C L A S S I C A L A L G E B R A

CHAPTER I

COMPLEX NUMBERS: DE’ MOIVRE’S THEOREM

A complex number z is an ordered pair of real numbers (a,b): a is called Real part of z, denoted by Re z and b is called imaginary part of z, denoted by Im z. If Re z=0, then z is called purely imaginary; if Im z =0, then z is called real. On the set C of all complex numbers, the relation of equality and the operations of addition and multiplication are defined as follows:

(a,b)=(c,d) iff a=b and c=d, (a,b)+(c,d)=(a+c,b+d), (a,b).(c,d)= (ac-bd,ad+bc)

The set C of all complex numbers under the operations of addition and multiplication as defined above satisfies following properties:

For z1,z2,z3∈C, (1) (z1+z2)+z3=z1+(z2+z3)(associativity), (2)z1+(0,0)=z1, (3) for z=(a,b)∈C, there exists –z=(-a,-b)∈C such that (-z)+z=z+(-z)=(0,0), (4)z1+z2=z2+z1.

For z1,z2,z3∈C, (1) (z1. z2).z3=z1.(z2.z3)(associativity), (2)z1.(1,0)=z1, (3)

for z=(a,b)∈C,z≠(0,0), there exists 1z∈C such that z.

1z=

1z .z=1,

(4)z1.z2=z2.z1. For z1,z2,z3∈C, z1.(z2+z3)=(z1.z2)+(z1.z3).

Few Observations

(1)Denoting the complex number (0,1) by i and identifying a real complex number (a,0) with the real number a, we see z=(a,b)=(a,0)+(0,b)=(a,0)+(0,1)(b,0) can be written as z=a+ib.

(2)For two real numbers a,b , a2+b2=0 implies a=0=b; same conclusion need not follow for two complex numbers, for example, 12+i2=0 but 1≡(1,0)≠

(0,0) ≡0 and i=(0,1)≠(0,0) (≡ denotes identification of a real complex number with the corresponding real number).

(3)For two complex numbers z1,z2, z1z2=0 implies z1=0 or z2=0.(4) i2=(0,1)(0,1)=(-1,0) ≡-1.(5) Just as real numbers are represented as points on a line, complex numbers

can be represented as points on a plane: z=(a,b)↔P: (a,b). The line containing points representing the real complex numbers (a,0), a real, is called the real axis and the line containing points representing purely

1

imaginary complex numbers (0,b) ≡ib is called the imaginary axis. The plane on which the representation is made is called Gaussian Plane or Argand Plane.

Definition 1.1Let z=(a,b) ≡a+ib. The conjugate of z, denoted by z , is (a,-b) ≡a-ib.

Geometrically, the point (representing) z is the reflection of the point (representing) z in the real axis. The conjugate operation satisfies the following properties:

(1) z=z , (2) z1+ z2=z1+z2, (3) z1 z2=z1+z2, (4)( z1

z2)= z1

z2

, (4) z+z=2 Rez, z-z=2i

Im(z)

Definition 1.2Let z=(a,b) ≡a+ib. The modulus of z, written as |z|, is defined as √a2+b2.

Geometrically, |z| represents the distance of the point representing z from the origin (representing complex number (0,0) ≡0+i0). More generally, |z1−z2| represents the distance between the points z1 and z2. The modulus operation satisfies the following properties:

(1) |z1+z2|≤|z1|+|z2|, (2) |z1 . z2|=|z1|.|z2|(3)|z1

z2|=|z1|

|z2| (4) ||z1|−|z2||≤|z1−z2|

GEOMETRICAL REPRESENTATION OF COMPLEX NUMBERS: THE ARGAND PLANE

Let z=a+ib be a complex number. In the Argand plane, z is represented by the point whose Cartesian co-ordinates is (a,b) referred to two perpendicular lines as axes, the first co-ordinate axis is called the real axis and the second the imaginary axis. Taking the origin as the pole and the real axis as the initial line, let (r,θ) be the polar co-ordinates of the point (a,b). Then a=r cosθ, b=r sinθ. Also r=√a2+b2=|z|. Thus z=a+ib=|z|(cosθ+isinθ): this is called modulus-amplitude form of z. For a given z≠0, there exist infinitely many values of θ differing from one another by an integral multiple of 2π: the collection of all such values of θ for a given z≠0 is denoted by Arg z or Amp z. The principal value of Arg z , denoted by arg z or amp z, is defined to be the angle θ from the collection Arg z that satisfies the condition -π<θ≤π . Thus Arg z={arg z+2nπ: n an integer}. arg z satisfies following properties: (1) arg(z1z2)=argz1+argz2+2kπ, where k is a suitable integer from the set{-1,0,1] such that -π<¿ argz1+argz2+2k

2

π ≤π, (2) arg( z1

z2)=¿ argz1-argz2+2kπ, where k is a suitable integer from the set{-

1,0,1] such that -π<¿ argz1-argz2+2kπ ≤π.

Note An argument of a complex number z=a+ib is to be determined from the relations cosθ=a/|z|, sinθ= b/|z| simultaneously and not from the single relation tan θ=b/a.

Example1.1Find arg z where z=1+i tan3π5 .

»Let 1+itan3π5 =r(cos θ+i sin θ). Then r2= sec2 3π

5 . Thus r=- sec3π5 >0.Thus cos θ

=- cos3π5 , sin θ=-sin

3π5 . Hence θ=π+

3π5 . Since θ>π,arg z=θ-2π=-

2π5 .

Geometrical representation of operations on complex numbers:

Addition Let P and Q represent the complex numbers z1=x1+iy1 and z2=x2+iy2

on the Argand Plane respectively. It can be shown that the fourth vertex R of the parallelogram OPRQ represents the sum z1+z2 of z1 and z2.

Product Let z1=|z1|¿) and z2=|z2|¿) where –π<θ1,θ2 ≤π . Thus z1z2=|z1||z2|{cos(θ1+θ2

)+isin(θ1+θ2)}. Hence the point representing z1z2 is obtained by rotating line segment OP{ where P represents z1} through arg z2 and then dilating the resulting line segment by a factor of |z2|. In particular , multiplying a complex

number by i=cos( π2 )+isin( π2 ) geometrically means rotating the line segment by

π2

.

T h e o r e m 1 . 1 (De Moivre’s Theorem) If n is an integer and θ is any real

number, then (cosθ+i sinθ)n= cos nθ+i sin nθ. If n=pq , q natural, p integer, |p|and

q are realtively prime, θ is any real number, then (cosθ+i sinθ)n has q number of values, one of which is cos nθ+i sin nθ.

Proof: Case 1 Let n be a positive integer.

Result holds for n=1: (cosθ+i sinθ)1= cos 1θ+i sin 1θ. Assume result holds for some positive integer k: (cosθ+i sinθ)k= cos kθ+i sin kθ.Then (cosθ+i sinθ

)k+1=(cosθ+i sinθ)k(cosθ+i sinθ)=( cos kθ+i sin kθ)( cosθ+i sinθ)= cos(k+1)θ+isin(k+1)θ . Hence result holds by mathematical induction.

Case 2 Let n be a negative integer, say, n=-m, m natural.

