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TRANSCRIPT
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Graduate Course
B.A. (PROGRAMME) 1 YEARALGEBRA AND CALCULUS
(PART-A : ALGEBRA)
CONTENTS
Lesson 1 : Complex Numbers
Lesson 2 : De Moivre’s Theorem
Lesson 3 : Applications of De Moivre’s Theorem
Lesson 4 : Applications of De Moivre’s Theorem to Summation of Seric
Lesson 5 : Theory of Equations–I
Lesson 6 : Theory of Equations–II
Editor:Dr. S.K. Verma
SCHOOL OF OPEN LEARNINGUNIVERSITY OF DELHI
5, Cavalry Lane, Delhi-110007
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Session 2012-2013 (1000 Copies)
© School of Open Learning
Published By: Executive Director, School of Open Learning, 5, Cavalary Lane, Delhi-110007Laser Composing by : M/s Computek System (2012–13)
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LESSON 1
COMPLEX NUMBERS
1.1 Complex Number
Definition : An ordered pair (x, y) of real numbers x, y is called complex number.
Illustration : (1, 6) (π, –2), (0, 1), (–1, 0), ( 2,1) , (1/2, 3) are complex numbers.
Two complex numbers : (x, y) and (p, q) are said to be equal if x = p, y = q.In symbols, (x, y) = (p, q) ⇔ x = p and y = q.
1.2 Addition of Complex Numbers
If α = (a, b) and β = (c, d) be to complex numbers, we define their sum α + β to be the complexnumber (a + c, b + d). Symbolically
(a, b) + (c, d) = (a + c, b + d)Illustrations : (3, 7) + (4, –1) = (7, 6)
(2p, q) + (0, –2q) = (2p – q)
1.3 Multiplication of Complex Numbers
If α = (a, b) and β = (c, d) be two complex numbers, we define the product αβ to be the complexnumber (ac – bd, ad + bd).
In symbols (a, b) (c, d) = [(ac – bd), (ad + bc)]Illustrations : (i) (2, 4) (–1, 3) = [2(–1) –4.3, 2.3 + 4 (–1)]
= (–14, 2)(ii) (1, 0) (2, –3) = [(1.2) – 0.(–3), 1.(–3) + 0.2)]
= (2, –3)
1.4 Division
If α, β, be any two complex number (β ≠ 0) then by definition α ÷ β = αβ–1. We shall alsouse the symbols α/β, to denote α ÷ β.
1.5 Usual Notation for Complex Numbers
We shall now see at to how we can express a complex number (a, b) in the usual way asa + ib. The mapping a → (a, 0) associates to each real number a, a complex number (a, 0) whosefirst number is ‘a’ and whose second number is ‘0’. This mapping is compatible with addition andmultiplication. That is, if ‘a’ and ‘b’ be any two numbers, then
a + b → (a + b, 0) = (a, 0) + (b, 0)ab → (ab, 0) = (a, 0) (b, 0)
The above relations mean the if we have any relation involving real numbers, then it remains trueif each real number ‘a’ in the relations is replaced by the complex number (a, 0). Because of this importantproperty we can write any relation involving complex numbers.
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Also (0, 1) (0, 1) = (–1, 0) = – 1
which suggest that we may write (0, 1) = i
If we agree to write ‘a’ for (a, 0) and i for (0, 1) then
(a, b) = (a, 0) + (0, b)
= (a, 0) + (0, 1) (b, 0)
= a + ib
Thus we find that if we set up the convention of writing ‘a’ for (a, 0) and i for (0, 1), then
(a, b) will be written as a + ib (where i = 1− )
We shall give below, some examples to illustrate the Algebraic operation on complex numbers,with usual Notation.
Example 1. Simplify (4 + 5i) + (2 – 7i)
Solution. (4 + 5i) + (2 – 7i) = (4 + 2) + (5i – 7i)
= 6 (5 – 7) i
= 6 – 2i
We note that to add two complex numbers, we add the real parts and the pure imaginary partsseparately.
With the new notation, the same example can be solved as follows :(4 + 5i) + (2 + 7i) = (4, 5) + (2, –7)
= (4 + 2, 5 – 7)
= (6, –2)
= 6 – 2i
Example 2. Simplify (2 + 3i) + (5 – 7i)
Solution. (2 + 3i) + (5 – 7i) = 2(5 – 7i) + 3i (5 – 7i)
= 10 – 14i + 15i – 21i2
= 10 + i (–14 + 15) –21(–1)
= 10 + 21 + i (15 – 14)
= 31 + i
With the new notation, we have
(2 + 3i) (5 – 7i) = (2, 3) (5, –7)
= [2.5 – 3 (–7), 2(–7) + 3.5]
= (10 + 21, –14 + 15)
= (31, 1)= (31 + i)
Note : By the above two examples, it is evident that any one of the two notations can be used.
1.6 Real and Imaginary Parts of a Complex Number
Let z = x + iy be any complex number. Then x is called the real part of z and y is called theimaginary part of z. We write R(z) = x, I(z) = y. We say that the complex number z is purely real ifI(z) = 0, and that z is purely imaginary if R(z) = 0. Also, the complex number x – iy is called the
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complex-conjugate of x + iy, the symbol z is usually used to denote the complex-conjugate of the complexnumber z. That is, if z = x + iy, then z = x – iy.
If z is the conjugate of z, then z is the conjugate of z that the relation of conjugacy betweencomplex number is a symmetric one. We therefore, generally say that z and z are conjugate complexnumbers instead of saying that z the complex-conjugate of z.
The following results follow directly from the defintiion.
(i) The sum of two conjugate complex numbers is purely real.
(ii) The difference to two conjugate complex numbers is purely imaginary.
(iii) The product of two conjugate complex numbers is a non-negative real number.
In order to see the truth of the above three statements, we have z = x + iy so that
z = x – i y i.e. z = (x, y) so that z = (x, – y), then
z z+ = 2x which is purely real [ ( , ) ( , ) (2 ,0)]z z x y x y x∴ + = + − =
z z− = 2iy, which is purely imaginary [ ( , ) ( , ) (0,2 )]z z x y x y y∴ − = − − =
zz = x2 + y2, which is a non-negative real number.
2 2[ ( , ) ( , ) ( ,0]zz x y x y x y= − = +
We give below some solved examples with the usual notation which would help us at a later state.
Example 3. Simplify a bic di
++
To divide two complex numbers, multiply both numerator and denominator of the fraction by theconjugate of the denominator.
a bic di
++
=( ) ( )( ) ( )a bi c dic di c di
+ −+ −
= 2 2( ) ( )ac bd bc ad i
c d+ + −
+
= 2 2 2 2ac bd bc ad
ic d c d
+ −+ + +
Example 4. Find the value of(2 + i) (3 – 2i) (8 + i)
Solution. Here (2 + i) (3 – 2i) = 6 + 3i – 4i – 2i2
= (6 + 2) + (3i – 4i) [∴ i2 = –1]= (8 + i)
∴ (8 – i) (8 + i) = 64 – i2 = 64 + 1 = 65Example 5. Find the value of (1 + i)3.Solution. (1 + i)3 = 1 + 3i + 3i2 + i3
= 1 + 3i – 3 – i= – 2 + 2i
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Example 6. Divide (1 + 3i) by (2 + i).
Solution.1 32
ii
++
=(1 3 ) ( 2 )(2 )(2 )
i ii i
+ −+ −
=(2 3) (–1 6)
4 1i+ + +
+
=5 5
15
ii
+ = +
Example 7. Express 5 23 4
ii
−− in the form x + iy
Solution. Here5 23 4
ii
−− =
(5 2 ) ( 3 4 )(3 4 ) ( 3 4 )
i ii i
− +− +
=(15 8)(20 6)
9 16i+ −
+
=23 14 3 4
25 25 25i
i+ 2 1 = +
Example 8. Find the roots of the quadratic equation
x2 – 6x + 10 = 0.
Solution. This can be written as (x2 – 6x + 9) + 1 = 0
or (x – 3)2 = –1 = i2
∴ x – 3 = ± i
or x = 3 ± i
Thus 3 + i and 3 – i are the roots of the quadratic equationx2 – 6x + 10 = 0
It is seen that the roots are conjugate. Verify that these values satisfy the equation. The same examplescan also be solved as
x =6 36 40 6 4
2 2± − ± −=
= 6 23
2i
i± = ±
1.7 Graphical Representation of a Complex Number
The complex number z = x + iy may be representated graphically by a point P whose rectangularco-ordinates are (x, y) i.e., we associate with each complex number z = x + iy the point of the planewhich has, with reference to a fixed rectangular system, the coordinates x and y and conversely witheach point having the co-ordinates (x, y) we associate the complex number z = x + iy.
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From each theorem involving complex numbers we can deduce a definite relationship betweenthe geometrical points of the Cartesian plane and conversely, the diagram showing points which representcomplex numbers geometrically is called Argand Diagram.
It may be noted that all points on the x-axis are of the form (x, 0) and are x + oi = x and socorresponds to real numbers x. Similarly points on the y-axis correspond to pure imaginary numbersyi.
In addition, the complex number may be repesented by the directed line segment or vector OP.
1.8 Trigonometric form of Complex Numbers
The complex number z = x + iy is represented by the vector OP. This vector and hence the complexnumber is described in terms of the length r of the vector and the angle θ which is vector makes withthe positive direction of the x-axis (real axis) measured positively.
Thus OP = 2 2r x y= + and tan q = y/x
Thus a complex quantity can always be put in the form r (cos θ + i sin θ) where r and θ areboth real.
Let x + iy = r (cos θ + i sin θ)Equating real and imaginary parts, we get
x = r cos θ, and y = r sin θ.By squaring and adding, we get
x2 + y2 = r2
r = 2 2x y+
Here, we take only positive sign of the square root.
Also by dividingyx
= sintan
cosrr
θ = θθ
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or θ = 1tan ( 0)y
xx
− ≠
2 2r x y= + is called the modulus or absolute value of the complex number x + iy and θ = 1tanyx
−
is called amplitude (or argument or phase).There are many values of θ satisfying the equation tan θ = y/x, the value of θ, such that –π <
θ ≤ π is called the principal value of the amplitude. We shall generally take the principal value of θ.In symbols the modulus of a complex number z = x + iy is also denoted by | z | or | x + iy |i.e., | z | = | x + iy |
= 2 2x y+
That amplitude is denoted by the symbols amp z or arg z.amp (x + iy) = amp z = θ
here the value of θ is so choosen that it is the principal value i.e.,–π < θ ≤ π
The form r(cos θ + i sing θ) is called the standard or the polar form of the complex numberx + iy, ∈ R, y ∈ R.
Example 1. Express 1 + i in the standard form.Solution. Let 1 + i = r (cos θ + i sin θ)Equating real and imaginary parts, we get
1 = r cos θ; 1 = r sin θ
Then r2 = 2 ∴ r = 2
Also tan θ = 1 i.e., θ =4π
Here cos θ is positive and sin θ is positive, and hence θ lies in the first quadrant.
∴ θ = .4π
Hence 1 + i = 2 cos sin4 4
iπ π +
The modulus of the complex number is 2 and the amplitude is .4π
Example 2. Express 1 3i− in the polar form.
Solution. Let 1 3i− = r (cos θ + i sin θ)
so that 1 = r cos q,
3− = ρ sin θ
∴ r = 2, tan θ = 3−
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Here cos θ is +ve and sin θ is –ve.∴ θ lies in the fourth quadrantAlso since –π < θ ≤ π
θ =3π−
Hence 1 3i− = 2 cos sin3 3
i π π − + −
Example 3. Express sin α + i cos α in the standard form.Solution. Let sin α + i cos α = r (cos θ + i sin θ)
so thatsin coscos sin
rr
α = θα = θ
⇒2 1
tan cotr =
θ = α
∴ r = 1, tan θ = cot α
= tan2π − α
∴ θ =2π − α
Hence sin α + cos α = cos sin .2 2
iπ π − α + − α
EXERCISE–1
1. Simplify the following :(i) (2 – 5i) + (–3 + 4i) + (8 – 3i)
(ii)5
3 2ii
+−
2. Evaluate the following :(i) (–7 – 2i) (–1 – 5i)
(ii)2
2(2 4 )(3 )
ii
+−
3. Solve the following equations :(i) [(x + 2y), (2x –y – 6) = (3, 2)
(ii) (x – y) + i (x + y) = 2 31
ii
− ++
4. Express the following complex number in the polar form
(i) 1 3i− +
10
(ii) 2 12i− −
5. Find the trigonometric representation of(i) sin α – i cos α(ii) 1 + cos α + i sin α
6. Show that | cos θ + i sin θ | = 1
7. Express the complex number 52
ii
++ in the form (x + iy) where x and y are real. Find its modulus
and amplitude.
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LESSON 2
DE MOIVRE’S THEOREM
In this lesson, we shall discuss an important theorem which is used in finding the roots of a complexnumber, solution of equations, expansion of trigonometric function etc. This theorem is known as DeMoivre’s Theorem.
2.1 De Moivre’s Theorem
Statement :(i) If n is an integer, positive, or negative, then (cos θ + i sin θ) = cos nθ + i sin nθ.
(ii) If n is a rational number, then one of the values of (cos θ + i sing θ)n isProof : Part (i) case (a) : Let n be a positive integerBy simple multiplication(cos α + i sin α) (cos β + i sin β) = (cos α cos β – sin α sin β)
+ i (sin α cos β + cos α sin β)= cos (α + β) + i sin (α + β)
Again multiplying the above results by (cos γ + i sin θ), we have(cos α + i sin α) (cos β + i sin β) (cos γ + i sin θ)
= [cos (α + β) + i sin (α + β)] (cos γ + i sin γ)= cos (α + β + γ) + i sin (α + β + γ)
This process can be continued to any number of factors so that(coa α + i sin α) (cos β + i sin β) (cos γ + i sin γ) ...... to n factors
= cos (α + β + γ + ....... to n terms)+ i sin (α + β + γ + ...... to n terms)
In this expression, put α = β = γ = ........ = θ.
So that we have (cos θ + i sin θ)n = cos n θ + i sin n θ.Case (b) : Let n be a negative integer. Suppose n = – m where is a positive integer.Then (cos θ i sin θ)n = (cos θ + i sin θ)–m
=1
(cos sin )miθ + θ(by the law of indices)
= 1(cos sin )m i mθ + θ
(by case (a))
= (cos sin )(cos sin )(cos sin )
m i mm i m m m
θ − θθ + θ θ − θ
= 2 2(cos sin )
cos sin(cos sin )
m i mm i m
m mθ − θ = θ − θθ + θ
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= cos( ) sin( )m i m− θ + − θ
= con n θ + i sin n θThus (cos θ + i sin θ)n = cos n θ + i sin n θ for integral values of n, whether positive or negative
Proof of (ii) : If n is (a rotational number, positive or negative), let ( 0)p
n qq
= ≠
We shall take q as a positive integer and p as a positive or a negative integer.
We haver cos sinq
iq q
θ θ+
= cos sinq i qq q
θ θ+
= cos θ + i sin θ
i.e., qth power of cos siniq q
θ θ+
is cos θ + i sin θ
i.e., cos siniq q
θ θ+
is one of the qth roots of (cos θ + i sin θ)
i.e., cos siniq qθ θ+ is one of the values of
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(cos sin )qiθ + θ
Raising each of these quantities of pth power, we have the results that one of the values of
/(cos sin ) is cos sinp
p qi iq q
θ θθ + θ +
i.e., cos sinp p
iq q
θ = θ
Hence cos n θ + i sin θ is one of the values of (cos θ + i sin θ)n if n is a rational number positiveor negative.
Corollary 1. cos n θ – i sin n θ is one of the values of (cos θ – i sin θ)n for all rational valuesof n.
Illustratives : With the help of the De Moivre’s Theorems, we see that1. (cos θ + i sin θ)4 = cos 4 θ + i sin 4 θ
2.7
cos sin2 2
iθ θ +
= 7 7
cos sin2 2
iθ θ+
3. (cos θ + i sin θ)–7 = cos (–7θ) + i sin (–7θ)= cos 7θ – i sin 7θ
4. Since (cos θ + i sin θ) (cos θ – i sin θ)= (cos2 θ + sin2 θ) + i (sin θ cos θ – sin θ cos θ) = 1 + i0 = 1
∴ 1cos sin
cos sini
iθ − θ =
θ + θ
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5. sin θ + i cos θ = cos sin2 2
iπ π − θ + − θ
6. (sin θ + i cos θ)n = cos sin2 2
n i nπ π − θ + − θ
e.g., (sin θ + i cos θ)3 =3
cos sin2 2
i π π − θ + − θ
= cos3 sin3 02 2
iπ π − θ + −
=3 3
cos 3 sin 32 2
iπ π − θ + − θ
= – sin 3θ – i cos 3θ
Remarks : (i) As any complex number x + iy can be put in the form r (cos θ + i sin θ), therefore,De Moivre’s Theorem may be written as
( )nx iy+ = (cos sin )n nr iθ + θ
= (cos sin )nr n i nθ + θ
where r is modulus of the complex number and θ is the amplitude.(ii) The product and quotient of any two complex numbers can be obtained by using this method.Let any two complex number be given by
r1 (cos θ1 + i sin θ1) and r2 (cos θ2 + i sin θ2),then their product is
r1r2[cos (θ1 + θ2) + i sin (θ1 + θ2)]i.e., the product of two complex numbers is a complex number whose modulus is the product of
the modulli and whose amplitude is the sum of the amplitudes two complex numbers.The quotient of these two complex numbers, r1(cos θ1 + i sin θ1) and r2(cos θ2 + i sin θ2) is
1 1 1
2 2 2
(cos sin )(cos sin )
r ir i
θ + θθ + θ
= 11 1 2 3
2(cos sin )(cos sin )
ri i
rθ + θ θ − θ
= 11 2 1 2
2[cos( ) sin( )]
ri
rθ − θ + θ − θ
Solved Examples
Example 1. Simplify 4 5
3 7(cos2 sin2 ) (cos sin )
.(cos6 sin6 ) (cos4 sin4 )
i ii i
θ + θ θ + θθ + θ θ − θ
Solution. By De Moivre’s Theorem, we havecos 2θ + i sin 2θ = (cos θ + i sin θ)2
cos θ – i sin θ = (cos θ + i sin θ)–1
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cos 6θ + i sin 6θ = (cos θ + i sin θ)–1
cos 4θ – i sin 4θ = (cos θ + i sin θ)–4
Now, 4 5
3 7(cos2 sin 2 ) (cos sin )
(cos6 sin6 ) (cos4 sin4 )i i
i iθ + θ θ − θ
θ + θ θ − θ
=8 5
18 –28(cos sin ) (cos sin )
(cos sin ) (cos sin )i
i i
−θ + θ θ + θθ + θ θ + θ
= (cos θ + i sin θ)8 – 5 –18 + 28
= (cos θ + i sin θ)36 – 23
= cos θ + i sin θ)13 = cos 13θ + i sin 13θ.
