 cos α 4 4(cos α + i si α) 5 4 cos α Equatig the imagiary parts o both sides, we get S 4 si α 5 4 cos α Example 8. Fid the sum to of the sides Let C S Here C + is (Summig the G.P. to ) It follows that C 0 provided si α 0 i.e., α kπ where k is a iteger cos α cos α + cos α cos α + cos α cos α +K cos α cos α + cos α cos α + cos α cos α +K cos α si α + cos α si α + cos α si α +K iα iα iα cos α e + cos α e + cos α e +... iα iα iα cos α e [ + cos α e + (cos α e ) +K ] iα cos α e iα cos α e iα iα cos α e [ cos α e ] iα iα ( cos α e ) ( cos α e ) iα cos α e cos α iα iα ( e e ) cos α + + cos α cos α [cosα + i si α] cos α cos α + cos α cos α + i si α cos α cos α i si α cos α si α cos i.e., The sum of the give ifiite series is zero, provided α is ot a multiple for π. α 44

45 EXERCISES Fid the sum to terms of the followig series :. cos θ + cos θ + cos θ si θ cos θ + si θ cos θ + si θ cos 4θ si α + si(α + β) + si (α + β) cos θ + cos 4θ + cos 6θ +... Fid the sum to ifiity of the followig series : cos θ cos 4θ + cos 6 θ si θ si θ + si θ si α si α + si α si α + si α si α si α si α + si α cos β cos β + cos β +... cos α cos α cos α cos β + cos β cos β +... si( α + β) si ( α + 4 β) si α

46 LESSON 5 THEORY OF EQUATIONS I 5. I this lesso we shall deal with the theory of equatios. We presume that you are all familiar with the theory of quadratic equatios. I the preset course of study of theory of equatios, we shall cofie ourselves to cubic ad biquadratic equatios. 5. Polyomial ad Equatios We kow that a fuctio p(x) of the form p(x) 0 0 a x + a x + a x... + a, a 0, where the coefficiets a 0, a, a,... a are all complex umbers (a 0 0) is called a polyomial of degree, beig a positive iteger. For example, is a polyomial of degree. The expasio p(x) 0 is called a th degree equatio, that is, p(x) 0 0 a x + a x + a x... + a 0, a 0...() The fudametal theorem of Algebra states that every equatio of the form (), has at least oe complex umber as a root. The proof of this result is beyod the scope of the preset course of study ad hece is omitted. I the discussio thereafter, it will be presumed that the coefficiets like a 0 a... a are all complex umbers. Now discuss the followig theorem which is kow as Factor theorem. We already kow that if α is a root of a equatio, say, a 0 x + a x + a 0, a 0 0 the x α is a factor of the correspodig polyomial a 0 x + a x + a. For example, is a root of the equatio x x 6 0, the it is easy to verify that x is a factor of the polyomial x x 6. We geeralize this idea ito the followig theorem : 5. Factor Theorem A umber α is a root of the equatio a0x a x + K + a 0...() if ad oly if (x α) is a factor of the polyomial p(x) 0 a x + a x + K + a...() where a 0 0, a, a..., a are complex umbers ad is a positive iteger. Proof. I the first part, it is give that α is a root of the equatio () ad we have to prove that (x α) is a factor of (). We divide the polyomial p(x) of degree by a polyomial (x α) of degree oe. The, the quotiet will be a polyomial of degree ( ). If the remaider is zero, the obviously (x α) is a factor of p(x). If the remaider is ot zero, the the remaider is a costat ad let it be R. Also let the quotiet be deoted by The we have K + + a x b x b x b x b 0 K 0 K a x + a x + + a x + a ( x α ) [ a x b x + + b x + b ] + R 46

47 Puttig x α o both sides, we get I other words, 0 a α + α α + K + a α + a 0 + R R R 0 a α + α α + K + a α + a...() Now agai, sice α is a root of the equatio (), therefore by puttig x α i (), we get 0 a a α + α + K + α α + α 0...(4) From () ad (4), it follows that R 0 so that (x α) is a factor of p(x). I the secod part, it is give that (x α) is a factor of the polyomial p(x) give by (), we have to show that α is a root of the equatio (). Sice (x α) is a factor of the polyomial p(x), therefore whe we divide p(x) by (x α), we must get R 0, ad cosequetly from (), we get a α + a α + K + a α + a 0 0 which implies that a satisfies the equatio α x + a x + K + a x + a 0 0 i.e., α is a root of the equatio (). This completes the proof of the theorem. Now we discuss the theory of cubic equatios. 5.4 Cubic Equatio The geeral cubic equatio is of the form a 0 x + a x + a x + a 0, a (5) where a 0, a, a, a are give complex umbers. By the fudametal theorem of algebra, this equatio (5) has atleast oe root. Let this root be deoted by α. The by Factor Theorem (x α) is a factor of the polyomial a 0 x + a x + a x + a. Divide the polyomial by x α ad let the quotiet be deoted by a 0 x + b x + b. The we have a 0 x + a x + a x + a (x α) (α 0 x + b x + b )...(6) Now cosider the quadratic equatio a 0 x + b x + b 0...(7) Agai by the fudametal theorem of algebra equatio (6) must have at least oe root, say, β ad accordigly (x β) must be a factor of the polyomial a x + b x + b Dividig a 0 x + b x + b by (x β), we get the quotiet, say, (a 0 x + c ) so that we have a x + b x + b ( x β ) ( a0 x + c )...(8) Also a 0 x + c c a0 x + a0( x γ) a0...(9) 47