3

(cosθ+i sinθ)n=(cosθ+i sinθ)-m=1

(cosθ+isinθ )m= 1

cosmθ+i sinmθ (by case 1) = cos mθ

-isin mθ=cos(-m)θ+isin(-m)θ= cos nθ+i sin nθ.

Case3 n=0: proof obvious.

Case 4 Let n=pq , q natural, p integer, |p|and q are realtively prime.

Let (cosθ+i sinθ)pq= cos φ+i sinφ. Then (cosθ+i sinθ)p= (cos φ+i sinφ)q. Thus cos pθ

+isin pθ= cos qφ+i sin qφ. Thus qφ=2kπ+pθ, that is, φ=2k π+ pθ

q . Hence

(cosθ+i sinθ)pq= cos(

2k π+ pθq )+isin(

2k π+ pθq

¿, where k=0,1,…,q-1 are the distinct

q values.

S o m e A p p l i c a t i o n s o f D e ’ M o i v r e ’ s T h e o r e m

(1)Expansion of cos nθ, sin nθ and tan nθ where n is natural and θ is real.Cos nθ+ i sin nθ = (cosθ+isinθ)n = cosnθ + C1

n cosn-1θ(isinθ) + C2n cosn-2θ ¿

)+…+insinnθ= (cosnθ- C2n cosn-2θ sin2 θ+…)+i( C1

n cosn-1θsinθ- C3n cosn-3θ sin3 θ+

…). Equating real and imaginary parts, cos nθ= cosnθ- C2n cosn-2θ sin2 θ+

………………………………. and sin nθ = C1n cosn-1θsinθ- C3

n cosn-3θ sin3 θ

+…………………….(2)Expansion of cosnθ and sinnθ in a series of multiples of θ where n is

natural and θ is real.Let x = cos θ+isinθ. Then xn=cos nθ+isin nθ, x-n= cos nθ-isin nθ. Thus (2

cosθ)n=(x+1x¿n

=(xn+1

xn¿+ C1

n(xn-2+

1

xn−2 )+…=2 cos nθ+ C1n (2 cos(n-2)θ)+…

Similarly, expansion of sinnθ in terms of multiple angle can be derived.

(3)Finding n th roots of unityTo find z satisfying zn=1=cos(2kπ)+isin(2kπ ¿, where k is an integer. Thus

z=[ cos(2kπ)+isin(2kπ ¿]1/n=cos(2kπn

¿+isin ( 2kπn ), k=0,1,…,n-1; replacing k

by any integer gives rise to a complex number in the set A = { cos(2kπn

¿+isin ( 2kπn )/ k=0,1,…,n-1}. Thus A is the set of all nth roots of unity.

4

Example1.2Solve x6+x5+x4+x3+x2+x+1=0

»We have the identity x6+x5+x4+x3+x2+x+1= x7−1x−1

. Roots of x7-1=0 are cos

2kπ7

+isin2kπ

7, k=0,1,…,6. Putting k=0, we obtain root of x-1=0. Thus the roots

of given equation are cos2kπ

7+isin

2kπ7 , k=1,…,6.

Example1.3 Prove that the sum of 99 th powers of all the roots of x7-1=0 is zero.

»The roots of x7-1=0 are {1,α ,α2,…,α6}, where α=cos2π7 +isin

2π7 . Thus sum of

99th powers of the roots is 1+α99+(α 2¿¿99+…+¿1+α99+(α 99¿¿2+…+¿=1−α 99.7

1−α 99 =0,

since α 99.7=1 and α 99≠1.

Example1.4If |z1|=|z2| and arg z1+argz2=0, then show that z1=z2.

»Let |z1|=|z2|=r, arg z1=θ, then argz2=-θ. Thus z1=r(cos θ+isinθ)=r(cosθ-isinθ)=z2.

Example1.5For any complex number z, show that |z|≥|ℜ z|+|ℑ z|

√2.

»Let z=x+iy.2(x2+y2)-(x+y)2=(x-y)2≥0. Thus x2+y2≥(x+ y)2

2 and so |z|=

√ x2+ y2 ≥|ℜ z|+|ℑ z|

√2.

Example1.6Prove that if the ratio z−iz−1 s purely imaginary, then the point

(representing )z lies on the circle whose centre is at the point 12(1+i) and radius

is 1

√2.

»Let z=x+iy. Then z−iz−1

= x2−x+ y2− y(x−1)2+ y2 +i

1−x− y(x−1)2+ y2 . By given condition ,

x2−x+ y2− y=0, that is, (x−12 )

2

+( y−12 )

2

=( 1√2 )

2

.Thus z lies on the circle whose

centre is at the point 12(1+i) and radius is

1

√2.

Example1.7 If the amplitude of the complex number z−iz+1 is

π4 , show that z

lies on a circle in the Argand plane.

5

»Let z=x+iy. Then z−iz+1=

x2+x+ y2− y(x+1)2+ y2 +i

y−x−1(x+1)2+ y2. By given condition,

y−x−1

x2+x+ y2− y=1. On simplification, (x+1)2+(y-1)2=1. Hence z lies on the circle

centred at (-1,1) and radius 1.

Example1.8 If z and z1 are two complex numbers such that z+z1 and zz1 are both real, show that either z and z1 are both real or z1=z.

Example1.9 If |z1+z2|=|z1−z2|, prove that arg z1 and argz2 differ by π2 or

3π2 .

»|z1+z2|2=|z1−z2|

2 Thus (z1+z2)(z1+ z2 ¿= (z1-z2)(z1−z2¿ Or, z1 z2+z1 z2=0(1). Let z1=r1(cosθ1+isinθ1), z2=r2(cosθ2+isinθ2). From (1), cos(θ1-θ2)=0 proving the result.

Example1.10 If A,B,C represent complex numbers z1,z2,z3 in the Argand plane and z1+z2+z3=0 and |z1|=|z2|=|z3|, prove that ABC is an equilateral triangle.

»z1+z2=-z3. Hence |z1+z2|2=|z3|

2, that is, |z1|2+|z2|

2+2z1.z2=|z3|2. By given condition,

|z1||z2|cos θ=|z1|2, where θ is the angle between z1 and z2.Thus cos θ=-

12, that is, θ

=1200. Hence the corresponding angle of the triangle ABC is 600. Similarly other angles are 600.

Example1.11 If (x,y) represents a point lying on the line 3x+4y+5=0, find the minimum value of |x+iy|.

Example1.12 Let z and z1 be two complex numbers satisfying z=1+z1

1−z1 and |z1|

=1. Prove that z lies on the imaginary axis.

» z1=z−1z+1 . By given condition, 1=|z−1

z+1 |=|z−1||z+1|. If z=x+iy, x=0. Hence.

Example1.13If z1, z2 are conjugates and z3,z4 are conjugates, prove that argz1

z4

=¿ argz3

z2.

» Since z1, z2 are conjugates, arg z1+arg z2=0. Since z3,z4 are conjugates, arg z3+arg z4=0. Thus arg z1+arg z2=0= arg z3+arg z4. Hence arg z1-arg z4=arg z3-arg z2; thus result holds.

Example1.14 complex numbers z1,z2,z3 satisfy the relation z12+z2

2+z32-z1z2-

z2z3-z3z1=0 iff |z1−z2|=|z2−z3|=|z3−z1|.