Example 2. Simplify 3 5
8 2(cos2 sin2 ) (cos sin )
.(cos3 sin3 ) (cos5 sin5 )
i ii i
θ + θ θ − θθ − θ θ + θ
Solution. By De Moiver’s Theorem, we have
3 5
8 2(cos2 sin2 ) (cos sin )
(cos3 sin3 ) (cos5 sin5 )i i
i iθ + θ θ − θ
θ − θ θ + θ =6 5
24 –10(cos sin ) (cos sin )
(cos sin ) (cos sin )i
i i
−
−θ + θ θ + θ
θ + θ θ + θ
= (cos θ + i sin θ)6 – 5 + 24 – 10
= (cos θ + i sin θ)15
= cos 15θ + i sin 15θ
Example 3. Obtain the quotient of 1 31
ii
− ++ by using De Moivre’s Theorem.
Solution.1 31
ii
− ++
=
1 3 2 22 2 cos sin2 2 3 31 1
2 cos sin24 42 2
ii
ii
π π − + + =π π ++
=2 2
2 cos sin3 4 3 4
i π π π π − + −
=5 5
2 cos sin .12 12
iπ π +
Example 4. Obtain ( 3 ) .i 4+
Solution. Here ( 3 )i 4+ =4
3 12 2 cos sin
2 2 6 6i
4 π π + = +
=4 4
16 cos sin16 6
iπ π +
15
=2 2
16 cos sin3 3
iπ π +
=1 3
162 2
i − +
= 8 8 3 i− + ⋅
Example 5. Prove that
1 sin cos1 sin cos
nii
+ φ + φ + φ − φ
= cos sin2 2
n nn i n
π π − φ + − φ Solution. Let
1 + sin φ + i cos φ = r (cos φ + i sin θ).Then equating real and imaginary part, we have
1 + sin φ = r cos θ, cos φ = r sin θ
Hence, tan θ = cos1 sin
φ+ φ
=sin
2
1 cos2
π − φ π + − φ
=sin
,where1 cos 2
α πα = − φ+ α
=2
2sin cos2 2 tan
22 cos2
α αα=α
∴ θ =2 4 2α π φ= −
Also 1 + sin φ – i cos φ = r(cos θ – sin θ)∴ The given expression
=(cos 1sin ) (cos sin )(cos sin ) (cos sin )
n n n
n n nr ir i i −
θ + θ θ + θ=
θ − θ θ + θ
= (cos θ + i sin θ)n (cos θ + i sin θ)n
= (cos θ + i sin θ)2n
= (cos 2nθ + i sin 2nθ)
= cos2 s in24 2 4 2
n i nπ φ π φ − + −
= cos sin2 2
n nn i n
π π − φ − φ
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Example 6. Prove that :
(a + bi)2m/n + (a – bi)m/n = 2 2 /2 12( ) cos tanm n m ba b
n a− +
Solution. Here put a + bi = r (cos θ + i sin θ)Then a = r cos θ and b = r sin θ∴ a2 + b2 = r2(cos2 θ + sin2 θ) ⇒ r2 = a2 + b2
and tan θ =ba
Hence r = 2 1and =tanb
a ba
2 −+ θ
Also a – bi = r cos θ – i sin θ).Thus the given expression can be written as
(a + bi)m/n + (a – bi)m/n = / /[ (cos sin )] [ (cos sin )]m n m nr i r iθ + θ + θ − θ
=/ /cos sin cos sinm n m nm m m m
r i r in n n nθ θ θ θ + + −
= /2 cosm n mr
nθ
= 2 2 /2 12( ) cos tanm n m ba b
n a− +
Example 7. If x = cos θ + i sin θ, then show that
1mmx
x+ = 2 cos mθ
and 1mmx
x− = 2 i sin mθ.
Solution. Here xm = (cos θ + i sin θ) = cos mθ + i sin mθ,
and1mx
= (cos θ + i sin θ)–m
= cos mθ – i sin mθ.
Hence1mmx
x+ = 2 cos mθ
Also,1mmx
x− = 2 i sin mθ
Example 8. If 1 12cos ,showthat 2cos .n
nx x nx x
+ = θ + = θ
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Solution. Here, it is given that
1x
x+ = 2 cos θ
or x2 – 2x cos θ + 1 = 0or x2 – 2x cos θ + (cos2 θ + sin2 θ) = 0or (x2 – 2x cos θ + cos2 θ) = –sin2 θor (x – cos θ)2 = ( i sin θ)2
∴ x – cos θ = ± i sin θ.i.e., x = cos θ ± i sin θLet x = cos θ + i sin θ (taking the + ve sign).
Then,1x
= 1cos sin
cos sini
i= θ − θ
θ + θ
∴1nnx
x+ = (cos θ + i sin θ)n + (cos θ – sin θ)n
= (cos nθ + i sin nθ) + (cos nθ – i sin nθ)= 2 cos nθ
If x cos θ – i sin θ, then 1
cos sin .ix
= θ + θ Then also the result can be proved.
Example 9. If sin α + sin β + sin γ = 0 and cos α + cos β + cos γ = 0. Prove that
cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)
Solution. Let x = cos α + i sin αy = cos β + i sin βz = cos γ + i sin γ
Then x + y + z = (cos α + cos β + cos γ) + i (sin α + sin β + sin γ)
= 0 + i0 = 0
⇒ x3 + y3 + z3 = 3xyz
⇒ (cos α + i sin α)3 + (cos β + i sin β)3 + (cos γ + i sin γ)3
= 3(cos α + i sin α) (cos β + i sin β) (cos γ + i sin γ)
⇒ (cos 3α + i sin 3α) + (cos 3β + i sin 3β) + (cos 3γ + i sin 3γ)
= 3[cos(α + β + ψ) + i sin (α + β + ψ)]
⇒ (cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ)
= 3 cos(α + β + γ) + 3 i sin (α + β + γ)Equating real and imaginary parts, we get
cos 3α + cos 3β + cos 3γ = 3 cos(α + β + γ)and sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)Hence the results.
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Example 10. Show that for an integer n
1 ( )(1 cos sin ) (1 cos sin ) 2 cos cos
2 2n n n n n
i i + π θ+ θ + θ + θ − θ =
Solution. We have (1 + cos θ + i sin θ)n + (1 + cos θ – i sin θ)n
= 2 22cos 2 sin cos 2cos 2 sin cos2 2 2 2 2 2
n n
i iθ θ θ θ θ θ + + −
= 2 cos cos sin 2 cos cos sin2 2 2 2 2 2
n nn n n ni i
θ θ θ θ θ θ + + −
= 2 cos cos sin 2 cos cos sin2 2 2 2 2 2
nn n n nn n
i i θ θ θ θ θ θ + + − + −
= 2 cos cos sin cos sin2 2 2 2 2
n n n n n ni i
θ θ θ θ θ + + − + −
= 2 cos cos sin cos sin2 2 2 2 2
n n n n n ni i
θ θ θ θ θ + + −
= 2 cos 2 cos2 2
n n n θ θ
= 12 cos cos2 2
n n n+ θ θ
Example 11. If sin α + sin β + sin γ = 0 = cos α + cos β + cos γ. Prove that
(i) cos(β + γ) + cos(γ + α) + cos(α + β) = 0
(ii) sin(β + γ) + sin(γ + α) + sin(α + β) = 0
Solution. Let x = cos α + i sin α, y = cos β + i sin β, z = cos γ + i sin γso that x + y + z = (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i0 = 0
Also1x
=1 1
cos sin , cos sin , cos sini i iy z
α − α = β − β = γ − γ
∴1 1 1x y z
+ + = (cos cos cos ) (sin sin sin )iα + β + γ − α + β + γ
= 0 – i0 = 0⇒ yz + zx + xy = 0⇒ xy + yz + zx = 0⇒ (cos α + i sin α) (cos β + i sin β) + (cos β + i sin β) (cos γ + i sin γ)
+ (cos γ + i sin γ) (cos α + i sin α) = 0
19
⇒ [cos (α + β) + i sin (α + β)] + [cos (β + γ) + i sin (β + γ)]+ [cos (γ + α) + i sin (γ + α)] = 0
⇒ [cos (α + β) + cos (β + γ)] + cos (γ + α)] + i [sin (α + β) + sin (β + γ)]+ sin (γ + α)] = 0
Equating real and imaginary parts on both sides, we get
cos (α + β) + cos (β + γ) + cos (γ + α) = 0
sin (α + β) + sin (β + γ) + sin (γ + α) = 0
Hence the result.
Example 12. If Z = cos θ + i sin θ, show that
2
211
n
nZZ
−+
= i tan nθ, n being an integer.
Solution.2
211
n
nZZ
−+
=n n
n nZ ZZ Z
−
−−+
(Dividing numerator and denominator by Zn)
=(cos sin ) (cos sin )(cos sin ) (cos sin )
n n
n ni ii i
−
−θ + θ − θ + θθ + θ + θ + θ
=(cos sin ) (cos sin )(cos sin ) (cos sin )
n i n n i nn i n n i nθ + θ − θ − θθ + θ + θ − θ
=2 sin
tan proved2cos
i ni n
nθ = θθ
Example 13. If (cos θ + i sin θ) (cos 2θ + i sin 2θ) ...... (cos nθ + i sin nθ) = 1 prove that
θ =4
,where is an integer.( 1)
kk
n nπ+
Solution. We have 1 = cos 0 + i sin 0
= cos 2kπ + i sin 2 kπ ......, where k is an integer. ...(1)
Also (cos θ + i sin θ) (cos 2θ + i sin 2θ) ....... (cos nθ + i sin nθ) = 1
L.H.S. = cos (θ + 2θ + ...... + nθ) + i sin (θ + 2θ + ...... + nθ)
=( 1) ( 1)
cos sin2 2
n n n ni
+ +θ + θ ( 1)
1 2 3 ......2
n nn
+ + + + + = Q
Thus( 1) ( 1)
cos sin2 2
n n n ni
+ +θ + θ = 1 = cos 2 kπ + i sin 2kπ by (1)
This gives us( 1)
2n n + θ =
42
( 1)k
kn n
ππ ⇒ θ =+
Hence the result.
20
Example 14. If (1 + x)n = 20 1 2 ,n n n n n
xC C x C x C x n+ + + +KK is a positive integer. Prove that
(i) 20 4 8
22 2 cos
2 4n n n n n nC C C − − π+ + + = +KK
(ii) /20 2 4 6 2 cos
4n n n n n nC C C C
π− + − + =KK
(iii) /21 3 5 7 2 sin
4n n n n n nC C C C
π− + − − =KK
Solution. We are given that
(1 + x)n = 2 3 40 1 2 3 4
n n n n nC C x C x C x C x+ + + + + KK
Putting x = 1, –1, i, – i in succession, we get
2n = 0 1 2 3 4n n n n n n
nC C C C C C+ + + + +KK ...(1)
0 = 0 1 2 3 4– – –n n n n nC C C C C+ + KK ...(2)
(1 + i)n = 0 1 2 3 4– –n n n n nC i C C C i C− + +KK ...(3)
= 0 2 4 1 3 5( – ) ( )n n n n n nC C C i C C C+ + − + KK ...(4)
(1 – i)n = 0 1 2 3 4–n n n n nC i C C i C C− + + KK ...(5)
Adding (1), (2), (4) and (5) we get
2 (1 ) (1 )n n ni i+ + + − = 0 4 84( .......)n n nC C C+ + + ...(6)
Now (1 + i)n = 2 cos sin4 4
n
i π π +
= /22 cos sin4 4
n n ni
π π +
...(7)
Similarly, (1– i)n = /22 cos sin4 4
n n ni
π π −
...(8)
Adding (7) and (8) and substituting in (6), we get
0 4 84( )n n nC C C+ + + KK = /22 2.2 cos4
n n nπ+
=1
22 2 cos4
nn n
+ π+
⇒ 0 4 8n n nC C C+ + + KK =
2 cos2 2 42 2 Provedn n
n− π
− +
21
(ii) From (4) and (7)
0 4 8 1 3 5( ) ( ......)n n n n n nC C C i C C C+ + + + − +KK
= /22 cos sin4 4
n n ni
π π +
Equating real and imaginary parts, we get
0 2 4n n nC C C− + KK = 2n/2 cos nπ/4
1 3 5n n nC C C− + KK = /22 sin / 4n nπ
EXERCISE 2.1
1. Simplify 9
9 8(cos sin ) (cos2 sin 2 )(cos2 sin2 ) (cos sin )
i ii i
8θ + θ θ − θθ θ θ − θ
2. Evaluate:
10 10
6
cos sin cos sin15 15 15 15
cos sin3 3
i i
i
π π π π + + −
π π +
3. Use De’ Moivre’s Theorem to simplify 4
3(cos sin )(sin cos )
ii
θ + θθ + θ
4. Prove that (sin x – i cos x)n = cos sin2 2
n x i n xπ π − − −
5. Obtain the value of (3 + 4i)3 with the help of De Moivre’s Theorem.6. If n is a positive integer, show that :
( ) ( ) 13 3 2 cos .6
n n n ni i − π+ + − =
7. Prove that :[cos θ + cos φ) + i (sin θ + sin φ)]n + [cos θ + cos φ) – i sin θ + sin φ)]n
= 1 ( )2 cos cos
2 2n n n+ θ − φ θ + φ
(cos cos (sin sin )
2 cos cos 2sin cos2 2 2 2
2 cos cos sin2 2 2
i
i
i
θ + φ + θ + φ θ + φ θ − φ θ + φ θ − φ = +
θ − φ θ + φ θ + φ = +
Hint:
22
8. Show that[(cos θ – cos φ) + i (sin θ – sin φ)]n + [(cos θ – cos φ) – i(sin θ – sin φ)]n
= 12 sin cos2 2
n n n+ θ − φ π + θ + φ
(cos cos (sin sin )
2sin sin 2 sin cos2 2 2 2
2sin sin cos2 2 2
2sin cos sin2 2 2 2 2
i
i
i
i
θ − φ + θ − φ θ + φ θ − φ θ − φ θ + φ = + θ − φ θ + φ θ + φ = − + θ − φ π θ + φ π θ + φ = + + +
Hint:
EXERCISE 2.2
1. If 2cos θ = 1
xx
+ , 2 cos φ =1
yy
+
Show that
1m nm nx y
x y+ = 2 cos (mθ + nθ)
m n
n mx yy x
+ = 2 cos (mθ – nθ)
where m and n are integers. What happens if m and n are rational numbers?2. If x = cos α + i sin α, y = cos β + i sin β and z = cos γ + i sin γ such athat x + y + z
= 0 then prove that :
1 1 1x y z
+ + = 0
and1
xyzxyz
+ = 2 cos (α + β + γ)
3. Let the complex numbers x, y, z be given respectively by cos a + i sin a, cos b + i sin band cos c + i sin c, then prove that for any integers p, q, r.
1p q rp q rx y z
x y z+ = 2 cos (pa + qb + rc)
4. Let x1 + iy1, x2 + iy2 ......, xn + iyn be any n complex number and A + iB be some other complexnumber, such that(x1 + iy1) (x2 + iy2) ...... ( xn + iyn ) = A + iB
23
Then show thats
(i) 1 1 1 11 2
1 2tan tan ...... tan tann
n
y y y Bx x x A
− − − −+ + =
(ii) 2 2 2 2 2 2 2 21 1 2 2( ) ( ) ( )n nx y x y x y A B+ + + = +KK
5. Expand (1 + i)n in two different ways and find the sum of the two series,
2 4 6 1 3 5(1 )and )n n n n n nC C C C C C− + − + − +KK KK
[Hint : First expand (1 + i)n by Boinomial theorem and then expand it by help of De Moivre’sTheorem.
6. If sin α + sin β + sin γ = cos α + cos β + cos γ = 0. Prove that(i) sin 2α + sin 2β + sin 2γ = 0
(ii) cos 2α + cos 2β + cos 2γ = 0(iii) cos2 α + cos2 β + cos2 γ = 3/2
24
LESSON 3
APPLICATIONS OF DE MOIVRE’S THEOREM
In this lesson, we shall discuss certain applications and uses of De Moivre’s Theorem. As mentionedearlier, it helps in finding the roots of a number, solution of some equations and expressions of trigonometricfunctions, etc.