48 Combiig (6), (8) ad (9), we get 0 c γ, a0 0 a 0 a x + a x + a x + a a 0 (x α) (x β) (x γ)...(0) Substitutig x α, β, γ successively o both sides of (9), we observe that each of the values, α, β, γ satisfies the equatio (5). Thus α, β, γ are the root of the equatio (5). Also it may be oted that o umber differet from α, β, γ ca be roots of the equatio (5). For, if possible let δ, a umber differet from each of α, β, γ be a root of this equatio (5). The by substitutig x δ o both sides of (0), we fid that while the left had side is zero by virtue of the suppositio that δ is a root but the right had side is o-zero ad thereby we arrive at a cotradictio. Hece, α, β, γ are the oly roots of the equatio (5). Thus we have proved that a cubic equatio has three ad oly three roots. These roots may be all distict (uequal), may be repeated (equal) or may be that oly two are repeated. From the above discussio, it is clear that if α, β, γ are the three roots of the a cubic equatio, the the correspodig factors are (x α), (x β), (x γ) ad hece the equatio is give by or (x α) (x β) (x γ) 0 x ( α + β + γ ) x + ( αβ + βγ + γα) x aβγ 0. We discuss the followig examples : Example. Fid the equatio whose roots are,,. Solutio. The required equatio is give by (x ) (x ) (x ) 0. or x 6x + x 6 0 [By puttig α, β, γ i equatio () above] Example. Fid the equatio whose roots are,,. Solutio. (x ) (x ) (x + ) 0. or x 4x x Reactios betwee the Roots ad Co-efficiet of a Cubic Equatio Let α, β, γ be the roots of a cubic equatio The we have the idetify which implies a x a x a x a a0 x + a x + ax + a a0 ( x α) ( x β) ( x γ ) 0 a x + a x + a x + a a0 [ x ( α + β + γ ) x + ( αβ + βγ + γα) x αβγ ] Equatig the co-efficiets powers of x, i.e., x, x, x ad the costat terms, we get a 0 a 0 a a 0 (α + β + γ) a a 0 (αβ + βγ + γα)ss 48

49 These give α a0( αβγ ) a0αβγ a α + β + γ a a αβ + βγ + γα a 0 0 a αβγ a I other words, we ca write these relatios as Sum of the roots Sum of the products of roots take i pairs 0 coefficiet of x coefficiet of x costat term coefficiet of x 49 coefficiet of x coefficiet of x Thus we ote that though we are ot able to fid the values of α, β, γ separately, yet we have bee able to express the three coditios combiatios amely α + β + γ, αβ + βγ + γα, αβγ i terms of coefficiets α 0 0, α, α, α. 5.6 Symmetric Fuctios Cosider the followig expressios : (i) (ii) a + b +γ ( α + β ) + ( β + γ ) + ( γ + α ) (iii) α β + α γ + β α + β γ + γ α + γ β Each of the above is a fuctio of α, β, γ with the property that if ay two of α, β, γ are iterchaged the fuctio remai ualtered. Such fuctios are called symmetric fuctios. More precisely a fuctio f (α, α,... α ) of variables is said to be symmetric fuctio of α, α,... α if it remais ualtered by iterchagig ay two of Thus, for example α, α... α. α + β + γ, αβ + βγ + γα, αβγ are symmetric fuctios of α, β, γ. I the case of a cubic equatio whose roots are α, β, γ the fuctios α + β + γ, αβ + βγ + γα, αβγ are called basic symmetric fuctios. Sigma Notatio for Symmetric Fuctios It is ofte coveiet to describe a symmetric fuctio by idicatig oly oe or more terms i such a maer that the other terms are obtaiable from the give term o replacig the roots therei by other roots i all possible ways. Also the we put dow the sig Σ(sigma) before the give term. Thus for a cubic equatio