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» 0=z12+z2

2+z32-z1z2-z2z3-z3z1=(z1+wz2+w2z3)(z1+w2z2+wz3), where w stands for

an imaginary cube roots of unity. If z1+wz2+w2z3=0, then (z1-z2)=-w2(z3-z2); hence |z1−z2|=|w2||z2−z3|=|z2−z3|.similarly other part.

Conversely, if |z1−z2|=|z2−z3|=|z3−z1|, then z1,z2,z3 represent vertices of an equilateral triangle. Then z2-z1=(z3-z1)(cos 600+isin600), z1-z2=(z3-z2)( cos 600+isin600 ); by dividing respective sides, we get the result.

Example1.15 Prove that |z1+z2|2+|z1−z2|

2=2¿), for two complex numbers z1,z2.

»|z1+z2|2=(z1+z2)(z1+z2)=|z1|

2+|z2|2+2 z1z2; similarly |z1−z2|

2=|z1|2+|z2|

2-2 z1z2; Adding we get the result.

Example1.16 If cosα+cos β+cos γ=sinα+sinβ+sinγ=0, then prove that (1) cos 3α+cos3β+cos 3γ=3cos(α+β+γ), (2) ∑ cos2 α=∑ sin2 α=3/2.

»Let x=cos α+i sinα , y= cos β+i sinβ, z= cos γ+i sinγ . Then x+y+z=0. Thus x3+y3+z3=3xyz. By De’ Moivre’s Theorem, (cos 3α+ cos 3β+cos 3γ)+i(sin 3α+ sin 3β+sin 3γ)=3[ cos(α+β+γ)+isin(α+β+γ ¿¿. Equating, we get result.

Let x=cos α+i sinα , y= cos β+i sinβ, z= cos γ+i sinγ . Then x+y+z=0. Also 1x+ 1

y+ 1

z=0 ;hence xy+yz+zx=0. Thus x2+y2+z2=0. By De’ Moivre’s Theorem,

cos 2α+cos2β+cos2γ=0. Hence∑ cos2 α=3/2. Using sin2α=1- cos2α , we get other part.

Example1.17 Find the roots of zn=(z+1)n, where n is a positive integer, and show that the points which represent them in the Argand plane are collinear.

Let w=z+1z

.Then z= 1w−1

.Now zn=(z+1)n implies wn=1.Thus, w=cos2kπn

+isin2kπn

,k=0,…,n-1.

So z=1

cos2kπn

+isin2kπn

, k=1,…,n-1

=−12

− i2

cotkπn . Thus all points z satisfying zn=(z+1)n lie on the line x=-

12.

Functions of a complex variableExponential function

7

For a complex number z=x+iy, exp(z)=ex(cos y +i sin y).

Note

(1) If z=x+i0 is purely real, exp(z)=ex.(2) If z = 0+iy is purely imaginary, exp(z)=exp(iy)=cos y+i sin y.(3)Since 0≠ex¿|exp z|, thus exp(z) is a non-zero complex

number for every complex number z.(4)For every non-zero complex number u+iv=r(cosθ+isinθ),

exp(ln r+iθ)=eln r(cosθ+isinθ)=u+iv. Thus exp:C→C-{0+i0}.(5)exp(z1).exp(z2)=exp(z1+z2)

» Let z1=x1+iy1,z2=x2+iy2. exp(z1)exp(z2) =ex1+ x2(cosy1+isiny1)(cosy2+isiny2) =ex1+ x2[cos(y1+y2)+isin(y1+y2)]= exp(z1+z2).

(6)exp(z1-z2)=exp (z1)exp ¿¿

; in particular exp(-z)=1

exp (z ).

(7) If n be an integer, (exp z)n=exp(nz) (follows from property(5) and (6))

(8) If n be an integer, exp(z+2nπi)=exp(z).(follows from (5), sinceexp(2nπi)=1)

Example1.18Find all complex number z such that exp(z)=-1

» Let z=x+iy. Then ex(cos y+isiny)=-1. Thus excos y=-1, exsin y=0. Squaring, adding gives e2x=1 whence ex=1(since ex>0); thus x=0. cosy=-1 and siny=0 gives y=(2n+1)π. Thus z=(2n+1)πi, n integer.

Logarithmic function

Let z be a non-zero complex number. Then there exists a complex number w such that exp(w)=z; further if exp(w)=z, then exp(w+2nπi)=z. For a given non-zero complex number z, we define Log z={w∈C/ exp(z)=w} and call the set Logarithm of z.

Let z=r(cos θ+isinθ),-π<θ≤π . Let w=u+iv be a logarithm of z(that is, a member of Log z). Then exp(u+iv)= r(cos θ+isinθ). Thus eucos v=r cosθ, eusin v=r sinθ. Hence eu=r (by squaring and adding last two relations), cos v=cosθ, sinv=sinθ. So u=ln r,v=θ+2nπ ,ninteger . Thus Logz=ln|z|+i(arg z+2nπ).

Note

Putting n=0 in the expression for Log z, we obtain principal value of Log z, denoted by log z. Thus log z= ln|z|+iarg z.

8

(1)exp(Log z)=z for z≠0; one of the values of Log(expz) is z, the other values are z+2nπi,n integer.

(2)For two distinct non-zero complex numbers z1,z2, Log(z1z2)=Log(z1)+Log(z2); log(z1z2)≠log(z1)+log(z2): take z1=i,z2=-1.» Let z1=r1(cosθ1+isinθ1), z2= r2(cosθ2+isinθ2). Then Log z1=lnr1+i(θ1+2nπ

), Log z2=lnr2+i(θ2+2mπ),Log(z1z2)= z1=ln(r1r2)+i(θ1+θ2+¿2kπ) where n, m, k are integers. Log(z1)+Log(z2)=ln(r1r2)+i(θ1+θ2+2qπ ¿, where q=m+n. Since p,q are arbitrary integers, Log(z1z2)=Log(z1)+Log(z2) holds.

(3)For two distinct non-zero complex numbers z1,z2, Log( z1

z2)=¿Log(z1)-

Log(z2). log( z1

z2)≠log(z1)-log(z2): take z1=-1,z2=-i.

» Proof similar.(4) If z≠0 and m be a positive integer, Log zm≠mLogz: take z=i,m=2.

Example1.19Verify Log(-i)1/2=1/2 Log(-i)

» -i=cos( π2 )+isin( π2 ). 12Log(-i)=

12[(2nπ− π

2¿ i]=( nπ− π

4 )I, where n is an integer.

Two values of (-i)1/2 are cos (−π4 )+isin(−π

4 ) and cos ( 3π4 )+isin( 3π

4 ). Now Log[

cos (−π4 )+isin(−π

4 )]=(2mπ-π4

)i and Log[cos ( 3π4 )+isin( 3π

4 )]= (2pπ+ 3π4

)i=[(2p+1)

π− π4

]I, where m,p are integers.Thus the values of Log(-i)1/2 can be expressed as

(nπ- π4 )I, where n is an integer. Hence the result.

Example1.20Prove that sin[iloga−iba+ib

¿=2ab

a2+b2.