We have seen that De Moivre’s Theorem can be stated as
cos sinq i qq qθ θ+ = cos sin
q
iq q
θ θ+
, where q is a positive integer..
i.e., cos θ + i sin θ = cos sinq
iq q
θ θ+
∴ cos siniq qθ θ+ is a qth root of cos θ + i sin θ
i.e., cos siniq qθ θ+ is one of the values of of (cos θ + i sin θ)1/q
Now the questions arises :‘What about the other values of (cos θ + i sin θ)n for rational values of n ?We shall now determine all the values of (cos θ + i sin θ)1/q, where θ is a positive integer.We know that
(cos θ + i sin θ) = cos (2πr + θ) + i sin (2πr + θ)∴ (cos θ + i sin θ)1/q = [cos (2πr + θ) + i sin (2πr + θ)]1/q where q is any integer.Now by De Moiver’s Theorem one of the values of (cos θ + i sin θ)1/q is
2 2cos sin
r ri
q qπ + θ π + θ+
By giving ‘r’ various values, we get different values of (cos θ + i sin θ)1/q.Let us give r in succession the values 0, 1, 2, ...... q – 1 and we see that each of the following
quantities
cos siniq qθ θ+
2 2cos sini
q qπ + θ π + θ+
4 4cos sini
q qπ + θ π + θ+
....... ....... ....... ........ ......
25
2( 1) 2( 1)cos sin
q qi
q q− π + θ − π + θ+
is equal to one of the values of (cos θ + i sin θ)1/q. We note here that if we give to ‘r’ integral valuesgreater than (q – 1) viz., the values q, (q + 1), (q + 2), ...... etc. then we do not obtain any new valuesof (cos θ + i sin θ)1/q.
For example, for r = q we obtain the value
2 2cos sin
q qi
q qπ + θ π + θ+
i.e., cos 2 sin 2iq q
θ θπ + + π +
i.e., cos sin .iq qθ θ+
which is the same values as obtained by putting r = 0.Thus the greatest values we need to assign to r is q – 1 for the values q, q + 1, q + 2 ...... will
be found to give the same result as the values, 0, 1, 2, ...... q – 1, for ‘r’.Also no two of the quantities obtain by giving ‘r’, the values 0, 1, 2, ...... q – 1 will be the same
for all the angles involved there in differ from one another by less than 2π and no two angles differingby less than 2π have their consines the same and also sines the same.
Thus the expression cos (2 ) (2 )
sin ,r r
iq q
π + θ π +θ+ where r = 0, 1, 2, ...... (q – 1) gives q and
only q different values of (cos θ + i sin θ)1/q. We may extend the above result to (cos θ +i sin θ)1/q where p and q are integers, q being taken as positive.
Thus (cos θ + i sin θ)1/q has q and only q different values and the these are given by
(2 ) (2 )cos , 0,1,2,......( 1).
r p r piain wherer q
q qπ + θ π + θ+ = −
Then, the θ values are
cos sin ,when 0p p
i rq qθ θ+ =
(2 ) (2 )cos sin ,where 1
p pi r
q qπ + θ π + θ+ =
(4 ) (4 )cos sin ,where 2
p pi r
q qπ + θ π + θ+ =
.......................................................................
[2 ( 1) ] [2 ( 1) ]cos sin ,when 1.
q p q pi r q
q qπ − + θ π − + θ+ = −
Note. We may note here that the ‘q’ distinct values of (cos θ + i sin θ)p/q can be arranged ina geometrical progression, for if
26
µ =2 2
cos sin and cos sin .p p p p
i a iq q q qπ π θ θ+ = +
then the q values of (cos θ + i sin θ)p/q can be arranged as a, aµ, aµ2, ...... aµq – 1 which is a G.P.The student should note that
(2 ) (2 ) 2 2cos sin cos sin cos sin
rp r p r p p p p
i i iq q q q q q
π + θ π + θ π π θ θ+ = + +
3.1 Extraction of Roots of Complex Quantity
We know thatx + iy = r (cos θ + i sin θ),
where r = 2 2 1and tan .y
x yx
−+ θ =
Hence (x + iy)1/q = r1/q[cos θ + i sin θ]1/q
=1/ 1 /[cos(2 ) sin(2 )]q qr n i nπ + θ + π + θ
= 1/ (2 ) (2 )cos sinq n n
r iq q
π + θ π + θ+
when ‘n’ is given in succession the values, 0, 1, 2, ......, q – 1.
This gives the required ‘q’th roots of x + iy.
It may be noted that even if we write x + iy = [cos(2 ) sin(2 )],r k i k kπ + θ + π + θ being an integer,,the same values are obtained.
Example 1. Find the values of (1 + i)1/6.
Solution. We know that cos4π
=1
sin .42π
∴ We can write (1 + i) =1 1
22 2
i +
=122 cos sin
4 4i
π π +
=122 cos 2 sin 2
4 4n i n
π π π + + π +
(1 + i)1/6 =
112 2 /4 2 /4
2 cos sin6 6
n ni
π + π π + π + when n is given the six values 0, 1, 2, 3, 4, and 5.
27
The six values of (1 + i)1/6, are therefore
1 112 12 9 9
2 cos sin , 2 cos sin ,24 24 24 24
i iπ π π π + +
1 112 1217 17 25 25
2 cos sin , 2 cos sin ,24 24 24 24
i iπ π π π + +
1 112 1233 33 41 41
2 cos sin , 2 cos sin24 24 24 24
i iπ π π π + +
It may be observed that the 4th value, viz.,
1
12 25 252 cos sin
24 24i
π π +
= 1
122 cos sin24 24
i π π π + + π +
= 1
122 cos sin ,24 24
iπ π − +
since cos (π + θ) = – cos θ and sin (π + θ) = – sin θ. Similarly the 5th and 6th values can simplified.The six value of (1 + i)1/6 can then be written as
1122 cos sin ,
24 24i
π π ± +
112 9 9
2 cos sin ,24 24
iπ π ± +
112 17 17
2 cos sin ,24 24
iπ π ± +
or simply as
1122 cos sin ,
24 24r r
iπ π ± +
where r = 1, 9 and 17.
Example 2. Find all the values of 3/4
1 32 2
i
+ and show that the continued product of all the
values is 1.
Solution. Let1 32 2
i+ = (cos sin ),r iθ + θ
28
then12
= r cos θ
and3
2 = r sin θ
so that r2 =1 3
14 4
+ =
and tan θ = 3 tan3π=
or r = 1 and 3πθ =
Now3/4
1 32 2
i
+ =
13 4
cos sin3 3
i π π +
= (cos π + i sin π)1/4
= [cos (2kπ + π) + i sin (2kπ + π)]1/4
= cos(2 1) sin ( 2 1) , 0,1,2,34 4
k i k kπ π+ + + =
Hence the required four values are
when k = 0 cos sin ,4 4
iπ π+
k = 13 3
cos sin ,4 4
iπ π+
k = 25 5
cos sin ,4 4
iπ π+
k = 37 7
cos sin ,4 4
iπ π+
And the continued product of four values
= 3 3 5 7 7
cos sin cos sin cos sin cos sin4 4 4 4 4 4 4 4
i i i iπ π π π π 5π π π + + + +
=3 5 7 3 5 7
cos sin4 4 4 4 4 4 4 4
iπ π π π π π π π + + + + + + +
= cos 4π + i sin 4π= 1 + i.0 [∴ cos 4π = 1, sin 4π = 0]= 1.
29
Example 3. Find the seven 7th roots of unity and prove that the sum of their nth powers alwaysvanishes unless n be a multiple of 7, n being integer and that the sum is 7 when n is multiple of 7.
Solution. We have to find the 7 values of (1)1/7,
We know that cos 0 = 1 and sin 0 = 0
We can write 1 = cos 0 + i sin 0= cos 2nπ + i sin 2nπ
∴ (1)1/7 =2 2
cos sin7 7n n
iπ π+
where n = 0, 1, 2, 3, 4, 5, 6.The 7 roots are, therefore (cos 0 + i sin 0) i.e., 1,
2 2cos sin
7 7i
π π+ = ω (say)
224 4 2 2
cos sin . ., cos sin7 7 7 7
i i e iπ π π π + + = ω
and similarly for the other roots.Thus the seven 7th roots of unity are
1, ω, ω2, ω3, ω4, ω5, ω6
where ω =2 2
cos sin .7 7
iπ π +
The sum of the nth power of the roots
= (1)n +s (ω)n + (ω2)n + (ω3)n + (ω4)n + (ω5)n + (ω6)n
= 1 + ωn + ω2n + ω3n + ω4n + ω5n +ω6n
(which is geometric progression)
=7 71 ( ) 1 ( ) 1 (1 ) 1 1
01 1 1 1
n n n
n n n n− ω − ω − −
= = = =− ω − ω − ω − ω
Since ω7 = 1
Also ωn ≠ 1, since n is not a multiple of 7.
If ‘n’ is a multiple of 7, i.e., if n = 7 m say where m is an integer, then
ωn =7
2 2 2 2cos sin cos sin
7 7 7 7
n m
i iπ π π π + = +
= cos 2πm + i sin 2πm
= 1.
Hence when n is a multiple of 7, the sum of the nth power of the 7th roots of unity is
= 1 + 1 + 1 +1 +1 +1 + 1 = 7
30
Example 4. Prove that n nx iy x iy+ + − has ‘n’ real values.
Solution. Let x = r cos θ and y = r sin θ. Thenx + iy = r[cos (2kπ + θ) + i sin (2kπ + θ)],
where k is an integer.∴ (x + iy)1/n + (x + iy)1/n
= 1/ 1/ 1/{[cos(2 ) sin(2 )] [cos(2 ) sin(2 )] }n n nr k i k k i kπ + θ + π + θ + π + θ − π + θ
= 1/ 22 cosn k
rn
π + θ, where k = 0, 1, 2, ...... (n – 1)
Here r2 = 2 2,andtan .y
x yx
+ θ =
Hence (x + iy)1/n + (x – iy)1/n =1
2 2 1/22 tan
2( ) cosn
yk
xx yn
−π ++
where k = 0, 1, 2, ...... (n – 1).This is real and has n distinct values.
3.2 Solution of Equations
De Moivre’s Theorem can be used to obtain solutions of certain types of equations. This will beillustrated by a few solved examples. Conversely De Moivre’s Theorem can be used to find the equationwhose roots are given as trigonometical functions.
Example 5. Solve the equation zn = 1, where n is a positive integer.Solution. Here we write
zn = 1 = cos (2kπ + 0) + i sin (2kθ + 0)
∴ z = 1/ 2 2(1) cos sinn k k
in nπ π = +
where k = 0, 1, 2, ..... (n – 1)Hence the nth roots of unity are
(cos 0 + i sin 0), i.e., 1, 2 2cos sini
n nπ π +
4 4cos sini
n nπ π +
=
22 2
cos sinin nπ π +
...............................................................
...............................................................
12( 1) 2( 1) 2 2cos sin cos sin
nn ni i
n n n n
− − π − π π π + = +
31
It may be noted that if we write ω = 2 2
cos sin ,in nπ π+
then the nth root of ‘1’ can be written as 1, ω, ω2, ...... ωn – 1. Thus we get an interesting result thatthe ‘n’ nth roots of unity form a G.P. Also it is seen that each of these roots can be expressed as apower of another.
Exmple 6. Solve the equation z9 – z5 + z4 – 1 = 0.Solution. Here (z9 – z5 + z4 – 1) = 0⇒ z5 (z4 – 1) + (z4 – 1) = 0i.e., ⇒ (z5 – 1) (z4 – 1) = 0∴ Either (i) z4 = 1 or (ii) z5 = – 1since 1 = cos 2 πn + i sin 2 πnand – 1 = cos π + i sin π
= cos (2nπ + π) + i sin (2nπ + π)= cos (2n + 1) π + i sin (2n + 1) π.
From (i), we have
(cos 2nπ + i sin 2nπ)1/4 =2 2
cos sin4 4n n
iπ π+
z = cos sin ,where 0,1,2,3,2 2
n ni n
π π+ =
and form (ii), we havez = (–1)1/5 = [cos (2n + 1) p + i sin (2n + 1) π]1/5
=(2 1) (2 1)
cos sin ,when 0,1,2,3,4.5 5
n ni n
+ π + π + = These (4 + 5) = 9 values give the complete solution of the given equations.Example 7. Solve the expression z5 + 1 = 0.Solution. z5 = –1 = [cos (2r + 1) π + i sin (2r + 1)π]where r is an integer.
∴ z =(2 1) (2 1)
cos sin ,where 0,1,2,3,4.5 5
r ri r
+ π + π + =
When r = 0, cos sin .5 5
z iπ π= +
When r = 1,3 3
cos sin5 5
z iπ π= +
And for r = 2, z = cos π + i sin π = –1.For r = 3, the value of z is
7 7cos sin
5 5i
π π+
32
=3 3
cos 2 sin 25 5
iπ π π − + π −
=3 3
cos sin5 5
iπ π − + −
=3 3
cos sin5 5
iπ π−
For r = 4, the value of z is
=9 9
cos sin5 5
iπ π+
= cos 2 sin 25 5
iπ π π − + π −
= cos sin5 5
iπ π − + −
= cos sin5 5
iπ π−
Thus the 5 roots of the equation z5 + 1 = 0 are
3 3–1,cos sin ,cos sin .
5 5 5 5i i
π π π π± ±
Example 8. Find the roots of the equation z6 + z3 + 1 = 0Solution. On solving the equation a quadratic in z3, we find
z3 = 1 3 1 32 2 2
i− ± = − ±
= 2cos sin
3 3i
2π π±
=2 2
cos 2 sin 23 3
r i rπ π π + ± π +
=2
cos (3 1) sin (3 1)3 3
r i rπ 2π + ± +
∴ z =
132 (3 1) (3 1)
cos sin3 3r r
iπ + 2π + ±
=2 (3 1) (3 1)
cos sin , 0,1,2.9 9r r
i rπ + 2π + ± =
33
On giving r the values, 0, 1, 2, the six roots of the given equation can be written as
2 2 8 8 14 14cos sin ,cos sin ,cos sin
9 9 9 9 9 9i i i
π π π π π π± + ±
Now14 14 4 4
cos sin cos 2 sin9 9 9 9
i iπ π π π ± = π − ± 2 π −
=4 4 4 4
cos sin cos sin9 9 9
i ip
π π π π − + − = ±
∴ The roots are 2 2 4 4 8 8cos sin ,cos sin ,cos sin
9 9 9 9 9 9i i i
π π π π π π± ± ±
Example 9. Solve the equation z5 + z4 + z3 + z2 + z + 1 = 0 multiply throught by z – 1 we getz6 – 1 = 0.
Solution. z5 + z4 + z3 + z2 + z + 1 = 0Now z6 – 1 = 0⇒ z6 = 1 = cos 2kπ + i sin 2kπ, where k is an integer
⇒ z =cos2 2
sin6 6
k ki
π π+
= cos sin ,where 0,1,2,3,4,5.3 3
k ki k
π π+ =
⇒ z = 2 2 2
cos0 sin0,cos sin ;cos sin sin ;cos sin3 3 3 3 3
i i i i iπ π π π π+ + + + π + π
4 4 5 5cos sin ;cos sin ;
3 3 3 3i i
π π π π+ +
=1 3 1 3 1 3 1 3
1, , , 1, ,2 2 2 2 2 2 2 2
i i i i+ − + − − − −
=1 3
1,2 2
i± ± +
The root 1 corresponds to the factor (z – 1)
The remaining roots 1 3
–1,2 2
i± ± are the roots of the given equation
Exercise
1. Find all the values of(i) (1 + i)1/3
(ii) 1 4(–1 3)i+
(iii) (–1)1/6
34
(iv) (1 + i)2/3
2. Solve the equation:(i) z7 = 1
(ii) z7 + z = 03. Find the nth roots of unity and show that from a series in G.P. whose sum is zero.
Also prove that the sum of their pth power always vanishes unless p be a multiple of n, pbeing an integer and that then the sum is n.
4. Solve the equation : z7 + z4 + z3 + 1 = 0.[Hint : z7 + z4 + z3 + 1 = (z4 + 1) (z3 + 1) = 0]
5. Use De Moivre’s Theorem to solve : z10 + 11z5 + 1 = 0.[Hint : Put z5 = y]
6. Solve the equation : z4 – z3 + z2 – z + 1 = 0.[Hint : Multiply the equation by (z + 1)]
3.3 Expansion of sin n θ and cos n θ
De Moivre’s Theorem may be applied to express trignometric functions of multiple angles in termsof the trignometric function of the angles. The theorem can be employed to express powers of sinesand cosines of angle in terms of trignometric functions of multiple angles.
First we shall learn to express cos nθ, sin nθ, tan nθ in terms of cos θ, sin θ, tan θ, etc.For example, by De Moivre’s Theorem
cos 2θ + i sin 2θ = (cos θ + i sin θ)2 = cos2 θ + 2 i sin θ cos θ + i2 sin2 θ= (cos2 θ – sin2 θ) + 2i cos θ sin θ)
Equating the real and imaginary parts, we getcos 2θ = cos2 θ – sin2 θ,
sin 2θ = 2 cos θ sin θGenerally
(cos nθ + i sin nθ) = (cos θ + i sin θ)n, n being a positive integer.By the Binomial Theorem for a positive integral index, we have
(cos θ + i sin θ)n = 1 2 2( 1)cos cos ( sin ) cos ( sin )
2n n nn n
n i i− −−θ + θ θ + θ θ
3 3( 1) ( 2)cos ( sin ) ...... ( sin )
3n nn n
i i−+ −+ θ θ + θ
But i2 = –1, i4 = 1, i3 = – 1 etc.Hence we can write
cos nθ + i sin nθ = 2( 1)cos cos sin
2n nn n − 2 −
θ − θ θ
4 4( 1)( 2)( 3)cos sin ......
4nn n n n − − − −
+ θ θ +
35
1 3 3( 1)( 2)cos sin cos sin
3n nn n n
i n − − − −+ θ θ − θ θ +
KK
By equating the real and imaginary parts, we get
cos nθ = 2 2( 1)cos cos sin
2n nn n −−θ = θ θ
4 4( –1)( 2)( 3)cos sin
4nn n n n −− −+ θ θ + KK
sin nθ = 1 3 3( 1)( 2)cos sin cos sin
3n nn n n
n − −− −θ θ − θ θ + KK
By using the binomial coefficients, we can also express these results as follows
cos nθ = 2 2 4 42 4cos cos sin cos sin ......n n n n nC C− −θ − θ θ + θ θ +
= 2 42 4cos [1 tan tan ......]n n nC Cθ − θ + θ + ...(i)
sin nθ = 1 3 31 3cos sin cos sin ......n n n nC C− −θ θ + θ θ +
= 31 3cos [ tan tan ......]n nn C Cθ θ − θ + ...(ii)
Divinding (ii) by (i), we have
tan nθ =3 5
1 3 52 4
2 4
tan tan tan1 tan tan
n n n
n nC C C
C C qθ − θ + θ −
− θ + −KK
KK
We have thus obtained the values of cos nθ, sin nθ, tan nθ in terms of cos θ, sin θ, tan θ.
Example 10. Obtain the values of cos 5θ, sin 5θ, and tan 5θ.Solution. We can either apply the formulae or proceed as below:
cos 5θ + i sin 5θ = (cos θ + i sin θ)5
= cos5 θ + 5 cos4 θ (i sin θ) + 10 cos3 θ(i2 sin2 θ)+ 10 cos2 θ (i3 sin3 θ) + 5 cos θ (i4 sin4 θ)
+ i5 sin5 θThus equating real and imaginary parts,
cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cos θ sin4 θ= cos5 θ – 10 cos3 θ (1 cos2 θ) + 5 cos θ (1 – cos2 θ)2
= 16 cos5 θ – 20 cos3 θ + 5 cos θ(Note that cos 5θ has been expressed in powers of cos θ only).
Also tan 5θ =4 2 3 5
3 2 4sin5 5cos sin 10cos sin sinc o s 5 cos5 10cos sin 5cos sin
θ θ θ − θ θ + θ=
θ θ − θ θ + θ + θ
=3 5
2 55tan 10tan tan
1 10tan 5tanθ − θ + θ
− θ + θ(Note that tan 5θ has been expressed in powers of cos θ only.
36
Note. We can apply a similar method to obtainsin (α + β + γ + ......), cos (α + β + γ) + ......), tan (α + β + γ + .......)
and thus obtain expression for the sine, cosine and tangent of the sum of any numbers of angles interms of trigonometric ratio of individual angles.
We havecos (α + β + γ + ......) + i sin (α + β + γ + ......)
= (cos θ + i sin α) (cos β + i sin β) (cos γ + i sin γ) .......Since
cos α + i sin α = cos α(1 + tan α)
cos β + i sin β = cos β(1 + i tan β)cos ψ + i sin γ = cos γ(1 + i tan γσ) etc.
We havecos (α + β + γ + ......) + i sin (α + β + γ + ......)
= cos α cos β cos cos γ ...... [(1 + i tan α) (1 + i tan β)
(1 + i tan γ) ......= cos α cos β cos γ ...... [1 + i(tan α + tan β + ......)
+ i2(tan α tan β + tan β tan γ + ......]
+ i3 (tan α tan β tan γ + ...... ) + ......]
= cos α cos β cos γ ...... [1 + iS1 – S2 – iS3 + ......]where S1 = tan α + tan β cos γ ......
= Σ tan α,S2 = tan α tan β + tan β tan γ + ......
= Σ tan α tan βS3 = tan α tan β tan γ + .....
= Σ tan α tan β tan γ,
i.e., S1 indicates the sum of the tangents of the angles, S2 indicates the sum of the products ofthe tangents of the angles taken two at a time and S3 indicates the sum of the products taken threeat a time and so on.
Equating real and imaginary parts, we have
cos (α + β + γ ......) = cos α cos β cos γ ...... [1 – S2 + S4 – S6 + ......]sin (α + β + γ ......) = cos α cos β cos γ ...... [S1 – S3 + S5 ......]
Also, by division, we get
tan (α + β + γ) = 1 3 5
2 4 6
......1 ......
S S SS S S− +
− + −
3.4 Expansion of sin θ and cos θ in Powers of θ
We know that for a positive integer n and an angle α,
cos nα = 2 2( 1)cos cos sin
2n nn n −−α − − α α
37
4 4 4( 1)( 2)( 3)cos sin sin ......
4nn n n n −− − −+ α α α
Now put nα = θ, then n = θα
and
cos θ =2
2( ) sincos cos
2n n −θ θ − α α α − α α
42( ) ( 2 ) ( 3 ) sin
cos4
θ θ − α θ − α θ − α α + α α KK
Now if α is mode indefinitely small, θ remaining constant and consequently n becoming indefinitely
large, cos α and sin α
α both tend to unity, and hence in the limit when a → 0 we have
cos θ =2 4 7
1 ......2 4 6
θ θ θ− + − +
Similarly by writing the value of sin nα and proceeding as before, we can obtain.
sin θ =2 4 7
......3 5 7
θ θ θθ − + − +
The method we have applied here is only an indication ot get the values of cos θ and sin θ anddoes not give a rigorous proof.
We shall now obtain a series for tan θ in terms of θ by using the series for cos θ and sin θ.
Thus tan θ =sincos
θθ
=
3 5
2 43 5
12 4
θ θθ − + −
θ θ− + −
KK
KK
=13 5 2 4
16 120 2 24
− θ θ θ θ θ − + − − − − KK KK
Expanding the second bracket by the Binomial theorem, we have
tan θ =3 5 2 4 2 2
16 120 2 24 2 24
2 θ θ θ θ θ θ θ − + − + − + − + KK K K K
=3 5 2 4 4
16 120 2 24 4
q θ θ θ θθ − + − + − + K K
38
=3 5 2
451
6 120 2 24q
θ θ θθ − + − + + K K
=3 3 5 5
552 6 24 12 120
θ θ θ θθ + − + θ − + K
=3
523 15
θθ + + θ + K
Thus tan θ =3
523 15
θθ + + θ + K
If powers of θ more than five are neglected, i.e., for small values of θ, we have
sin θ =3
,6θθ −
cos θ =2 4
1 ,2 24
θ θ− +
tan θ =3
,3
θθ +
and if power more than 2 are to be neglected then
sin θ = θ, cos θ = 2
1 ,tan2
θ− θ = θ
It is understood that the angle θ is expressed in radians.
Example 11. Find the value of θ when
cos θ = 91009
Solution. Here cos θ is nearly 1 and θ is small, Hence we can take
cos θ = 12
2θ−
i.e.,2
2θ =
99 11 cos 1
100 100− θ = − =
∴ θ =2 1.41
.141radians10 10
= =
or θ =180
0.141 degrees π
= 0.8 degrees approx.
39
Example 12. Find the value of sin 3º correct to theree places of decimals.
Solution. We have 3º =3
radians180
π
Now3
sin180
π =
313 1 3...
180 3 180π π − +
=3
nearly=180 60
π π
∴ sin 3º =3.14159
0.0523approx.60 60π = =
3.5 Expression of the Products of the Form cosm θ sinn θ Terms of sines or cosines ofMultiples of θ.
De Moivre’s Theorem can be employed to express a product of the form cosm θ sinm θ wherem and n are positive integers in terms of sines or consines of multiples or θ.
Let z = cos θ + i sin θ
then1z
=1
cos sincos sin
ii
= θ − θθ + θ
Hence1
zz
+ = 2 cos θ
1z
z− = 2 i sin θ
Also from De Moivre’s Theorem, if r is an integer.
(i) zr = cos r θ = i sin r θ
(ii)1
cos sinr r i rz
= θ − θ
and, therefore, by adding and subtraction (i) and (ii), we get
2 cos rθ =1rrz
z+
and 2i sin rθ =1rrz
z−
We know that
(2 cos θ)m (2i sin θ)n =1 1
m m
z zz z
+ − The right hand side of this equation is expanded of z and terms with indices equal but opposite
40
in sign are grouped together. The resulting sum is made up, if n is even, of term of the form 1rrz
z+
and if n is odd, of term of the form 1.r
rzz
−
Also we can express these terms 1 1
andr rr rz z
z z + −
as 2 cos rθ and 2i sin rθ respectively..
The following examples will illustrate the method.
Example 13. Express sin6 θ in terms of cosines of multiples of θ.Solution. Let z = cos θ + i sin θ, so that
1z
z+ =
12 cos , z
zθ − = 2 i sin θ,
1rrz
z+ =
12 cos , 2 sinr
rr z i rz
θ − = θ
r being any positive integer.
Now (2i sin θ)6 =6
6 4 6 2 6 0 6 2 6 4 6 41 2 3 4 5 6
1z z C z C z C z C z C z C z
z6 − − − − = − + + + − +
⇒ 64 i6 sin6 θ = 6 4 2 2 4 66 15 20 15 6z z z z z z− − −− + − + − +
⇒ sin6 θ = 6 4 26 4 2
1 1 1 16 15 20
64z z z
z z z − + − + + + −
=1
[2cos6 6.2cos4 15.2cos2 2 ]64
− θ − θ + θ − θ
= 1[cos6 6cos4 15cos2 20]
32− θ − θ + θ −
Example 14. Express sin4 θ cos2 θ in cosines of multiples of θ.
Solution. Let z =1
cos sin then cos sin .i iz
θ + θ = θ − θ
so that1
zz
+ = 2 cos θ and 1
2 sinz iz
− = θ
Also1rrz
z+ =
12 cos and 2 sinr
rr z i rz
θ − = θ
2 (i sin θ)4 (2 cos θ)2 =4 2
1 1z z
z z − +
41
=2 2 2
1 1 1z z z
z z z − − +
=2 2
21 1z z
z z2 − −
=2 4
2 41 1
2 2z zz z
− + − +
=2
4 26 4 2
1 1 12 4z z z
z z z6 + − + + + +
i.e., 64 sin4 θ cos2 θ = 2 cos 6θ – 4 cos 4θ + 2 cos 2θ + 4.
Hence sin4 θ cos2 θ =1
(cos6 2cos4 cos2 2)32
θ − θ + θ +
EXERCISE
1. Express sin 5θ in terms of sin θ only.2. Express cos 6θ in terms of cos θ only.3. Express tan 7θ in terms of tan θ only.4. Find the value of sin 1º approximately.
5. Find the value of θ, where sin 1013
.1014
θ =θ
6. Show that 16 sin5 θ = sin 5θ – 5 sin 3θ + 10 sin θ.7. Express sin7 θ cos2 θ in terms of sines of multiples of θ.8. Express cos6 θ in terms of cosines of multiples of θ.
42
LESSON 4
APPLICATIONS OF DE MOIVRE’S THEOREM
In this lesson, we shall study the application of De Moivre’s theorem to find the sum of certaintypes of trigonometric series. The series may be finite or infinite. The general method is known as theC + iS method. If we are given as series of cosines such as C = cos θ + cos 2 θ + ... cos nθ, weconsider a similar series of sines i.e. S = sin θ + sin 2θ + ... sin nθ.
The new series C + iS so formed can be summed up by different methods. Equating real and imaginaryparts we obtain the values of C and S.
The following formulae are mainly used in the summation of series.
(i) 2 1 11 , 1
1
nn x
x x x xx
− −+ + + = ≠
−K
(ii) 21 2(1 ) 1n n n n
nx C x nC x C x+ = + + +K
(iii) Euler’s Formula: If z is complex variable, then the complex exponential function ez is defined
by 2 3
12! 3!
z z ze z= + + + + ∞K
Replacing z by i θ we have
eiθ =2 3 4 5( ) ( ) ( ) ( )
12 3 4 5
i i i ii
θ θ θ θ+ θ + + + + + K
=2 3 4 5
1 ...2 3 4 5
i i iθ θ θ θ
+ θ − − + + +
=2 4 3
1 ...2 4 3
i θ θ θ− + + θ − +
K
= cos θ + i sin θSimilarly e–iθ = cos θ – i sin θ∴ eiθ + e–iθ = 2 cos θand eiθ – e–iθ = 2 i sin θ
Also (eiθ)n = eniθ = cos nθ + i sin nθwhere n is an integer.We shall illustrate the method C + iS in the following solved examples.Example 1. Find the sum to n terms of the series.1 + x cos θ + x2 cos 2θ + ... + xn – 1 cos (n – 1)θLet C = 1 + x cos θ + x2 cos 2θ + ... + xn –1 sin (n – 1) θ
S = x sin θ + x2 sin 2θ + ... + xn – 1 sin (n – 1)θ
43
∴ C + iS = 1 + x(cos θ + i sin θ) + x2(cos 2θ + i sin 2θ)+ ... + xn – 1 [cos(n – 1)θ + i sin (n – 1)θ] + i sin (n – 1) θ)]
= 1 + xeiθ + x2e2iθ + ... + xn – 1 e(n – 1)iθ
=1
,1
n ni
ix e
xe
θ
θ−
− summing the GP to n terms
= (1 )(1 )(1 ) ( )
n ni i
i ix e xe
xe i xe
θ − θ
θ − θ− −
− −
[Multiply the numerator and denominator by 1 – xe–iθ]
=1 ( 1)
21
1 ( )
n ni i n n i
i ix e xe x e
x e e x
θ − θ + − θ
θ θ− − +
− + +
= 1
21 [cos sin ] (cos sin ) [cos( 1) sin( 1) ]
1 2 cos
n nx n i n x i x n i nx x
+− θ + θ − θ − θ + − θ + − θ− θ +
(Using Euler’s Formulae)Equating the real part we get
C =1
21 cos cos cos( 1)
1 2 cos
n nx n x x nx x
+− θ − θ + − θ− θ +
Example 2. Find the sum to n term of the seriessin 2θ + sin 4θ + sin 6θ + ...
Let C = cos 2θ + cos 4θ + cos 6θ + ... + cos 2nθS = sin 2θ + sin 4θ + sin 6θ + ... + sin 2nθ
C + iS = e2iθ + e4iθ + ... e2niθ
= 2 2 2( 1)[1 ]i i n ie e eθ θ − θ+ +K
= 2 1[1 ]nx x x x −+ + +K where x = e2iθ
C + iS =(1 )1
nx xx
−−
=2 2
2[1 ]
1
i ni
ie e
e
θ θ
θ−
−
=2 2 2
2 2(1 ) (1 )
(1 )(1 )
i ni i
i ie e e
e e
θ θ − θ
θ − θ− −
− −
(Multiplying numerator and denominator by the conjugate of the denominator)
=2 2 2 2( 1)
2 2[1
1 ( ) 1
i ni i n i
i ie e e e
e e
θ θ − θ − θ
θ − θ− − +− + +
44
=2 2( 1) 21
1 2 cos2 1
i n i nie e eθ + θ θ− − +− θ +
=(cos2 sin2 ) [cos2( 1) s in2 ( 1) ] 1 [cos2 sin2 ]
2(1 cos2 )i n i n n i nθ + θ − + θ + + θ − + θ θ + θ
− θ
Equating the imaginary parts on both sides, we get
S =sin2 sin2( 1) s in2
2(1 cos2 )n nθ − + θ + θ
− θ
= 22sin cos [sin2( 1) sin2 )]
4sinn nθ θ − + θ − θ
θ
= 22sin cos 2cos(2 1) sin
4sinnθ θ − + θ θ
θ
= 22sin [cos cos(2 1) sin ]
4sinnθ θ − + θ θ
θ
= ( )2sin 1 sin2sin
n n+ θ θθ
=sin( 1) sin
sinn n+ θ π
θwhere sin θ ≠ 0 i.e., θ ≠ nπ
Example 3. Find the sum of the series
1 2cos cos( ) cos( 2 ) ... cos( )n n nnC C C nθ θ + φ + θ + φ + + θ + φ
Let C = 1 2cos cos( ) cos( 2 ) ... cos( )n n nnC C C nθ θ + φ + θ + φ + + θ + θ
and S = 1sin sin( ) ... sin( )n nnC C nθ + θ + φ + + θ + θ
Here C + iS = ( ) ( 2 ) ( )1 2 ...i n i n i n i n
ne C e C e C eθ θ + φ θ + φ θ + φ+ + + +
= 21 2[1 ...i n i n i n ni
ne C e C e C eθ φ φ φ+ + + +
= [1 ]i i ne eθ φ+
= [1 cos sin ]i ne iθ + φ + φ
= 22cos 2 sin cos2 2 2
nie iθ φ φ φ +
= 2cos cos2 2 2
n nie iθ φ φ φ +
45
= 2cos [cos sin ] cos sin2 2 2
n n ni i
φ φ φ θ + θ +
= 2cos cos sin2 2 2
n n ni q
φ φ φ θ + + +
Equating the real part on both sides, we get
C = 2cos cos2 2
nnφ φ θ +
Sometimes the sum to n terms of a trigonometric series can be obtained by using simple trigonometricformulae viz.