»Let a+ib=r(cosθ+isinθ),-π<θ≤π . Then a=r cosθ, b=r sinθ. Thus z=cos(-2θ

)+isin(-2θ). Thus arg z=-2θ+2kπ, where k is an integer such that -π<-2θ+2kπ ≤π.

So log z=(-2θ+2kπ)i. hence sin(ilog z)=sin(2θ-2kπ)=sin2θ=2ab

a2+b2.

Exponent function

If w be a nonzero complex number and z be any complex number, then wz=exp(z Log w) . Since Log is many-valued, exponent is many-valued; principal value of wz is defined as exp(z log w).

9

NoteIf z1,z2,w are complex numbers, w≠0, then w z1+z 2≠w z1 w z2 but p.v.w z1+z 2=(p.v.w z1)(p.v.w z2).

Example1.21Find values of ii.

» ii=exp[iLogi]=exp[i(2nπ+ π2

¿i]=e−(4n+1) π

2 , n any integer.

Example1.22If a,b are the imaginary cube roots of unity, prove that (1) p.v.

of aa+p.v. of bb=e−π√3 , (2) p.v. of ab+p.v. of ba =e

π√3.

p.v. of aa=exp[a log a]=exp[a{ln1+i2π3 ]], if a=-1

2+i √3

2

=exp[-π

√3− i

π6 ]=e

−π√3 [cos

π6−isin

π6 ]. Similarly, p.v. of bb=e

−π√3 [cos

π6

+isinπ6 ].Hence

(1) is proved. Similarly (2) can be proved.

Example1.23 If iz=i, show that z is real and the general values of z are given

by z=4m+14n+1 , m,n are integers.

Example1.24Find all the values of √2√3 and show that the values lie on a circle in the Argand plane.

Trigonometric function

From definition of exponential function, for real y, exp(iy)=cos y+isin y, exp(-

iy)=cosy-isiny. Thus for real y, cos y=exp (iy )+exp (−iy)

2, sin y=

exp (iy )−exp (−iy)2i

.

Backedup by theseresults, for complex z, we define cos z=exp ( iz )+exp (−iz)

2, sin

z=exp (iz )−exp (−iz)

2i.

(1)cos2z+sin2z=1(2) sin(z1+z2)=sinz1cosz2+cosz1sinz2, cos(z1+z2)=cosz1cosz2-

sinz1sinz2.(3) sin(z+2π)=sin z, sin(z+π ¿=-sin z(4) if x,y are real, sin(x+iy)=sin x cosh y+icos xsinh y, cos(x+iy)=cosx cosh

y-isin xsinh y, where cosh x=exp ( x )+exp (−x )

2 and sinh x=

exp ( x )−exp (−x)2

.

Thus |sin (x+iy)|2=sin2x+sinh2y and |cos (x+iy)|2=cos2x+sinh2y: since sinh

10

y is an unbounded function, sin z and cos z are unbounded in absolute value. But if x is real, sin x and cos x are bounded functions.

(5)cos(iz)=cosh z, sin(iz)=isinh z, cosh(iz)=cos z, sinh(iz)=i sin z, where

cosh z=exp ( z )+exp (−z )

2 and sinh z=

exp ( z )−exp (−z)2

.

Example1.25 Find all values of z such that cos z=0.

»Let z=x+iy. cos z=0 implies cos xcosh y=0, sin xsinh y=0. Since cosh y≠0, cos

x=0. Thus x=(2n+1)π2 , n integer. Thus sin x=sin[(2n+1)

π2 ]≠0. Hence sinh y=0.

Thus y=0. Hence z==(2n+1)π2 , n integer.

Example1.26 If x=log tan(π4+ θ

2¿, where θ is real, prove that θ=-iLog tan

( π4 +ix2 ).

»Since θ is real, x real. ex=tan(π4+ θ

2¿=

1+tanθ2

1−tanθ2

. Therefore, tanθ2=

ex−1ex+1

. Thus

t−t−1

t+t−1 =ie

x2−e

− x2

ex2 +e

−x2

, where t=exp( iθ2 ). Hence t2−1t 2+1

=isinh

x2

coshx2

=sin

ix2

cosix2

=tanix2 . Thus t2=

1+tanix2

1−tanix2

. So exp(iθ)=tan( π4 + ix2 ). θ=-iLog tan( π4 +i

x2 ).

Inverse Trigonometric functions

Let z be a given complex number and w be a complex number such that sin w=z. Then cos w=±√1−z2. Thus exp(iw)=iz±√1−z2 or, w=-iLog(iz±√1−z2). Since Log is a multi-valued function, Sin-1z=w=-iLog(iz±√1−z2) is a multiple valued function of z. The principal value of Sin-1z is obtained by choosing cos w=+√1−z2 and by taking the principal value of the logarithm and is denoted by sin-1z.

Example1.27Find Cos-1(2), cos-1(2).

»Let Cos-1(2)=z. Then cos z=2. sin z=±√3i. Thus exp(iz)=cos z+I sin z=2=±√3. When exp(iz)=2+√3, z=-iLog(2+√3)=-i[log(2+√3¿+2nπi]=2nπ-i log(2+√3).

11

When exp(iz)=2-√3, z=-iLog(2-√3)=-i[log(2-√3¿+2nπi]=2nπ-i log(2-√3) =2nπ+i log(2+√3¿. Thus Cos-1(2)= 2nπ ±i log(2+√3¿. Hence cos-1(2)=ilog(2+√3).

Example1.28 If tan-1(x+iy)=a+ib, where x,y,a,b are real and (x,y)≠(0,±1), prove that (1) x2+y2+2xcot 2a=1, (2) x2+y2+1-2ycoth 2b=0.

»tan (a+ib)=x+iy; thus tan(a-ib)=x-iy. Thus tan 2a=tan[(a+ib)+(a-ib)]=2 x

1−x2− y2 .

Hence the first part. Again tan(2ib)=tan[(a+ib)-(a-ib)]=2iy

1+ x2+ y2 . But tan(2ib)=

sin (2ib )cos (2ib)

=isinh2bcosh 2b

=i tanh 2b. hence the second part.

CHAPTER II

THEORY OF EQUATIONS

An expression of the form a0xn+a1xn-1+…+an-1x+an, where a0,a1,…,an are real or complex constants, n is a nonnegative integer and x is a variable (over real or complex numbers) is a polynomial in x. If a0≠0, the polynomial is of degree n and a0xn is the leading term of the polynomial. A non-zero constant a0 is a polynomial of degree 0 while a polynomial in which the coefficients of each term is zero is said to be a zero polynomial and no degree is assigned to a zero polynomial.

Equality two polynomials a0xn+a1xn-1+…+an-1x+an and b0xn+b1xn-1+…+bn-1x+bn

are equal iff a0=b0,a1=b1,…,an=bn.

AdditionLet f(x)= a0xn+a1xn-1+…+an-1x+an, g(x)= b0xn+b1xn-1+…+bn-1x+bn. the sum of the polynomials f(x) and g(x) is given by

f(x)+g(x)= a0xn+…+an-m-1xm+1+(an-m+b0)xm+…+(an+bm), if m<n

= (a0+b0)xn+…+(an+bn), if m=n

= b0xm+…+bm-n-1xn+1+(bm-n+a0)xn+…+(bm+an), if m>n.