2 sin A sin B = cos (A – B) – cos (A + B)2 cos A sin B = sin (A – B) – sin (A + B)
These formulae can be useful when the nth term of a series can be expressed at the differenceof two consines of sines. The following example will illustrate this method.
Example 4. Find the sum to n terms of the series.sin θ + sin 3θ + sin 5θ + ...
Let S = sin θ + sin 3θ + sin 5θ + ... + sin (2n – 1)θMultiplying both sides by 2 sin θ (≠ 0) we get,
2 sin θ · S = 2 sin2 θ + 2 sin θ · sin 3θ + 2 sin θ · sin 5θ + ... + 2 sin θ sin (2n – 1)θ= [1 – cos 2θ] + [cos 2θ – cos 4θ] + [cos 4θ – cos 6θ] ...
+ [cos 2(n –1) θ – cos 2 nθ)= 1 – cos2 nθ= 2 sin2 nθ
S =22sin
sin cosec2sin
nn
2 θ= θ θ
θ
provided sinθ ≠ 0 i.e., θ ≠ nπExample 5. Find the sum of the infinite series.
sin 3 s in5sin
3 5θ θθ − + +K
Let S = sin3 sin 5sin
3 5θ θθ − + + K
C = cos3 cos5cos
3 5θ θθ − + + K
and we have
C + iS =3 5
3 5
i ii e e
eθ θ
θ − + −K
46
=3 5
3 5x x
x − + −K where x = eiθ
= sin x
= sin [cos θ + i sin θ]
= sin (cos θ) · cos (i sin θ) + cos(cos θ) · sin (i sin θ)
= sin (cos θ) · cos h (sin θ) + i cos(cos θ) · sin h (sin θ)
Equating imaginary parts on both sides, we get
S = cos (cos θ) · sin h (sin θ) [Q cos iθ = cos hθ, sin iq = i sin hθ]
from the definition of Hyperbolic functions)Example 6. Sum of the series
1 1.3 1.3.51 cos cos2 cos3
2 2.4 2.4.6− θ + θ − θ + ∞K
Let C = 1 1.3
1 cos cos22 2.4
− θ + θ + K
S = 1 1.3sin sin2
2 2.4− θ + θ + K
Here C + iS = 2 31 1.3 1.3.51 ...
2 2.4 2.4.6i i ie e eθ θ θ− + − +
=12(1 )ie
−θ+
=12[1 cos sin ]i
−+ θ + θ
=
1222cos 2 sin cos
2 2 2i
−θ θ θ + ⋅
=
12
2cos cos sin2 2 2
i−
θ θ θ +
=
12
2cos cos sin2 4 1
i−θ θ θ −
Equating real parts on both sides we get,
C =cos
4
2cos2
θ
θ
47
Example 7. Find the sum to infinitely of the series
21 1
sin sin2 sin3 ...2 2
α + α + α +
Let C = 21 1
cos cos2 cos3 ...2 2
α + α + α +
S = 21 1
sin sin2 sin3 ...2 2
α + α + α +
∴ C + iS =2
21 1
...2 2
i i ie e eα α α+ + +
= 22
1 11 ...
2 2i i ie e eα α α + + +
=11
12
i
ie
e
α
α⋅
−
[Summing the G.P. to ∞]
=
11
21 1
1 12 2
i i
i i
e e
e e
α − α
− α − α
− − −
=
12
1 11 ( )
2 4
i
i i
e
e e
α
α − α
−
− + +
=
1cos sin
2
cos4
iα + α −
5 − α
=4(cos sin ) 2
5 4cosiα + α −
− α
Equating the imaginary parts on both sides, we get
S =4sin
5 4cosα
− α
Example 8. Find the sum to ∞ of the sides
2 2 3cos cos cos cos 2 cos cos3α ⋅ α + α ⋅ α + α ⋅ α +K
48
Let C = 2 2 3cos cos cos cos 2 cos cos3α ⋅ α + α ⋅ α + α ⋅ α +K
S = 2 3cos sin cos sin 2 cos sin3α ⋅ α + α α + α ⋅ α +K
Here C + iS = 2 2 3 3cos cos cos ...i i ie e eα α αα ⋅ + α ⋅ + α ⋅ +
= 2cos [1 cos (cos ) ]i i ie e eα α αα ⋅ + α ⋅ + α ⋅ + K
= 1cos
1 cosi
iee
αα
α ⋅
− α ⋅
(Summing the G.P. to ∞)
= ( )( )cos [1 cos ]
1 cos 1 cos
i i
i i
e ee e
α α
α − αα − α ⋅
− α ⋅ − α ⋅
= ( )2
2
cos cos
1 cos cos
i
i i
e
e e
α
α − α
α − α
− α + + α
=2
2 2cos [cos sin ] cos
1 2cos cosiα α + α − α
− α + α
=2 2
2cos sin cos cos
1 cosiα + α ⋅ α − α
− α
= 2sin cos
sini α ⋅ α
α
It follows that C = 0provided sin2 α ≠ 0i.e., α ≠ kπ where k is an integeri.e., The sum of the given infinite series is zero, provided α is not a multiple for π.
EXERCISES
Find the sum to n terms of the following series :1. cos θ + cos 2θ + cos 3θ + ...2. sin θ · cos 2θ + sin 2θ · cos 3θ + sin 3θ · cos 4θ + ...3. sin α + sin(α + β) + sin (α + 2β) + ...4. 2 cos 2θ + 22 cos 4θ + 23 cos 6θ + ...
Find the sum to infinity of the following series :
5.1 1 .3
1 cos2 cos4 cos6 ...2 2.4 2.4.6
1+ θ − θ + θ +
49
6. 1 1sin sin2 sin3 ...
2 3θ − θ + θ +
7. sin α · sin α + sin2 α · sin 2α + sin3 α · sin 3α + ...
8.2 3sin sin
1 sin cos cos2 cos3 ...2 3
α α+ α ⋅ β β + β +
9.2 3cos cos
1 cos cos cos2 cos3 ...2 3
α α− α ⋅ β + β − β +
10. sin( 2 ) sin( 4 )sin ...
3 4α + β α + βα − + +
50
LESSON 5
THEORY OF EQUATIONS I
5.1 In this lesson we shall deal with the theory of equations. We presume that you are all familiarwith the theory of quadratic equations. In the present course of study of theory of equations, we shallconfine ourselves to cubic and biquadratic equations.
5.2 Polynomial and Equations
We know that a function p(x) of the form
p(x) = 1 20 1 2 0... , 0,n n n
na x a x a x a a− −+ + + ≠
where the coefficients a0, a1, a2, ... an are all complex numbers (a0 ≠ 0) is called a polynomial
of degree n, n being a positive integer. For example, 3 20 1 2 3a x a x a x a+ + + is a polynomial of degree 3.
The expansion p(x) = 0 is called an nth degree equation, that is,
p(x) = 1 20 1 2 0... 0, 0n n n
na x a x a x a a− −+ + + = ≠ ...(1)
The fundamental theorem of Algebra states that every equation of the form (1), has at least onecomplex number as a root. The proof of this result is beyond the scope of the present course of studyand hence is omitted.
In the discussion thereater, it will be presumed that the coefficients like a0a1 ... an are all complex
numbers. Now discuss the following theorem which is known as Factor theorem. We already knowthat if α is a root of an equation, say, a0x
2 + a1x + a2 = 0, a0 ≠ 0 then x – α is a factor of the correspondingpolynomial a0x
2 + a1x + a2. For example, 3 is a root of the equation x2 – x – 6 = 0, then it is easyto verify that x – 3 is a factor of the polynomial x2 – x – 6. We generalize this idea into the followingtheorem :
5.3 Factor Theorem
A number α is a root of the equation1
0 1 0n nna x a x a− + + =K ...(1)
if and only if (x – α) is a factor of the polynomial
p(x) = 10 1
n nna x a x a−+ + +K ...(2)
where a0 ≠ 0, a1, a2 ..., an are complex numbers and n is a positive integer.Proof. In the first part, it is given that α is a root of the equation (1) and we have to prove that
(x – α) is a factor of (2).We divide the polynomial p(x) of degree n by a polynomial (x – α) of degree one. Then, the quotient
will be a polynomial of degree (n – 1). If the remainder is zero, then obviously (x – α) is a factorof p(x).
If the remainder is not zero, then the remainder is a constant and let it be R. Also let the quotientbe denoted by
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1 2 30 1 2 2 1
n n nn na x b x b x b x b− − −
− −+ + + + +K
Then we have
1 1 20 1 1 0 1 2 1( ) [ ]n n n n
n n n na x a x a x a x a x b x b x b R− − −− − −+ + + + ≡ − α + + + +K K
Putting x = α on both sides, we get
10 1 1
n nn na a a−
−α + α α + + α +K ≡ 0 + R = R
In other words,
R = 10 1 1
n nn na a a−
−α + α α + + α +K ...(3)
Now again, since α is a root of the equation (1), therefore by putting x = α in (1), we get
10 1 1
n nn na a −
−α + α + + α α + αK = 0 ...(4)
From (3) and (4), it follows that R = 0 so that (x – α) is a factor of p(x).
In the second part, it is given that (x – α) is a factor of the polynomial p(x) given by (2), wehave to show that α is a root of the equation (1).
Since (x – α) is a factor of the polynomial p(x), therefore when we divide p(x) by (x – α), wemust get R = 0, and consequently from (3), we get
10 1 1 0n n
n na a a a−−α + α + + α + =K
which implies that a satisfies the equation1
0 1 1 0n nn nx a x a x a−
−α + + + + =K
i.e., α is a root of the equation (1).
This completes the proof of the theorem.
Now we discuss the theory of cubic equations.
5.4 Cubic Equation
The general cubic equation is of the form
a0x3 + a1x
2 + a2x + a3 = 0, a0 ≠ 0 ...(5)
where a0, a1, a2, a3 are given complex numbers.
By the fundamental theorem of algebra, this equation (5) has atleast one root. Let this root bedenoted by ‘α’. Then by Factor Theorem (x – α) is a factor of the polynomial a0x
3 + a1x2 + a2x +
a3. Divide the polynomial by x – α and let the quotient be denoted by a0x2 + b1x + b2.
Then we have
a0x3 + a1x
2 + a2 x + a3 ≡(x – α) (α0x2 + b1x + b2) ...(6)
Now consider the quadratic equation
a0x2 + b1x + b2 = 0 ...(7)
Again by the fundamental theorem of algebra equation (6) must have at least one root, say, β and
52
accordingly (x – β) must be a factor of the polynomial 20 1 2 0.a x b x b+ + = Dividing a0x
2 + b1x + b2
by (x – β), we get the quoetient, say, (a0x + c1) so that we have
20 1 2a x b x b+ + = 0 1( ) ( )x a x c− β + ...(8)
Also a0 x + c1 = 10 0
0( )
ca x a x
a
+ = − γ
...(9)
γ = 10
0, 0
ca
a− ≠
Combining (6), (8) and (9), we get
3 20 1 2 3a x a x a x a+ + + ≡ a0 (x – α) (x – β) (x – γ) ...(10)
Substituting x = α, β, γ successively on both sides of (9), we observe that each of the values,α, β, γ satisfies the equation (5). Thus α, β, γ are the root of the equation (5). Also it may be notedthat no number different from α, β, γ can be roots of the equation (5).
For, if possible let δ, a number different from each of α, β, γ be a root of this equation (5). Thenby substituting x = δ on both sides of (10), we find that while the left hand side is zero by virtueof the supposition that δ is a root but the right hand side is non-zero and thereby we arrive at a contradiction.Hence, α, β, γ are the only roots of the equation (5). Thus we have proved that a cubic equation hasthree and only three roots. These roots may be all distinct (unequal), may be repeated (equal) or maybe that only two are repeated.
From the above discussion, it is clear that if α, β, γ are the three roots of the a cubic equation,then the corresponding factors are (x – α), (x – β), (x – γ) and hence the equation is given by
(x – α) (x – β) (x – γ) = 0
or 3 2( ) ( ) 0.x x x a− α + β + γ + α β + β γ + γ α − βγ =
We discuss the following examples :
Example 1. Find the equation whose roots are 1, 2, 3.Solution. The required equation is given by
(x – 1) (x – 2) (x – 3) =0.
or x3 – 6x2 + 11x – 6 =0[By putting α = 1, β = 2, γ = 3 in equation (11) above]
Example 2. Find the equation whose roots are 3, 3, –2.Solution. (x – 3) (x – 3) (x + 2) =0.or x3 – 4x2 –3x + 18 = 0
5.5 Reactions between the Roots and Co-efficients of a Cubic EquationLet α, β, γ be the roots of a cubic equation
3 20 1 2 3 0a x a x a x a+ + + =
Then we have the identify
53
3 20 1 2 3 0 ( ) ( ) ( )a x a x a x a a x x x+ + + ≡ − α − β − γ
which implies
3 20 1 2 3a x a x a x a+ + + ≡ 3 2
0 [ ( ) ( ) ]a x x x− α + β + γ + αβ + β γ + γ α −αβγ
Equating the co-efficients powers of x, i.e., x3, x2, x and the constant terms, we geta0 = a0
a1 = –a0 (α + β + γ)
a2 = a0 (αβ + βγ + γα)ss
α3 = 0 0( )a a−αβγ = − αβγ
These give
α + β + γ =1
0
aa
−
αβ + βγ + γα =2
0
aa
αβγ = 3
0
aa
−
In other words, we can write these relations as
Sum of the roots =2
3coefficientof
–coefficientof
xx
Sum of the products of roots taken in pairs = 3coefficientof
coefficientofx
x
= 3– constantterm
coefficientof x
Thus we note that though we are not able to find the values of α, β, γ separately, yet we havebeen able to express the three conditions combinations namely α + β + γ, αβ + βγ + γα, αβγ in termsof coefficients α0 ≠ 0, α1, α2, α3.
5.6 Symmetric FunctionsConsider the following expressions :
(i) 2 2 2a b+ +γ
(ii) 2 2 2( ) ( ) ( )α + β + β + γ + γ + α
(iii) α2β + α2γ + β2α + β2γ + γ2α + γ2βEach of the above is a function of α, β, γ with the property that if any two of α, β, γ are interchanged
the function remain unaltered. Such functions are called symmetric functions. More precisely a function
54
f (α1, α2, ... αn) of n variables is said to be symmetric function of α1, α2, ... αn if it remains unalteredby interchanging any two of
α1, α2 ... αn.Thus, for example
α + β + γ, αβ + βγ + γα, αβγ are symmetric functions of α, β, γ.In the case of a cubic equation whose roots are α, β, γ the functions
α + β + γ, αβ + βγ + γα, αβγ are called basic symmetric functions.
Sigma Notation for Symmetric FunctionsIt is often convenient to describe a symmetric function by indicating only one or more terms in
such a manner that the other terms are obtainable from the given term on replacing the roots thereinby other roots in all possible ways. Also then we put down the sign Σ(sigma) before the given term.
Thus for a cubic equation
Σα = (α + β + γ)Σαβ = (αβ + βγ + γα)
Σα2β = α2β + αβ2 + β2γ + β2gγ + αγ2 + α2γΣα2 = α2 + β2 + γ2
Σα3 = α3 + β3 + γ3 and so on.
It is interesting to know that every symmetric function of α, β, γ can be expressed in terms ofthe symmetric function α + β + γ, αβ + βγ + γα, and αβγ whose values can be written down fromthe cubic equation whose roots are α, β, γ. We illustrate this fact by the following examples.
Example 3. If α, β, γ be the roots of the equationx3 + px2 + qx + r =0
find the value of(i) Σ(α + β)2
(ii) (β + γ) (γ + α) (α + β)
Solution. From the given equation, we have
(α + β + γ) =2
3co-efficient
1co-efficient x p
px
− = − = − s
(αβ + βγ + γα) = 3co-efficient of
1co-efficient of xx q
p= =
αβγ = 3constantterm
1co-efficient of r
rx
− = − = −
(i) 2( )Σ α − β = 2 2 2 2 2( 2 ) 2 2 2Σ α + β − αβ =Σα + Σβ − Σαβ= Σα − Σαβ
= 2 2 22[( ) 2 ] 2 2( ) 6 2 6a p qΣ − Σαβ − Σαβ= Σα − Σαβ= −
(ii) We shall now express the symmetric function, (β + γ) (γ + α) (α + β) in terms of the basicsymmetric functions.
55
We have, by actual multiplication
( ) ( ) ( )β + γ γ + α α + β = 2( ) ( ]aβ + γ α γ + β γ + + αβ
= 2αβγ + (α2β + β2γ + γ2α + αβ2 + βγ2 + γα2)Also we have (α + β + γ) (αβ + βγ + γα)
= (α2β + αβ2 + β2γ + βγ2 + αγ2 + γα2) + 3αβγthat (α2β + αβ2 + β2γ + βγ2 + αγ2 + γα2)
= (α + β + γ) (αβ + βγ + γα) –3αβγThus we have (β + γ) (γ + α) (α + β)
= 2αβγ + {(α + β + γ) (αβ + βγ + γα) –3αβγ}
= (α + β + γ) (αβ + βγ + γα) –αβγ = – pq + r.
Another method: We have(β + γ) (γ + α) (α + β)
= (α + β + γ – α) (α + β + γ – β) (α + β + γ – γ)
= (–p – α) (–p – β) (–p – γ), [Q (α + β + γ = – p)]= – (p + α) (p + β) (p + γ)
= – [p3 + p2 (α + β + γ) + p) (αβ + βγ + γα) + αβγ]
= 2–[ ( ) ( )] .p p p pq r pq r3 + − + − = − +
Example 4. If α, β, γ be the roots of the equation x3 + px2 + qx + r = 0. Find the value of
(i) 2 2Σα β
(ii) α βΣ + β α
Solution. We have α + β + γ = – pαβ + βγ + γα = q
αβγ = – r
Now(i) (βγ + γα + αβ)2 = β2γ2 + γ2α2 + α2β2 + 2αβγ2 + 2α2βγ + 2αβ2γ
= (β2γ2 + γ2α2 + α2β2) + 2αβγ (α + β + γ)∴ (β2γ2 + γ2α2 + α2β2) = (βγ + γα + αβ)2 – 2αβγ (α + β + γ)
= q2 – 2 (–r) (–p)
= θ2 – 2rp.
(ii) α β
Σ + β α =
2 2α + βΣ
αβ
=2 2 2 2 2 2α + β β + γ γ + α
+ +αβ βγ αγ
56
=2 2 2 2 2 2( ) ( ) ( )γ α + β + α β + γ + β γ + α
αβγ
=2Σα β
αβγ
=3ΣαΣαβ− αβγ
αβγ
= ( ) ( ) 3( )p q rr
− − −−
=3pq r
r−
Example 5. If α, β, γ be the root of the cubic equation x3 + px2 qx + r = 0. Find the value
of 3 3 3( ).α + β + γ
Solution. We have(α + β + γ) (α2 + β2 + γ2)
= (α3 + β3 + γ3) + (α2β + αβ2 + β2γ + βγ2 + αγ2 + γα2)⇒ Σα Σα2 = Σα3 + Σα2 β∴ Σα3 = Σα Σα2 – Σα2 β ...(1)But as shown in Example 3.
(α2β + αβ2 + β2γ + βγ2 + γα2 + γ2α)
= (α + β + γ) (αβ + βγ + γα) – 3αβγ= – pq + 3r
Againα2 + β2 + γ2 = (α + β + γ)2 – 2(αβ + βγ + γα) = p2 – 2q.
Also
α + β + γ = – pHence from (1), we get
Σα3 = (α3 + β3 + γ3) = – p (p2 – 2q) – (–pq + 3r)= –p3 + 3pq – 3r.
Another method :
As α, β, γ are the roots of the equation x3 + px2 + qx + r = 0, we haveα3 + pα2 + qα + r = 0.
β3 + pβ2 + qβ + r = 0.γ3 + pγ2 + qγ + r = 0.
Adding these three equations we obtain
(α3 + β3 + γ3) + p (α2 + β2 + γ2) + (α + β + γ) + 3r = 0 ...(1)Now α + β + γ = – p
57
α2 + β2 + γ2 = (α + β + γ)2 – (αβ + βγ + γα)= p2 – 2q
Substituting in (1), we obtain(α3 + β3 + γ3) + p(p2 – 2q) – qp + 3r = 0
⇒ α3 + β3 + γ3 = – p3 + 3pq – 3r.Example 6. α, β, γ be the roots of the equation
x3 + qx + r = 0,
then prove that
5 5 5
5α + β + γ
=3 3 3 2 2 2
3 2b bα + + γ α + + γ×
Solution. Since α, β, γ are the roots of equation x3 + qx + r = 0
∴ α3 + qα + r = 0 ...(i)β3 + qb + r = 0 ...(ii)
γ3 + qγ + r = 0 ...(iii)Adding (i), (ii), (iii), we get
3 3q rΣα + Σα + = 0
AlsoΣα = 0, Σαβ = q, αβγ = –r
∴ Σα3 = –q Σα – 3r = – 3r ...(iv)Σα2 = (Σα)2 – 2Σαβ = 0 – 2q = – 2q ...(v)
Multiply (i) by α2, (ii) by β2 and (iii) by γ2 and adding we get
Σα5 + q Σα3 +sss r Σα2 = 0⇒ Σα5 = – q Σα3 – r Σα2 = –q(–3r) – r (–2q)
= 3qr + 2qr = 5qr by (iv) and (v)
∴ 515
Σα =3 21 1
3 2qr
= Σα Σα
Hence the result.
Example 7. Find the equation whose roots α, β, γ satisfy the relations :α + β + γ = 3
α2 + β2 + γ2 = 5
α3 + β3 + γ3 = 7Also find the value of α4 + β4 + γ4.
Solution. Let the required equation beax3 + bx2 + cx + d = 0
so that α + β + γ =ba
−
58
αβ + βγ + γα =ca
αβγ =da
−
We, therefore have 3 = α + β + γ = ba
− ...(ii)
and 5 = 2 2 2 ( ) 2 ( )a b 2+ + γ = α + β + γ − αβ + β γ + γ α
= 29
ca
−
Hence, from (ii), 3andfrom( ) 2.b c
iiia a
− =
Since α, β, γ, satisfy (i) we have
3 2a b c dα + α + α + = 0
3 2a b c dβ + β + β + = 0
3 2a b c dγ + γ + γ + = 0
Adding these, we obtaina(α3 + β3 + γ3) + b(α2 + β2 + γ2) + c (α + β + γ) + 3d = 0
i.e., 7a + 5b + 3c + 3d = 0
Dividing by ‘a’ we obtain
7 5 3 3b c da a a
+ + + = 0
Substituting the values of ba
and ca
obtained above, we obtain
7 15 6 3da
− + + = 0
i.e.,da
=23
Thus the required equation (i) on the diving by a is
3 2b c dx x x
a a a+ + + = 0
i.e., 3 2 23 2
3x x x− + + = 0
i.e., 3 23 9 6 2x x x− + + = 0
59
Since α, β, γ are the roots of this equation, they must satisfy the equation.∴ 3α3 – 9α2 + 6α + 2 =0
3β3 – 9β2 + 6β + 2 = 03γ3 – 9γ2 + 6γ + 2 = 0
Multiplying the above equations by α, β, γ respectively and adding we get,3Σα4 – 9Σα3 + 6Σα2 + 2Σα = 0
⇒ 3Σα4 = 9Σα3 – 6Σα2 – 2Σα
3Σα4 =1
[9.7 6.5 6]3
− −
= 21 – 10 – 2= 9
5.7 Biquadratic Equations
We consider the general biquadratic equation, i.e., the equation of degree four viz.,
4 3 20 1 2 3 4 0a x a x a x a x a+ + + + =
where a0 ≠ 0, a1, a2, a3, a4 are complex coefficients.Let ‘α’ be a root of this equation so that (x – α) is a factor of the polynomial.
4 3 20 1 2 3 4a x a x a x a x a+ + + +
Then we have a certain
4 3 20 1 2 3 4a x a x a x a x a+ + + + ≡ 3 2
0 1 2 3( ) ( )x a a x b x b x b− + + + ...(1)
where 3 20 1 2 3a x b x b x b+ + + is the quoedient.
As proved previously 3 20 1 2 3 0a x b x b x b+ + + = being a cubic equation, has 3 roots. Let these roots
be β, γ and δ.
Then, we have3 2
0 1 2 3a x b x b x b+ + + ≡ 0( ) ( ) ( )a x x x− β − γ − δ ...(2)
From (1) and (2) we obtain
4 3 20 1 2 3 4a x a x a x a x a+ + + + ≡ 0( ) ( ) ( ) ( )a x x x x ax− α − β − γ − δ ...(3)
Substituting x = α, β, γ, δ successively on the two dies of (3), we see that each of the values
α, β, γ, δ satisfies the equation 4 3 20 1 2 3 4a x a x a x a x a+ + + + = 0.
Again, if substitute for x, a value different from each of α, β, γ, δ on both sides of (3), we seethat since R.H.S. is not zero, L.H.S. is also not zero. Thus no value other than α, β, γ, δ is a rootof the equation.
We have thus shown that a biquadratic equation has four and four roots.The roots α, β, γ, δ may not be all different and we have the following different possibilities
(i) α = β ≠ γ ≠ δ Two roots equal.(ii) α = β ≠ γ ≠ δ Roots equal in pairs.
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(iii) α = β = γ ≠ δ Three roots equal.(iv) α = β = γ = δ All the four roots equal.
(v) α ≠ β ≠ γ ≠ δ No two roots equal.
The factors of a 4 3 20 1 2 3 4a x a x a x a x a+ + + + are
(a) (x – α)2 (x – γ) (x – δ) for the case (i)(b) (x – α)2 (x – γ)2 for the case (ii)
(c) (x – α)3 (x – δ) for the case (iii)(d) (x – α)4 for the case (iv)
(e) (x – α) (x – β) (x – γ) (x – δ) for the case (v)
5.8 Relation between the Roots and Coefficients of a Biquadratic Equation
4 3 20 1 2 3 4a x a x a x a x a+ + + + = 0
We have
4 3 20 1 2 3 4a x a x a x a x a+ + + + ≡ 0( ) ( ) ( ) ( )a x x x x ax− α − β − γ − δ
≡ 4 30[ ( )a x x− α + β + γ + δ
2( )x+ α β + α γ + α δ + β γ + γ δ + β δ
( ) ]x− αβγ + αβδ+αγδ+βγδ +α+βγδ
Equating the coefficients of x3, x2, x and the constant term, we have
a1 = 0– ( )a α + β + γ + δ
a2 = 0( )a α β + α γ + α δ + β γ + β δ + γ δ
a3 = 0– ( )a αβγ+αβδ+αγδ+βγδ
a4 = 0a ⋅αβγδ
These give
(α + β + γ + δ) =3
14
0
co-efficient of co-efficient of
a xa x
− = −
)( α β + α γ + α δ + β γ + β δ + γ δ =2
24
0
co-efficient of co-efficient
a xa x
=
⇒ (α + β) (γ + δ) + αβ + γδ = 2
0
aa
(αβγ+αβδ+αγδ+βγδ = 34
0
co-efficient of co-efficient
a xa x
− = −
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⇒ ( ) ( )αβ γ + δ +γδ α + β = 3
0
aa
−
αβγδ = 44
0
constanttermco-efficient of
aa x
=
Any function of α, β, γ, δ which does not change its value of inter-changing any two of themis a symmetric function.
It may be easily seen thatα + β + γ + δ (≡ Σα)
αβ + αγ + αδ + βγ + βδ + γδ (≡Σαβ)αβγ + αβδ + αγδ + βγδ (≡ Σαβγ)
α β γ δare four symmetric functions known as basic symmetric functions of the roots of the biquadratic.
Every symmetric function of α, β, γ, δ can be expressed in terms of the above symmetric functions.
EXERCISE
1. If α, β, γ, are the roots of the equation
x3 – 5x2 + x + 12 = 0Calculate the value of Σα2β.
2. If α, β, γ, be the roots of the equation x3 + qx + r = 0, prove that
2 53( ) ( )Σα Σα = 35( ) ( ).4Σα Σα
[Hint : Since, α, β, γ, are the roots of the given cubic equation]α3 + qα + r = 0
β3 + qβ + r = 0γ3 + qγ + r = 0
To find Σα4 multiply equations (1) α, β, γ, respectively and add.To find Σα5 multiply (1) α2, β2, γ3, respectively and add.
3. Find the sum of the cubes of the roots of the equation
x3 – 6x2 + 11x – 6 = 04. If α, β, γ, be the roots of the equation
x3 + 5x2 – 6x + 3 = 0
Find the values of 3−Σα
3 33
3 3 3 3 3 31 1 1− Σβ γΣα ≡ + + = α β γ α β γ
Hint:
Again 3 3 2 2 23( ) [ ( ) ]Σα β − αβγ =Σαβ Σα β − Σα αβγ
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5. If α, β, γ, be the roots of the cubic x3 + qx + r = 0 find the values of
(i)1
b cΣ
+
(ii)2 2b cb c
+Σ
+
(iii) Σ(b + c)2
6. If α, β, γ, are the roots of the equation
x3 – px2 + qx – r = 0.Find the value of
(i)2
1Σα
(ii) Σα2β
(iii) ( ) ( ) ( )α + β β + γ γ + α
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LESSON 6
THEORY OF EQUATIONS II
In this lesson, we shall discuss the methods of transforming a given equation into another equationor to solve the given equation under certain given conditions.
6.1 Transformation of Equations
We shall discuss methods of finding an equation whose roots are related to those of a given equation.The example below, will indicate and illustrate and methods which may be employed for the purpose.
We know that the cubic equation, with roots α, β and γ is given by
(x – α) (x – β) (x – γ) = 0i.e., x3 – x2 (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0i.e., x3 – x2Σα + x Σαβ – αβγ = 0
Also the biquadratic equation with roots, α, β, γ, δ is(x – α) (x – β) (x – γ) (x – δ) = 0
i.e., 4 3 2x x x x− Σα+ Σαβ− Σαβγ+αβγδ = 0
ExamplesExample 1. Find the equation whose roots are the negatives of the roots of the equation
5x3 – 3x2 + 7x + 2 = 0 ...(1)
Solution. If α, β, γ, are the roots of the given equation, then the roots of the new equation are–α, –β, –γ.
We write y = –x so that when x takes values, α, β, γ then y takes the values –α, –β, –γ respectively,
Write y = –x ⇒ x = – y ...(2)
Eliminating x between (1) and (2)
We get, 5(–y)3 – 3 (–y)2 + 7(–y) + 2 = 0
or –5y3 – 3y2 – 7y + 2 = 0If 5y3 + 3y2 + 7y – 2 = 0 is the required equation.
Example 2. Find the equation whose roots are the roots ofx4 – 5x3 + 7x2 – 17x + 11 = 0 ...(1)
each diminished by 5.
Solution. If α, β, γ, δ are the roots of the given equation we have to find the equation whoseroots are α – 4, β – 4, γ – 4, and δ – 4.
Write y = x – 4 ⇒ x = y + 4Eliminating x between (1) and (2) we get
(y + 4)4 – 5 (y + 4)3 + 7 (y + 4)2 – 17 (y + 4) + 11 = 0
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i.e., y4 + 16y3 + 96y2 + 256y + 256 – 5(y3 + 12y2 + 48y + 64)+7(y2 + 8y + 16) – 17 (y + 4) + 11 = 0
Simplyfying we get,y4 + 11y3 + 43y2 + 55y – 9 = 0
as the required equation.Example 3. Find an equation whose roots are three times those of the equation
2x3 – 5x2 + 7 = 0
Solution. If α, β, γ are the roots of given equation , then we write equation whose roots are 3α,3β, 3γ.
We write y = 3x
⇒ x = .3y
Substituting x =3y
in the given equation, we gets
3 2
2 5 73 3y y − +
= 0
⇒ 2y3 – 15y2 + 189 = 0which is the required equation.Example 4. Find the equation whose roots are the squares of the roots of the equation.
x3 + qx + r = 0Solution. We shall give two method for the solution.
First Method:Let α, β, γ, be the roots of the given equation so that we require the equation with roots α2, β2.
γ2 we writey = x2 ...(1)
so that when x takes up the vales α, β, γ, then y takes up the values α2, β2, γ2.
Aslo if x3 +qx + r = 0 ...(2)then x necessarily takes up the values, α, β, γ and these only.
Thus the required equations will be obtained by eliminating x between (1) and (2) with y indicatingthe unknown.
Thus we have
y y q y r+ + = 0
or ( 9)y y + = – r
Squaring, we gety (y + q)2 = r2
i.e., y3 + 2y2 + q + yq2 + r2 = 0which is the required equation.
65
Second Method:Indicating the unknown by y, the equation with roots α2, β2γ2, is
(y – α2) (y – β)2 (y – γ)2 = 0
i.e., y3 – y2 (α2 + β2 + γ2) + y (α2β2 + β2γ2 + γ2α2) – α2β2γ2 = 0 ...(3)
We have now to compute the three symmetric functions.
viz., (α2 + β2 + γ2), (α2β2 + β2γ2 + ψ2α2) and α2β2γ2
From the given equation, we have
(α + β + γ) =0, (αβ + βγ + γα) = θ, αβγ = – r
Hence (α2 + β2 + γ2) = (α + β + γ)2 – 2(αβ + βψ + γα) = – 2q
and (α2β2 + β2γ2 + β2α2) =[(αβ + βγ + γα)2 – 2αβγ (α + β + γ)] = q2.
Also α2β2γ2 = r2
Making substitutions in (3), we obtain the required equation.