MultiplicationLet f(x)= a0xn+a1xn-1+…+an-1x+an, g(x)= b0xn+b1xn-1+…+bn-1x+bn. the product of the polynomials f(x) and g(x) is given by

12

f(x)g(x)=c0xm+n+c1xm+n-1+…+cm+n, where ci=a0bi+a1bi-1+…+aib0. c0=a0b0≠0; hence

degree of f(x)g(x) is m+n.

Division Algorithm Let f(x) and g(x) be two polynomials of degree n and m respectively and n≥m. Then there exist two uniquely determined polynomials q(x) and r(x) satisfying f(x)=g(x)q(x)+r(x), where the degree of q(x) is n-m and r(x) is either a zero polynomial or the degree of r(x) is less than m. In particular, if degree of g(x) is 1, then r(x) is a constant, identically zero or non-zero.

T h e o r e m 1 . 2 ( Remainder Theorem)If a polynomial f(x) is divided by x-a, then the remainder is f(a).

»Let q(x) be the quotient and r (constant)be the remainder when f is divided by x-a. then f(x)=(x-a)q(x)+r is an identity. Thus f(a)=r.

T h e o r e m 1 . 3 ( Factor Theorem)If f is a polynomial, then x-a is a factor of f iff f(a)=0.

»By Remainder theorem, f(a) is the remainder when f is divided by x-a; hence, if f(a)=0, then x-a is a factor of f. Conversely, if x-a is a factor of f, then f(x)=(x-a)g(x) and hence f(a)=0.

Example1.29Find the remainder when f(x)=4x5+3x3+6x2+5 is divided by 2x+1.

»The remainder on dividing f(x) by x-(-12¿=x+

12 is f(-

12)=6. If q(x) be the

quotient, then f(x)=q(x)(x+12)+6=

q(x )2

(2x+1)+6. Hence 6 is the remainder when

f is divided by 2x+1.

Synthetic division

Synthetic division is a method of obtaining the quotient and remainder when a polynomial is divided by a first degree polynomial or by a finite product of first degree polynomials. Let q(x)=b0xn-1+b1xn-2+…+bn-1 be the quotient and R be the remainder when f(x)=a0xn+a1xn-1+…+an is divided by x-c. Then

a0xn+a1xn-1+…+an=( b0xn-1+b1xn-2+…+bn-1)(x-c)+R.

Above is an identity; equating coefficients of like powers of x, a0=b0, a1=b1-cb0, a2=b2-cb1,…,an-1=bn-1-cbn-2,an=R-cbn-1. Hence b0=a0,b1=a1+cb0,b2=a2+cb1,…,bn-

1=an-1+cbn-2,R=an+cbn-1.

The calculation of b0,b1,…,bn-1,R can be performed as follows:

13

a0 a1 a2 … an-1 an

cb0 cb1 … cbn-2 cbn-1

b0 b1 b2 …. bn-1 R

Example1.30Find the quotient and remainder when 2x3-x2+1 is divided by 2x+1.

» -12 2 -1 0 1

-1 1 -½

……………………………..

2 -2 1 ¿ ½

Thus 2x3-x2+1=(x+12)(2x2-2x+1)+

12=(2x+1)(x2-x+

12)+

12; hence the quotient is x2-

x+12 and the remainder is

12.

Example1.31Find the quotient and remainder when x4-3x3+2x2+x-1 is divided by x2-4x+3.

»x2-4x+3=(x-1)(x-3). We divide x4-3x3+2x2+x-1 by x-1 and then the obtained quotient again by x-3 by method of synthetic division. We get x4-3x3+2x2+x-1=(x-1)[x3-2x2+1]+0=(x-1)[(x-3){x2+x+3}+10]= (x2-4x+3)(x2+x+3)+10(x-1); hence the quotient is x2+x+3 and the remainder is 10(x-1).

Applications of the method

(1)To express a polynomial f(x)=a0xn+a1xn-1+…+an as a polynomial in x-c.Let f(x)=A0(x-c)n+A1(x-c)n-1+…+An. Then f(x)=(x-c)[A0(x-c)n-1+A1(x-c)n-

2+…+An-1]+An. Thus on dividing f(x) by x-c, the remainder is An and the remainder is q(x)= A0(x-c)n-1+A1(x-c)n-2+…+An-1. Similarly if q(x) is divided by x-c, the remainder is An-1. Repeating the process n times, the successive remainders give the unknowns An,…,A1 and A0=a0.

(2)Let f(x) be a polynomial in x. to express f(x+c) as a polynomial in x.Let f(x)=a0xn+a1xn-1+…+an=A0(x-c)n+A1(x-c)n-1+…+An; by method explained above , the unknown coefficients can be found out in terms of the known coefficients a0,…,an. Now f(x+c)=A0xn+…+An.

Example1.32Express f(x)=x3-6x2+12x-16=0 as a polynomial in x-2 and hence solve the equation f(x)=0.

14

» Using method of synthetic division repeatedly, we have

2 ¿ 1 -6 12 -16 Hence f(x)=(x-2)3-8=0.

2 -8 8 Thus x-2=2,2w,2w 2 (w is an imaginary cube roots of unity)

---------------------------- Hence x=4,2+2w,2+2w2.

1 -4 4 ¿-8

2 -4

-----------------------

1 -2 ¿ 0

2

------------

1 ¿ 0

Example1.33If f(x) is a polynomial of degree ≥2 in x and a,b are unequal,

show that the remainder on dividing f(x) by (x-a)(x-b) is( x−b ) f ( a )−( x−a ) f (b)

a−b.

» By division algorithm, let f(x)=(x-a)(x-b)q(x)+rx+s, where rx+s is the remainder. Replacing x by a and by b in turn in this identity, f(a)=ra+s, f(b)=rb+s; solving for r,s and substituting in the expression rx+s,we get required expression for remainder.

Example1.34 If x2+px+1 be a factor of ax3+bx+c, prove that a2-c2=ab. Show that in this case x2+px+1 is also a factor of cx3+bx2+a.

»Let ax3+bx+c=( x2+px+1)(ax+d) (*) (taking into account the coefficient of x3). Comparing coefficients, d+ap=0,a+pd=b,d=c whence the result a2-c2=ab follows. Replacing x by 1/x in the identity (*), we get cx3+bx2+a=( x2+px+1)(dx+a): this proves the second part.

If f(x) is a polynomial of degree n, then f(x)=0 is called a polynomial equation of degree n. If b is a real or complex number such that f(b)=0, then b is a root of the polynomial equation f(x)=0 or is a zero of the polynomial f(x). If (x-b)r is a factor of f(x) but (x-b)r+1 is not a factor of f(x), then b is a root of f(x) of multiplicity r: a root of multiplicity 1 is called a simple root. Thus 2 is a simple root of x3-8=0 but 2 is a root of multiplicity 3 of (x-2)3(x+3)=0.

15

Example1.35 Show that 1-x

1!+

x (x−1)2 !

+…+ (−1 )n x ( x−1 ) … ( x−n+1 )n!

=(−1 )n

n !( x−1 )… (x−n).