Example 5. If α, β, γ are the roots of the cubics x3 + px2 + qx + r = 0, find the equation whose
roots are 1 1 1
, and where , , 0.α − β − γ − α β γ ≠βγ γα αβ
Solution.1α −
βγ =
1 1 1 1and
r rr r
αβγ− αβγ− − − + α = α = α βγ αβγ − [Q αβγ = –r]
Similarly 1β −γα
=1r
r+ β
1α −αβ
= 1r
r+ γ
We write 1r
y xr+ =
so that as x takes α, β, γ then y takes up the values
1α −βγ
, 1β −γα
,
1γ −αβ
respectively. Eliminating x between the given equation and 1r
y xr+ =
we get3 2
1 1 1ry ry ry
p q rr r r
+ + + + + + = 0
i.e., 3 3 2 2 2 3( 1) ( 1) ( 1) 0r y pr r y qry r r r+ + + + + + =
2 3 2 2 3( 1) ( 1) ( 1) 0r y pr r y q r y r+ + + + + + =
Example 6. If α, β, γ are the roots of x3 – px2 + r = 0, form the equation whose roots are
, ,and .α + β β + γ γ + α
γ α β
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Solution. α, β, γ are the roots ofx3 – px2 + r = 0
∴ α + β + γ = p, αβ + βγ + γα = 0, αβγ = – r ...(1)
We haveα + β
γ =( ) pα + β + γ − γ − γ=
γ γ
Similarlyβ + γ
α = ,
p p− α γ + α − β=α β β
We write y =p x
x−
...(2)
so that when x takes the values α, β, γ; y takes the values , andp p p− α − β − γ
α β γ respectively..
The required equation is obtained by eliminating x between (1) and (2).
From (2) xy =p – x ⇒ x(y + 1) = p ⇒1
px
y=
+
∴ Putting the value of x in (1) we have
2
3 2(1 ) (1 )p p
p ry y
3− +
+ + = 0
or p3 – p3 (1 + y) + r (1 + y)3 = 0
i.e., ry3 + 3ry2 + (3r – p3) + y + r = 0
is the required equation.
Example 7. Find the equation whose roots are less by 2 less than the roots of the equation
x4 + 2x3 – 3x2 – 2x + 2 = 0 ...(i)
Solution. If a, b, c, d be the roots of the given equation
We require the equation with roots a – 2, b – 2, c – 2, d – 2,
We write y = x – 2 i.e., x = y + 2
and eliminate x between (ii) and the given equation (i).
The eliminant is
(y + 2)4 + 2(y + 2)3 – 3(y + 2)2 – 2(y + 2) + 2 = 0
i.e., (y4 + 8y3 + 24y2 + 32y + 16) + 2(y3 + 6y2 + 12y + 8)
– 3 (y2 + 4y + 4) – 2 (y + 2) + 2 = 0
i.e., y4 + 10y3 + 33y2 + 42y + 18 = 0
Example 8. Transform the equationax3 + by2 + cx + d = 0 ...(i)
into one, from which the second term is missing.
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Solution. Let α, β, γ be the roots of the given equation.We shall find another equation whose roots are
α + h, β + h, γ + hwhere h is a number to be so chosen that the new equation is free of the second term
We write y = x + h ⇒ x = y – h ...(ii)
so that the new equation will be obtained on eliminating x between (i) and (iii)
The eliminant is
a(y – h)3 + β(y – h)2 + c (y – h) + d = 0
i.e., ay3 + y2 (b – 3ah) + y(3ah2 – 2bh + c) + (–ah3 + bh2 – ch + d) = 0 ...(iii)
We choose h so that β – 3ah = 0, i.e., b/3a
This choice of h will make the coefficient of y2 zero, i.e., the second term missing.
Substituting the value of h (iii), we obtain
2 2 3 33
2 3 22
3 03 39 27 9
b b ab b bcay a c y d
a aa a a
−+ − + + + − + =
i.e.,2 2 3
32
3 27 9 20
3 27ac b a d abc b
ay ya a
− − ++ + =
i.e., 3 3 2 2 327 9 (3 ) (27 9 2 ) 0y a ac b y a d abc b+ − + − + =
which is the required equation.
EXERCISE
1. From an equation whose roots are negatives of the roots of the equationx3 + 6x2 + 8x – 9 = 0.
2. Form an equation whose roots are twice the roots of2x4 + 3x2 + 7x – 9 = 0.
3. Find the equation whose roots are the squares of the roots of the equation
x3 – x2 – 8x + 6 = 0.4. Find the equation whose roots are the reciprocals of the roots of the equation
ax3 + bx2 + cx + d = 05. If a, b, c are the roots of x3 + qx + r = 0, for the equation whose roots are
b2c2, c2a2, a2b2.
6. Find the equation whose roots are the roots of the equationx4 – 3x3 + 7x2 + 5x – 2 = 0
each diminished by 2.7. Transform the equation
x3 – 6x2 + 4x – 7 = 0
into one, from which the second term is missing.
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6.2 Roots Related by Some Conditions
Sometimes we are required to find conditions under which the roots of given equation may berelated in a given manner. We are also sometimes required to solve an equation it is given that theroots of the equation are related in some given manner. The examples given below will illustrate theprocedure to be adopted in solving the two types of problems referred to here.
Example 1. Solve the equation
x3 – 12x2 + 39x – 28 = 0
whose roots are given to be in arithmetical progression.
Solution. We put down a general set of three number in A.P. We could take this set simply as
a, a + d, a + 2d
But it would, however, be found more convenient to take this set as
a – d, a, a + d.
Let us suppose the roots of the given equation are
a – d, a, a + d.
So that we have,
(a – d) + a + (a + d) = sum of the roots = 12.
(a – d) a + a (a + d) + (a – d) (a + d) = sum of the product of the roots takes in pairs = 39.
(a – d) × a × (a + d) = product of roots = 28
Thus we
3a = 12 ...(i)
3a2 – d2 = 39 ...(ii)
a(a2 – d2) = 28 ...(iii)
From (i), we get a = 4
Substituting a = 4 in (ii), we get
d2 = 9; i.e., d = ±3
We may easily see that a = 4, d = 3 as also a = 4, d = –3 satisfy (iii).
Taking a = 4, d = + 3 and putting these values in a – d, d, a + d, we see that the required rootsare 1, 4, 7.
Again taking a = 4, d = – 3, we shall see that the roots are 7, 4, 1 which are the same as thoseobtained above except for the order.
Thus we have proved that the roots of the given cubic are 1, 4, 7.
Example 2. Solve
8x3 – 14x2 + 7x – 1 = 0
give that its roots are in geometrical progression.
Solution. We suppose that its roots are a/r, a, ar.
We havea
a arr
+ + = sum of the roots 14 78 4
=
69
( ) ( )a a
a a ar arr r
+ + +
= sum of the products of the roots taken in pairs = 78
( )a
a arr
= 1product of roots=
8
Thus we have
(1 )a r rr
2+ + =
74
...(i)
2(1 )a r rr
2+ + =
78
...(ii)
and a3 =18
...(iii)
Dividing (i) by (ii), we get
1a
= 2 so that a = 12
Substituting this value of a in (i), we get
21 r rr
+ + =
72
or 22 5 2r r− + = 0
or (2r – 1) (r – 2) = 0
The value 12
of a satisfied (iii).
Taking a = 12
, r = 2, we see that the roots 1 1, , are , ,1.
4 2a
a arr
Taking a = 1 1
,2 2
r = we obtain the same roots but in the reverse order..
Thus 1 1
1, , ,2 4
are the required roots.
Examples 3. Solve the equation 3 23 11 12 4 0,x x x+ + + = the roots being in H.P..
(Harmonoic Progression)
Solution. Let the roots of the given equation be α, β, γ then
α, β, γ =113
−
αβ + βγ + ψα = 4
70
αβγ =4
.3
−
Since α, β, γ are in H.P.
∴2β =
1 1+α γ
2αγα + γ
= β
⇒ β(α + γ) = 2αγ⇒ αβ + βγ + γα = 3αγFrom (ii) and (iv), we get
3αγ = 4
⇒ αγ =43
Substituting αγ =43
in (iii), we get
43
β =43
−
⇒ β = –1.Thus –1 is the roots of the given equation and (x + 1) is the factor of
3x3 + 11x2 + 12x + 4.
The given equation may be written as
3x2 (x + 1) + 8x(x + 1) + 4(x + 1) = 0
⇒ (x + 1) (3x2 + 8x + 4) = 0
Now the other two roots of the given equation are obtained by the solving the equation
3x2 + 8x + 4 = 0
⇒ x =8 64 48 – 8 4 2
, 26 6 3
− ± − ± −= = −
Hence the required roots are –1, – 2, –2/3.Note : If α, β, γ, δ are the roots of biquadratic equation
30 1 2 3 4 0a x a x a x a x a4 2+ + + + =
Then α + β + γ + δ = 1
0,
aa
−
(α + β) (γ + δ) + αβ + γδ =2
0,
aa
+
71
αβ (α + δ) + γδ(α + β) = 3 4
0 0and .
a aa a
− αβγδ=
Example 4. Solve the equation
4 3 28 14 8 15x x x x− + + − = 0
the roots being in A.P.Solution. Let the required roots be a – d, a, a + d, a + 2d.
Then (a – d) + a + (a + d) + (a + 2d) = 8 ⇒ 4a + 2d = 8 ⇒ d = 4 – 2a.
(a – d + a) (a + d + a + 2d) + a(a – d) + (a + d) (a + 2d) = 14
Substituting d = 4 – 2a in this equation, we get
(2a – 4 + 2a) (2a + 12 – 6a ) + a (a – 4 + 2a) + (a + 4 – 2a ) (a + 8 – 4a) = 14
⇒ (4a – 4) (12 – 4a) + a(3a – 4) + (4 – a) (8 – 3a) = 14
⇒ 48a – 48 – 16a2 + 16a + 3a2 – 4a + 32 – 8a + 12a + 3a2 = 14
⇒ – 10a2 + 40a – 30 = 0
⇒ a2 – 4a + 3 = 0
⇒ (a – 1) (a – 3) = 0 ⇒ a = 1, 3
Now d = 4 – 2a = 2 when a = 1
And d = 4 – 6 = – 2 when a = 3.
∴ a = 1, d = 2 and a = 3, d = –2.
Now when a = 1, d = 2 ; Roots are –1, 1, 3, 5.
And when a = 3, d = – 2 ; Roots are 5, 3, 1, –1. This is the same roots but in different order.
∴ Required roots are –1, 1, 3, 5.
Example 4. Solve the equation
x4 + 15x3 + 70x2 + 120x + 64 =0
the roots being in G.P.
Solution. We know that if four number be in G.P. then the product of the first and fourth is equalto the product of the second and third. Thus, if the roots are in G.P. then the product of the two ofthe roots must be equal to the product of the other two.
Let the roots be α, β, γ, δ then
αβ = γδ ...(i)
α + β + γ + δ = –15 ...(ii)
(α + β) (γ + δ) + αβ + γδ = 70 ...(iii)
αβ(γ + δ) + γδ(α + β) = – 120 ...(iv)
αβγδ = 64 ...(iv)
From (i) and (v), we get
(αβ)2 = 64 ⇒ αβ = 8
αβ = 8 = γδ
72
Substituting these values in (iii), we get
(α + β) (γ + δ) = 70 – 16
= 54 ...(vi)
From (ii) (α + β) (γ + δ) = –15 ...(vii)
From (vi) and (vii), we find that (α + β) and (γ + δ) are the roots of the equation
t2 + 15t + 54 = 0
⇒ t2 + 9t + 6t + 54 = 0
⇒ (t + 9) (t + 6) = 0
⇒ t = – 6, –9
Thus (α + β) = – 6 and γ + δ = – 9.
Also, αβ = 8 and γδ = 8
Thus α, β are the roots of equationy2 + 6y + 8 = 0
and γ, δ are the roots of equation
y2 + 9y + 8 = 0 ...(viii)Solving (viii), we get
y = – 2, – 4 ...(ix)
and solving (ix), we gety = –1, – 8
Thus roots of given equations are –1, – 2, – 4, – 8.
Example 5. The equation
x4 – 2x3 + 4x2 + 6x – 21 = 0
has two real roots whose sum is zero. Solve the equation.
Solution. Let the roots be α, β, γ, δ, where α + β = 0
We have, α + β + γ + δ = sum of the roots = 2
(αβ + αγ + αδ + βγ + βδ + γδ) = sum of the products of the roots taken in pairs = 4.
(αβδ + αβγ + αγδ + βγδ) = sum of the product of the roots taken in traids = –6.
αβγδ = product of the roots = –21.
We now make use of the fact that α + β = 0
Now αβ + αγ + αδ + βγ + βδ + γδ = αβ = α(γ + δ) + β (γ + δ) + γδ
= (αβ + γδ) + (α + β) (γ + δ)
= (αβ + γδ) + [α + β =0]
Also αβγ + αβδ + αγδ + βγδ = αβ(γ + δ) + γδ (α + β)
= αβ (ψ + δ), [α + β = 0]
Thus we haveγ + δ = 2 ...(i)
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αβ + γδ = 4 ...(ii)
αβ(γ + δ) = – 6 ...(iii)
αβγδ = – 21 ...(iv)
From (i) and (iii), αβ = – 3
Also α + β = 0 , i.e., β = –α
α2 = 3 or α = 3±
α2 = 3 , we have β = 3−
Putting αβ = –3 in (iii) and (iv), we obtain
γ + δ = 2
γδ = 7
These give, on elimination δ,γ (2 – γ) = 7
i.e., γ2 – 2γ + 7 = 0
∴ γ =2 4 28
1 62
i± −
= ±
It will be seen that other choices of the values α and γ give us the same of roots.
The roots are therefore, 3, 3,1 6 ,1 6 .i i− + −
Example 6. Solve the equation 4 3 16 4 48 0x x x x2+ − − + =
Give that the product of the two roots of the equation is 6.
Solution. Let α, β, γ, δ be the roots of the given equation.
Then α + β + γ + δ = – 1, ...(1)
(α + β) + (γ + δ) + αβ + γδ = – 16 ...(2)
αβ (γ + δ) + γδ (α + β) = 4 ...(3)
αβγδ = 48 ...(4)
It is given that αβ = 6. Then by (4), γδ = 8.
By (2), (α + β) (γ + δ) + 6 + 8 = – 16
⇒ (α + β) (γ + δ) = – 30 ...(5)
From (1) and (5) if follows that α + β and γ + δ are the roots of
y2 + y – 30 = 0
⇒ (y + 6) (y – 5) = 0
γ = – 6, 5
∴ α + β = 5,
γ + δ = – 6.
(α – β)2 = (α + β)2 – 4αβ = 25 – 24 = 1
74
⇒ α – β = ±1Also (γ – δ)2 = (γ + δ)2 – 4γδ = 36 – 4 × 8 = 4
⇒ γ – δ = ± 2Let us take α – β = 1 and γ – δ = 2
Also α + β = 5 and γ + δ = – 6∴ 2α = 6, 2γ = –4 ⇒ α = 3, γ = – 2and β = α – 1= 3 – 1= 2, δ = γ – 2 = – 4
∴ Required roots are 3, 2, –2, – 4.Example 7. Solve the equation
4 35 10 4 0x x x2− + + =given that the product of two of the roots of the equation is equal to the product of the other
two.
Solution. Let α, β, γ, δ the roots of the given equation,Then αβ = γδ ...(1)
(α + β) + (γ + δ) = 5 ...(2)
(α + β) + (γ + δ) + αβ + γδ = 10 ...(3)(αβ) (γ + δ) + γδ (α + β) = 0 ...(4)
αβγδ = 4 ...(5)Substituting αβ = γδ in (5), we get
α2β2 = 4
⇒ αβ = 2 = γδSubstituting in (3), we get
(α + β) + (γ + δ) + 2 + 2 = 10⇒ (α + β) (γ + δ) = 6Also (α + β) (γ + δ) = 5 From (2)
∴ α + β and γ + δ are the roots of equation
y2 – 5y + 6 = 0⇒ (y – 3) (y – 2) = 0
⇒ y = 3, 2
Then a + β = 3 and y + δ = 2∴ (α – β)2 = (α + β)2 – 4αβ
= 9 – 8 = 1⇒ α – β = ±1Also (γ – δ)2 = (γ + δ)2 – 4γδ
= 4 – 8 = – 4 = 4i2
⇒ ψ – δ = ± 2iLet α – β = 1 and γ – δ = 2iAlso α + β = 3 and γ + δ = 2∴ α = 2, β = 1, γ = 1 + i, δ = 1 – iRequired roots are 2, 1, 1 ± i.
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Examples 8. Find the condition that the roots of the given equationpx2 + qx2 + rx + s = 0
may be in arithmetical progression.
Solution. Let a – d, a, a + d be the roots of the given equation.
We have (a – d) + a + (a + d) =qp
i.e., 3a =qp
− ...(i)
(a – d) a + a (a + d) + (a + d) (a – d) =rp
i.e., 3a2 – d2 =rp
...(ii)
a (a + d) (a – d) =sp
−
a(a2 – d2) =sp
− ...(iii)
From (i) and (ii), d2 = 23r
ap
−
=2 2
2 23
3 3q r q pr
pp p−
− =
Substituting the values of a and d2 in (iii), we obtain
2 2
2 23
3 9 3q q q prp p p
−− − =
sp
−
⇒ 2 23 ( 3 9 )
27q
q q prp
− + =sp
−
⇒ q(9pr – 2q2) = 27p2swhich is the condition to be necessarily satisfied if the roots of the given equation are in A.P.
Note : You may verify that the equation in Ex. 1 does satisfy this condition.
Example 9. Find the condition for the equation 0x px q3 + + = to have a pair of equal roots.
Solution. Let the roots be α, α, β.We have 2α + β = 0 ...(i)
α2 + 2αβ = 0 ...(ii)
α2β = – q ...(iii)
76
Now (i) and (ii) give
α2 + 2α(–2α) = p i.e., α2 = 3p−
β = –2 23p−α −
Substituting in (iii), the values of α2 and β we have
23 3p p −− −
= – q
Squaring we obtain
427
p− = q2
i.e., 3 24 27p q+ = 0
which is the required condition.
EXERCISES
1. Solve the equation
3 26 3 10x x x− + + = 0
whose roots are in A.P.2. Solve the equation
3 23 26 52 24x x x− + − = 0 the roots being in G.P.
3. Solve the equation
3 228 39 12 1x x x− + − = 0
the roots being in Harmonical Progression.
4. Solve the equation
3 24 16 9 36x x x− − − = 0
where the sum of the two of the roots being equal to zero.5. Solve the equation
3 22 22 24x x x− − − = 0
two of its roots being in the ratio 3 : 4.6. Solve the equation
3 22 7 6x x x+ − − = 0
given that the difference of two roots is 3.
77
7. Solve the equation
3 25 2 24x x x− − + = 0
given that the product of two of its roots is 12.8. Solve the equation
4 3 23 40 130 20 27x x x x− + − + = 0
given that product of two of its roots is equal to the product of the other two.
9. Prove that the necessary condition for the roots of the equation
3 20 1 2 32 3 3a x a x a x a+ + + = 0
to be in G.P. is 20 3a a = 3
1 3a a
10. Prove that the necessary condition for the root of the equation
3 2x px qx r− + − = 0
to be in H.P. is 3 2 227 9 2r r pq rq− + = 0
11. Prove that the necessary condition for the sum of two roots of the equation
3 2x px qx rx s4 − + − + = 0
to be equal to the sum of the other two is
3 4 8p pq r− + = 0
12. Solve the equation
22 15 35 30 8x x x4 2− + − + = 0
given that the product of two roots is equal to the product of the other two.13. Solve the equation
4 3 28 21 20 5x x x x− + − + = 0
given that the sum of two roots is equal to the sum of the other two.
14. Solve the equation
4 22 15 35 30 8x x x x3− + − + = 0
given that the product of the two roots is equal to the product of the other two.
6.3 NOW WE DISCUSS CERTAIN IMPORTANT THEOREMS REGARDING THE COMPLEXROOTS AND SURD ROOTS OF AN EQUATION
Theorem 1. In an equation with real coefficient complex roots occur in conjugate pairs.Proof. Let f(x) = 0 ... (1)
be an equation, with real co-efficients. Let a + ib be a root of (1). Then we have to prove that a –ib is also a root of (1).
Since a + ib is a root of (1), therefore, the factor of f(x) corresponding to the root a + ib is[x – (a + ib)]
78
Now consider the product[x – (a + ib)] [x – (a – ib)]
or [(x – a) – ib] [(x – a) + ib]or [(x – a)2 + b2]
We divide f(x) by the second degree polynomial 2 2[( ) ].x a b− + The remainder will be a polynomialof degree one. Also since the dividend f(x) and the divisor (x – a)2 + b2 are polynomials with realcoefficient the quotient and remainder will also be polynomials with real coefficient. Let the quotientbe deonted by f(x) and the remainder by rx + rx′ where r and r′ are real numbers. We have numbers,thus
f(x) = 2 2[( ) ] ( ) ( )x a b x rx r′− + φ + + ...(i)
We put x = a + ib on the two dies and obtain
0 = 0 + (a + ib) r + r′or 0 = (ra + r′) + irbEquating the real and imaginary party to zero, we obtain
ra + r′ = 0 ...(ii)rb = 0 ...(iii)
Now b ≠ 0 for if b = 0, a + ib would be real so we can suppose that b ≠ 0.Therefore from (iii), r = 0.Finally from (ii), r′ = 0
Substituting r = 0, r′ = 0 in (i), we obtain
f(x) ≅ 2 2[( ) ] ( )x a b x− + φ
≅ [ ( )][ ( ) ( )]x a ib x a ib x− + − − φ
Putting x = a – ib, on the two sides, we see that the R.H.S is zero and hence the L.H.S. is zero.Thus a – ib is a root of f(x) = 0Hence the result.
Theorem 2. In an equation with rational coefficients, surd roots occur in pairs, i.e., if ,p q+ where
p and q are rational number and q is not the square of a rataional number, is a root of an equation
with rational coefficients then p – q is also a root of the same equation f(x) = 0 with rational numbers.
Proof. We have [ ( )][ ( )]x p q x p q− − −
= [( ) ] ][( ) )]x p q x p q− − − +
= [(x – p)2 – q]
We divide f(x) by [(x – p)2 – q]. Since the coefficients in the divident f(x) and the divisor[(x – p]2 – q)] are all the rational, the co-efficients in the quotient and the remainder will also be ratational.Moreover the remainder will be of the first degree.
We write
f(x) = [(x – p)2 – q] φ(x) + rx + r′. ...(i)
79
where r and r′ are rational numbers.
Putting x = p q+ on the two dies of (i), we obtain
0 = ( )r p q r′+ +
= ( )rp r r q′+ +
Equating the surd part and the non-surd part secondary to zero, we have
r q = 0 and rp + r′ = 0
From these, we deduce r = 0, r′ = 0 (Q q ≠ 0)Thus f(x) ≅ [(x – p)2 – q] φ(x)
≅ [ ( ) ( ( ) ( )]x p q x p q x− + − − φ
so that ( )x p q− − is a factor of f(x) and as such ( )p q− is a root of f(x) = 0.
Hence the result.
Example 1. From an integer with lowest degree whose roots are 2 , 3 4 .i+
Solution. Since in an equation with real co-efficients surd roots occur in pairs and complete rootsoccur in conjugate pairs.
∴ Roots of required equation are 2 , 3 4 .i± ±
Hence required equation is
( 2 ) ( 2 ) [ (3 4 ) [ (3 4 )x x x i x i− + − + − − = 0
⇒ (x2 – 2) [(x – 3)2] – (4i)2 = 0⇒ (x2 – 2) (x2 – 6x + 9 + 16) = 0
⇒ (x2 – 2) (x2 – 6x + 25) = 0⇒ x4 – 2x2 – 6x3 + 12x + 25x2 – 50 = 0⇒ x4 – 6x3 + 23x2 + 12x – 50 = 0
Example 2. Solve the equation
4 3 23 10 4 6 0x x x x− + − − =
some root being 1 3
2+ −
Solution. The coefficients of the given equation are all real.
Now 1 3 1 3
2 2 2i
+ − = + is a root and accordingly the conjugate complex of this root viz. 1 32 2
i−
is also a root.
Now1 3 1 32 2 2 2
i ix x
− + − −
=1 3 1 32 2 2 2
i ix x
− − − +
80
=2 2
21 3 1 3( 1)
2 2 2 4i
x x x x2
− − = − + = − +
Thus (x2 – x + 1) is a factor of3x4 + 10x3 + 4x2 – x – 6
Dividing (i) by (x2 – x + 1), we may see that
3x4 + 10x3 + 4x2 – x – 6 ≅ (x2 – x + 1) (3x2 – 7x – 6)
The roots of 3x2 – 7x – 6 = 0 are 7 49 72 7 11
, . .,3,6 6 3
i e± + ± 2= −
Hence the required roots are 2 1 33, , .
3 2 2i
− ±
81
B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 1.1 Part A (Section III)
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School of Open LearningUniversity of Delhi.
5, Cavalry Lanes,Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. Simplify the following:
(i) (2 – 5i) + (–3 + 4i) + (8 + 3i)
(ii)5
3 2ii
+−
2. Evaluate the following:(i) (–7 –2i) (–1 – 5i)
(ii)2
2(2 4 )(3 )
ii
++
3. Solve the following equations:(i) [(x + 2y), (2x – y – 6)] = (3, 2)
(ii)2 3
( ) ( )1
ix y i x y
i− +− + + =
+4. Express the following complex numbers in the polar form.
(i) –1 3i+
(ii) –2 12i−
5. Find the trigonometric representation of
(i) sin α – i cos α.(ii) 1 + cos α + i sin α.
6. Show that | cos θ + i sin θ | = 1.
7. Express the complex number 52
ii
++ in the form x + iy where and x and y are real.
Find its modulus and amplitude.
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 2.1 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise
1. Simplify 8 9
9 8(cos sin ) (cos2 sin2 )(cos2 sin ) (cos sin )
i ii i i
θ + θ θ − θθ + θ θ − θ
2.
10 10
6
cos sin cos sin15 15 15 15
cos sin15 15
i i
i
π π π π + + −
π π +
3. Use De Moivre’s theorem to simplify 4
5(cos sin )(sin cos )
ii
θ + θθ + θ
4. Prove that (sin cos ) cos sin2 2
nx i x n x n xπ π − = − − −
5. Obtain the value of (3 + 4i)3 with the help of De Moivre’s theorem.
6. In n is a positive integer, show that 1( 3 1) ( 3 ) 2 cos .6
n n n ni + π+ + − =
7. Prove that
[(cos cos ) (sin sin )] [(cos cos ) (sin sin )]n ni iθ + φ + θ + φ + θ + φ − θ + φ
= 1 ( )
2 cos cos2 2
n n n q− θ − φ + φ
[ (cos cos ) (sin sin )]iθ + φ + θ + θHint:
= 2 cos cos 2sin cos2 2 2 2
iθ + φ θ − φ θ + φ θ − φ +
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 2.2 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise
1. If 1 1
2 cos ,2cos show thatx yx y
θ = + φ = +
1m nm nx y
x y+ = 2 cos (mθ + nθ)
andm n
n mx yy x
+ = 2 cos (mθ – nθ),
where m and n are integer. What happens if m and n are rational numbers?
2. If x = cos α + i sin α, y = cos β + i sin β and z = cos γ + i sin γ, such that x + y + z =
0 then prove that 1 1 1
0x y z
+ + =
and1
xyzxyz
+ = 2cos (pa + qb + rc)
3. Let the complex numbers x, y, z be given respectively by cos a + i sin a, cos b + i sin b andcos c + i sin c, then prove that for any integers p, q, r
1p q rp q rx y z
x y z+ = 2 cos (pq + qb + rc)
4. 1 1 2 2Let ; ,...... n nx iy x iy x iy+ + + be any n complex number and A + Bi be some other complexnumber, such that
1 1 2 2( ) ( )......( )n nx iy x iy x iy A iB+ + + = +
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then show that
(i) 1 1 1 11 2
1 2tan tan tan tann
n
y y y Bx x x A
− − − −+ + + =KK
(iii) 2 2 2 2 2 2 2 21 1 2 2( ) ( ) ( )n nx y x y x y A B+ + + + = +KK
5. Expand (1 + i)n in two different ways and find the sum of the two series.
2 4 6 1 3 5(1– )and( )n n n n n nC C C C C C+ − + − +KK KK )
both the series being finite.[Hint : First expand (1 + i)n by Binomial theorem and then expand it by the help of De Moivre’sTheorem.]
85
B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 3.1 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. Find the values of
(i)13(1 )i+
(ii)14(–1 3)i+
(iii)16( 1)−
(iv) 2/3(1 )i+
2. Solve the equations (i) z7 = 1 (ii) z7 + z = 0.3. Find the nth roots of unity and show that they form a series in G.P. whose sum is zero. Also
prove that the sum of their pth power always vanishes unless p be a multiple of n, p beingan integer and that then the sum is n.
4. Solve the equation 7 1z z z4 3+ + + = 0.
[Hint : 7 4 3 41 ( 1) ( 1) 0]z z z z z3+ + + = + − + =
5. Use De Moivre’s theorem to solve
10 511 1z z+ + =0
[Hint : Put z5 = y]
6. Solve the equation
4 3 2 1z z z z− + − − =0
[Hint : Mutiply the equation by (z – 1)]7. Solve the equation z12 – 1 = 0 and find which of its roots satisfy the equation
4 1z z2+ + =0.
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 3.2 Part A (Section III)
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Response sheet no :
School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. Express sin 5θ in terms of sin θ only.
2. Express cos 6θ in terms of cos θ only.3. Express tan 7θ in terms of tan θ only.4. Find the value of sin 1º approximately.
5. Find the value of θ, where sin 1013
.1014
θ =θ
6. Show that sin 5θ = 16 sin 5θ – 5 sin 3θ + 10 sin θ.7. Express sin7 θ cos2 θ in terms of sines of multiples of θ.8. Express cos9 θ in terms of cosines of multiples of θ.
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 4 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. cos θ + cos 2θ + cos 3θ + ......
2. sin θ · cos 2θ + cos 2θ · sin 3θ · cos 4θ + ......3. sin α + sin (α + β) + sin (α + 2β) + ......4. 2 cos 2θ + 22 cos 4θ + 23 cos 6θ + ......
Find the sum to infinity of the following series.
5.1 1 1.3
1 cos2 cos4 cos6 ......2 2.4 2.4.6
+ θ − θ + θ +
6.1 1
sin sin2 sin3 .....2 3
θ − θ + θ +
7. sin α · sin α + sin2 α · sin 2α + sin3 α · sin 3α + ......
8.2 3sin sin
1 sin cos cos2 cos3 ......2 3
α α+ α β + β + β +
9.2 3cos cos
1 cos cos cos c o s 32 3
α α− α β + + β +KK
10. sin( 2 ) sin( 4 )sin
2 4α + β α + βα − + + KK
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 5 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. α, β, γ are the roots of the equation
3 25 12x x x− + + = 0
Calculate the value of Σα2β.
2. If α, β, γ are the roots of the equation x3 + qx + r = 0, prove that
2 53( )( )Σα Σα = 3 45( )( ).Σα Σα
[Hint: Since α, β, γ are the roots] of the given cubic equationα3 + qα + r = 0β3 + qβ + r =0
γ3 + qγ + r = 0To find Σα4 multiply equations (1) by α, β, γ respectively and add.
To find Σα5 multiply equation by (α2, β2, γ2 respectively and add.3. Find the sum of the cubes of the roots of the equation
x3 – 6x2 – 11x + 6 = 0
4. If α, β, γ be the roots of the equation
x3 +5x2 – 6x + 3 =0Find the values of Σα–3.
3 33
3 3 3 3 2 2
3 3 2 2 2
1 1 1
Again –3( ) = [ ]
− Σβ γΣα ≡ + + = α β γ α β γ Σα β αβγ Σαβ Σα β −αβγΣα
Hint:
5. If a, b, c, be the roots of the cubic x3 + qx + r = 0
(i) 1,
b cΣ
+
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89
(ii)2 2
,b cb c
+Σ
+
(iii) Σ(b + c)2
6. If α, β, γ, δ be the roots of the equation
4 3 2x px qx rx s+ + + + =0
Find the value of
(i) 2Σα β
(ii) 1.Σ
α7. If α, β, γ are the roots of the equation
3 2x px qx r− + − = 0
Find the value of
(i) 21
,Σα
(ii) Σα2β(iii) (α + β) (β + γ) (γ + α)
90
B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 6.1 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. From an equation whose roots are negatives of the roots of the equation
3 26 8 9x x x− + − = 0
2. From an equation whose roots are twice the roots of the equation
4 22 3 7 9x x x+ + − = 0
3. Find the equation whose roots are the squares of the roots of the equation
x3 – x2 – 8x – 6 =04. Find the equation whose roots are the reciprocals of the roots of the equation
ax3 + bx2 + cx + d = 0.
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 6.2 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. If α, β, γ are the roots of the equation
3 2x px qx r+ + + = 0
form the equation whose roots are
1 1, ,a
1− β − γ −βγ γα αβ
2. If a, b, c are the roots of 3 0,x qx r+ + = from the equation whose roots are
2 2 2 2, , .b c c a a b2 2
3. Find the equation whose roots are the roots of the equationx4 – 3x2 + 7x2 + 5x – 2 = 0
each diminished by 2.4. Transform the equation
3 26 4 7x x x− + − = 0
into one, from which the second term is missing.
5. If α, β, γ are the roots of the equation
3 2 0,where 0x px qx r r− + − = ≠
Find the equation whose roots are 1 1 1
, ,βγ + γ α + α β +α β γ
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 6.3 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
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Academic Session 2012–2013
Exercise1. Solve the equation
4 3 23 40 130 20 27x x x x− + − + = 0
given that product of two of its roots is equal to the product of the other two.
2. Prove that the necessary condition for the roots of the equation
3 20 1 23 3a x a x a x a+ + + = 0
to be in G.P. isa0 + a2
3 = a13a3
3. Prove that the necessary condition for the roots of the equationx3 – px2 + qx – r = 0
to be in H.P. is
2 327 9 2r pqr q− + = 0.
4. Prove that the necessary condition for the sum of two roots of the equation
x4 – px3 + qx2 – rx + s = 0to equal to the sum of the other two is,
3 4 8p pq r− + = 0.
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B.A. (Programme) I Year Discipline Course MathematicsS.R.S. 6.4 Part A (Section III)
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School of Open LearningUniversity of Delhi.5, Cavalry Lanes,
Delhi-11007 (India)
Academic Session 2012–2013
Exercise1. Solve the equation
3 26 3 10x x x− + + = 0
whose roots are in A.P.
2. Solve the equation
3 23 26 52 24x x x− + − = 0
the roots being in G.P.3. Solve the equations
28x3 – 39x2 + 12x – 1 = 0
the roots being in Harmonical Progression.4. Solve the equation
4x3 + 16x2 – 9x – 36 = 05. Solve the equation
2x3 – x2 – 22x – 24 = 0
two of its roots being in the ratio 3 : 46. Solve the equation
2x3 + x2 – 7x – 6 = 0given that the difference of two roots is 3.
7. Solve the equation
x3 – 5x2 – 2x + 24 = 0given that the product of two if roots is 12.
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Notes
95
Notes
96
Notes