»1,…,n are all zeros of the polynomial on the left; by factor theorem, each of (x-1),…,(x-n) are factors of the polynomial and hence (x-1)…(x-n) is a factor of the n th degree polynomial on the left; equating coefficient of xn from both side,

we get the constant of proportionality (−1 )n

n! on the right.

T h e o r e m 1 . 4 ( Fundamental Theorem of Classical Algebra)

Every polynomial equation of degree≥1 has a root, real or complex.

Corollary A polynomial equation of degree n has exactly n roots, multiplicity of each root being taken into account.

Corollary If a polynomial f(x) of degree n vanishes for more than n distinct values of x, then f(x) =0 for all values of x.

Example1.36x2-4=(x+2)(x-2) is an identity since it is satisfied by more then two values of x; in contrast (x-1)(x-2)=0 is an equality and not an identity.

T h e o r e m 1 . 5 If b is a multiple root of the polynomial equation f(x)=0 of multiplicity r, then b is a multiple root of f (1)(x)=0 of multiplicity r-1. Thus to find the multiple roots of a polynomial equation f(x)=0, we find the h.c.f. g(x) of the polynomials f(x) and f(1)(x). The roots of g(x)=0 are the multiple roots of f(x)=0.

Example1.37 Find the multiple roots of the equation x5+2x4+2x3+4x2+x+2=0.

Let f(x)=x5+2x4+2x3+4x2+x+2. Then f(1)(x)=5x4+8x3+6x2+8x+1. The h.c.f. of f and f(1) are obtained by method of repeated division(at each stage, we can multiply by a constant of proportionality to our convenience; that does not affect the outcome):

5 8 6 8 1 | 1 2 2 4 1 2

(x 5) 5 10 10 20 5 10

…5…………8…………6………8………1………….

16

2 4 12 4 10

(x5) 10 20 60 20 50

10 16 12 16 2

…………………………………………………..

4 48 4 48

(x1/4) 1 12 1 12

1 12 1 12 | 5 8 6 8 1

5 60 5 60

………………………………………….

-52 1 -52 1

-52 -624 -52 -624

……………………………………………

625 0 625

1 0 1 | 1 12 1 12

1 0 1

………………………….

12 0 12

12 0 12

……………………………..

Thus the h.c.f. is x2-1. Thus f(x)=0 has two double roots( that is, multiple roots of multiplicity 2) i and –i.

Polynomial equations with Real Coefficients

T h e o r e m 1 . 6 If a+ib is a root of multiplicity r of the polynomial equation f(x)=0 with real coefficients, then a-ib is a root of multiplicity r of f(x)=0.

N o t e 1+i is a root of x2-(1+i)x=0 but not so is 1-i.

17

Example1.38 Prove that the roots of 1

x−1+ 2

x−2+ 3x−3

=1x are all real.

» The given equation is 1

x−1+4

x−2+9

x−3=-5 (*). Let a+ib be a root of the

polynomial equation (*) with real coefficients. Then a-ib is also a root of

(*).Thus 1

(a−1 )+ib+4

(a−2 )+ ib+9

(a−3 )+ib=-5 and 1

(a−1 )−ib+4

(a−2 )−ib+9

(a−3 )−ib=-5.

Subtracting,-2ib[1

(a−1)2+b2+ 4

(a−2)2+b2+ 9

(a−3)2+b2]=0 which gives b=0. Hence

all roots of given equation must be real.

Example1.39 Prove that the roots of1

x+a1

+…+ 1x+an

= 1x+b are all real where

a1,…,an,b are all positive real numbers and b>ai for all i.

Example1.40 Solve the equation f(x)=x4+x2-2x+6=0 , given that 1+i is a root.

» Since f(x)=0 is a polynomial equation with real coefficients, 1-i is also a root of f(x)=0. By factor theorem,(x-1-i)(x-1+i)=x2-2x+2 is a factor of f(x). by division, f(x)=( x2-2x+2)(x2+2x+3). Roots of x2+2x+3=0 are -1±√2i. Hence the roots of f(x)=0 are 1±i, -1±√2i.

T h e o r e m 1 . 7 If a+√b is a root of multiplicity r of the polynomial equation f(x)=0 with rational coefficients, then a-√b is a root of multiplicity r of f(x)=0 where a,b are rational and b is not a perfect square of a rational number.

Since every polynomial with real coefficients is a continuous function from R to R, we have

T h e o r e m 1 . 8 (Intermediate Value Property) Let f(x) be a polynomial with real coefficients and a,b are distinct real numbers such that f(a) and f(b) are of opposite signs. Then f(x)=0 has an odd number of roots between a and b. If f(a) and f(b) are of same sign, then there is an even number of roots of f(x)=0 between a and b.

Example1.41Show that for all real values of a, the equation (x+3)(x+1)(x-2)(x-4)+a(x+2)(x-1)(x-3)=0 has all its roots real and simple.

»Let f(x)= (x+3)(x+1)(x-2)(x-4)+a(x+2)(x-1)(x-3). Then limx→−∞

f (x)=∞, f(-2)<0,

f(1)>0, f(3)<0,limx→∞

f (x)=∞. Thus each of the intervals (−∞,-2),(-2,1),(1,3),(3,∞)

contains a real root of f(x)=0. Since the equation is of degree 4, all its roots are real and simple.

18

T h e o r e m 1 . 9 (Rolle’s Theorem) Let f(x) be a polynomial with real coefficients . Between two distinct real roots of f(x)=0 ,there is at least one real root of f(1)(x)=0.

N o t e

(1)Between two consecutive real roots of f(1)(x)=0, there is at most one real root of f(x)=0.

(2) If all the roots of f(x)=0 be real and distinct, then all the roots of f(1)(x)=0 are also real and distinct.

Example1.42Show that the equation f(x)=(x-a)3+(x-b)3+(x-c)3+(x-d)3=0, where a,b,c,d are not all equal , has only one real root.

» Since f(x)=0 is a cubic polynomial equation with real coefficients, f(x)=0 has either one or three real roots. If α be a real multiple root of f(x)=0 with multiplicity 3, then α is also a real root of f(1)(x)=3[(x-a)2+(x-b)2+(x-c)2+(x-d)2]=0, and hence α=a=b=c=d (since α ,a,b,c,d are real), contradiction. If f(x)=0 has two distinct real roots, then in between should lie a real root of f (1)(x)=0, contradiction since not all of a,b,c,d are equal. Hence f(x)=0 has only one real root.

Example1.43 Find the range of values of k for which the equation f(x)=x4+4x3-2x2-12x+k=0 has four real and unequal roots.

» Roots of f(1)(x)=0 are -3,-1,1. Since all the roots of f(x)=0 are to be real and

distinct, they will be separated by the roots of f (1)(x)=0. Now limx→−∞

f (x)=∞,f(-

3)=-9+k,f(-1)=7+k,f(1)=-9+k,limx→∞

f (x)=∞. Since f(-3)<0, f(-1)>0 and f(1)<0, -

7<k<9.

Example1.44 If c1,c2,…,cn be the roots of xn+nax+b=0, prove that (c1-c2)(c1-c3)…(c1-cn)=n(c1

n-1+a).

» By factor theorem, xn+nax+b=(x-c1)(x-c2)…(x-cn).Differentiating w.r.t. x, n(xn-

1+a)= (x-c2)…(x-cn)+(x-c1)(x-c3)…(x-cn)+…+ (x-c2)(x-c3)…(x-cn).replacing x by c1 in this identity, we obtain the result.

Example1.45 If a is a double root of f(x)=xn+p1xn-1+…+pn=0, prove that a is also a root of p1xn-1+2p2xn-2+…+npn=0.

» Since a is a double root of f(x)=0, both f(a)=0 and f(1)(a)=0 hold. Thus an+p1an-

1+…+pn=0 (1) and nan-1+(n-1)p1an-2+…+pn-1=0(2). Multiplying both side of (1)

19

by n and both side of (2) by a and subtracting, we get p1an-1+2p2an-2+…+npn=0.Hence the result.

Example1.46 Prove that the equation f(x)=1+x+x2

2!+…+xn

n !=0 cannot have a

multiple root.

» If a is a multiple root of f(x)=0, then 1+a+ a2

2!+…+ an

n !=0 and 1+a+ a2

2!+…+

an−1

(n−1)!=0;it thus follows that a

n

n !=0, so that a=0; but 0 is not a root of given

equation. Hence no multiple root.

Descartes’ Rule of signs

T h e o r e m 1 . 1 0 The number of positive roots of an equation f(x)=0 with real coefficients does not exceed the number of variations of signs in the sequence of the coefficients of f(x) and if less, it is less by an even number.

The number of negative roots of an equation f(x)=0 with real coefficients does not exceed the number of variations of signs in the sequence of the coefficients of f(-x) and if less, it is less by an even number.

Example1.47 If f(x)=2x3+7x2-2x-3 , express f(x-1) as a polynomial in x. Apply Descartes’ rule of signs to both the equations f(x)=0 and f(-x)=0 to determine the exact number of positive and negative roots of f(x)=0.

» By using method of synthetic division, f(x)=2(x+1)3+(x+1)2-10(x+1)+4. Let g(x)=f(x-1)=2x3+x2-10x+4. By Descartes’ Rule, g(x)=0 has exactly one negative root, say, c. Thus g(c)=f(c-1)=0; hence c-1(<0) is a negative root of f(x)=0. Since there are 2 variations of signs in the sequence of coefficients of f(-x)and since c-1 is a negative root of f(x)=0, f(x)=0 has two negative roots. Also, f(x)=0 has exactly one positive root ,by Descartes’ rule.

Sturm’s Method of location of real roots of a polynomial equation with real coefficients

Let f be a polynomial with real coefficients and f1 be its first derivative. Let the operation of finding the h.c.f. of f and f1be performed with the following modifications: The sign of each remainder is to be changed before it is used as the next divisor and the sign of the last remainder is also to be changed. Let the modified remainders be denoted f2,…,fr. f,f1,f2,…,fr are called Sturm’s functions. During the process of finding Sturm’s functions, at any step, we can multiply by positive constant but not by a negative constant.

20

Sturm’s Theorem

T h e o r e m 1 . 1 1 Let f be a polynomial with real coefficients and a,b be real numbers, a<b. The number of real roots of f(x)=0 lying between a and b (a multiple root, if there be any, being counted only once) is equal to the excess of the number of changes of signs in the sequence of Sturm’s functions f,f1,…,fr

when x=a over the number of changes of signs in the sequence when x=b.

Example1.48Find the number and position of the real roots of the equation x3-3x+1=0.

Let f(x)=x3-3x+1. f(1)(x)=3(x2-1). f1(x)=x2-1.The remainder on dividing f by f1 is -2x+1. Thus f2=2x-1. Dividing 2f1 by f2 (and multiplying by 2 at an intermediate step), the remainder is -3; hence f3 is 1.

f f1 f2 f3 No.of changes of sign

-∞ - + - + 3

0 + - - + 2

∞ + + + + 0

Hence the equation has 3 real roots,(3-2=)1 negative root and (2-0=)2 positive roots.

Location of roots

f f1 f2 f3 No.of changes of sign

-2 - + - + 3

-1 + 0 - + 2(0 to be

treated as continuation)

0 + - - + 2

1 - 0 + + 1

2 + + + + 0

Thus one root lies between -2 and -1; one lies between 0 and 1 and one lies between 1 and 2.

NoteIf at any stage of finding out the sequence of Sturm’s sequence, we obtain a function all of whose roots are complex, then the h.c.f. process need not be continued further and the determination and location of real roots will be

21

possible from the set of functions f,f1,…,fs. This is because fsretains same sign for all values of x and no alteration in the number of changes of sign can take place in the sequence of functions beyond fs.

Example1.49Find the number and position of the real roots of the equation x4+4x3-x2-2x-5=0.

f(x)=x4+4x3-x2-2x-5, f1(x)=2x3+6x2-x-1, f2(x)=7x2+2x+9. We can verify all the roots of f2=0 are complex and leading coefficient of f2 is positive; hence f2(x)>0 for all real x. Hence the remaining Sturm functions need not be calculated.

f f1 f2 No. of changes of sign

-∞ + - + 2

0 - - + 1

∞ + + + 0

The equation has two real roots, one positive and one negative.

Relations between roots and coefficients

Let c1,…,cn be the roots of the equation a0xn+a1xn-1+…+an-1x+an=0. By factor theorem,

a0xn+a1xn-1+…+an-1x+an=a0(x-c1)(x-c2)…(x-cn).

Equating coefficients of like powers of x,a1=a0(−∑ c1), a2=a0∑ c1 c2,….,an=a0 (-1)nc1c2…cn. Hence

∑ c1=-a1

a0,∑ c1 c2=

a2

a0,…, c1c2…cn=(-1)n

an

a0.

Example1.50 Solve the equation 2x3-x2-18x+9=0 if two of the roots are equal in magnitude but opposite in signs.

» Let the roots be -a, a, b .Using relations between roots and coefficients, b=(-a)

+a+b=12 and –a2b=-

92 . Hence a2=9, that is, a=±3. Hence the roots are 3,-3,

12.

Example1.51 Solve x3+6x2+11x+6=0 given that the roots are in A.P.

Symmetric functions of roots

22

A function f of two or more variables is symmetric if f remains unaltered by an interchange of any two of the variables of which f is a function. A symmetric function of the roots of a plolynomial equation which is sum of a number of terms of the same type is represented by any one of its terms with a sigma notation before it: for example, if a,b,c be the roots of a cubic polynomial, then ∑ a2 will stand for a2+b2+c2.

Example1.52 If a,b,c be the roots of x3+px2+qx+r=0, find the value of (1)

∑ a2, (2)∑ a2 b, (3) ∑ a3,(4)∑ a2 b2,(5)∑1a ,(6)∑

1ab ,(7)∑

1

a2 .

» (1)∑ a2=(∑ a )2−2∑ ab=p2-2q, (2) ∑ a2 b=∑ a∑ ab -3abc=-pq+3r, (3)∑ a3=

∑ a2∑ a-∑ a2 b, (4) ∑ a2 b2=(∑ ab )2-2abc∑ a, (5) ∑

1a=∑ ab

abc, (6) ∑

1ab=∑ a

abc, (7)

∑1

a2 =(∑ 1a )

2

−2∑1ab

.

T h e o r e m 1 . 1 1 ( Newton) Let a1,…,an be the roots of xn+p1xn-1+p2xn-2+…+pn=0, sr=a1

r+…+anr where r is a non-negative integer. Then

(1) sr+p1sr-1+p2sr-2+….+pr-1s1+rpr=0, if 1≤r ≤n

(2) sr+p1sr-1+p2sr-2+….+pnsr-n=0, if r≥n

Example1.53 If a1,a2,a3,a4 be the roots of the equation x4+p2x2+p3x+p4=0, find the value of (1)∑ a1

3, (2)∑ a14,(3)∑ a1

6.

»(1)By Newton’s Theorem, s3+p2s1+3p3=0. Here s1=0.Thus s3=-3p3.

(2)By Newton’s Theorem, s4+p2s2+p3s1+4p4=0. Here s1=0 and s2=-2p2. Thus s4=2(p2

2-2p4)

(3)s6+p2s4+p3s3+p4s2=0. Hence s6=6p4p2+3p32-2p2

3.

Transformation of equations

When a polynomial equation is given, it may be possible , without knowing the individual roots, to obtain a new equation whose roots are connected with those of the given equation by some assigned relation. The method of finding the new equation is said to be a transformation. Study of the transformed equation may throw some light on the nature of roots of the original equation.

(1)Let c1,…,cn be the roots of a0xn+a1xn-1+…+an-1x+an=0; to obtain the equation whose roots are mc1,mc2,…,mcn. (m=-1 is an interesting case)

23

» Let d1=mc1. Since c1 is a root of a0xn+a1xn-1+…+an-1x+an=0, we have a0c1

n+a1c1n-1+…+an-1c1+an=0. Replacing c1 by d1/m, we get a0d1

n+ma1d1n-

1+m2a2d1n-2+…+mn-1an-1d1+mnan=0. Thus the required equation is

a0xn+ma1xn-1+…+mn-1an-1x+mnan=0.(2)Let c1,…,cn be the roots of a0xn+a1xn-1+…+an-1x+an=0 and let c1c2…cn≠0;

to obtain the equation whose roots are 1c1

,…,1cn

.

» Let d1=1c1

. So c1=1d1

. Substituting in a0c1n+a1c1

n-1+…+an-1c1+an=0, we get

a0+a1d1+a2d12+…+an-1d1

n-1+an d1n=0. Thus

1c1

,…,1cn

are the roots of anxn+an-

1xn-1+…+a1x+a0=0. (3)Find the equation whose roots are the roots of f(x)=x4-8x2+8x+6=0, each

diminished by 2.» f(x)=(x-2)4+8(x-2)3+16(x-2)2+8(x-2)+6=0(by method of synthetic division). Undertaking the transformation y=x-2, the required equation is y4+8y3+16y2+8y+6=0

Example1.54 If a,b,c be the roots of the equation x3+qx+r=0, find the equation whose roots are (1) a(b+c),b(c+a),c(a+b), (2) a2+b2,b2+c2,c2+a2, (3) b+c-2a,c+a-2b,a+b-2c.

(1)»a(b+c)=∑ ab−bc=q-abca =q+

ra . Thus the transformation is y=q+

rx .

sustituting x=r

y−q in x3+qx+r=0 and simplifying, we obtain the required

equation.(2)»a2+b2=∑a2-c2=-2∑ ab-c2=-2q-c2; hence the transformation is y=-2q-x2,

or, x2=-(y+2q). the given equation can be written as x2(x2+q)2=r2; thus the transformed equation is (y+2q)(y+q)2=- r2.

(3)»b+c-2a=∑ a-3a=-3a; the transformation is y=-3x.

Example1.55 Obtain the equation whose roots exceed the roots of x4+3x2+8x+3=0 by 1. Use Descartes’ Rule of signs to both the equations to find the exact number of real and complex roots of the given equation.

» Let f(x)= x4+3x2+8x+3=(x+1)4-4(x+1)3+9(x+1)2-2(x+1)-1 (by method of synthetic division). By Descartes’ rule, f(-x) has two variations of signs in its coefficients and hence f(x)=0 has either two negative roots or no negative roots; also f(x)=0 has no positive root(since there is no variation of signs in the sequence of coefficients of f). Undertaking the transformation y=x+1, f(x)=0 transforms to g(y)=y4-4y3+9y2-2y-1; considering g(-y), by Descrtes’ Rule,

24

g(y)=0 has a negeative root, say,a. Then f(x)=0 has a-1 as a negative root; the conclusion is f(x)=0 has two negative root, no positive root, does not have 0 as one of its roots and consequently exactly two complex conjugate roots (since coefficients of f are all real).

Example1.56 Find the equation whose roots are squares of the roots of the equation x4-x3+2x2-x+1=0 and use Descartes’rule of signs to the resulting equation to deduce that the given equation has no real root.

»The given equation, after squaring, can be written as (x4+2x2+1)2=x2(x2+1)2 ; undertaking the transformation y=x2, the transformed equation is (y2+2y+1)2=y(y+1)2, that is, y4+3y3+4y2+3y+1=0. The transformed equation, whose roots are squares of the roots of the original equation, has no nonnegative root by Descartes’ Rule of signs; hence the original equation has no real root.

Example1.57 The roots of the equation x3+px2+qx+r=0 are a,b,c. Find the equation whose roots are a+b-2c,b+c-2a,c+a-2b. Deduce the condition that the roots of the given equation may be in A.P.

»a+b-2c=∑ a−3c=-p-3c. We undertake a transformation y=-p-3x, or,x=− y+p

3 .

Thus the equation whose roots are a+b-2c,b+c-2a,c+a-2b is (y+p)3-3p(y+p)2+9q(y+p)-27r=0. If the roots of given equation are in A.P., at least one root of transformed equation is zero, that is , the product of all the roots of the transformed equation is 0. Hence the condition is 2p3-9pq+27r=0.

Example1.58 Find the equation whose roots are cube of the roots of the equation x3+4x2+1=0.

»Let the roots of the given equation be a,b,c. Then x3+4x2+1=(x-a)(x-b)(x-c). In this identity, we replace x by xw and xw2, where w is the imaginary cube roots of unity, to obtain

x3+1+4x2=(x-a)(x-b)(x-c)

x3+1+4x2w2=(xw-a)(xw-b)(xw-c)=(x-aw2)(x-bw2)(x-cw2)

x3+1+4x2w=(xw2-a)(xw2-b)(xw2-c)=(x-aw)(x-bw)(x-cw)

Multiplying respective sides,

(x3+1)3+(4x2)3=(x3-a3)(x3-b3)(x3-c3).

Undertaking the transformation y=x3, (y+1)3+4y2=(y-a3)(y-b3)(y-c3).

Thus the equation whose roots are a3,b3,c3 is (y+1)3+4y2=0, or, y3+7y2+3y+1=0.

25

Cardan’s Method of solving a cubic equation

Example1.59 Solve the equation:x3-15x2-33x+847=0.

Step 1 To transform the equation into one which lacks the second degree term.

Let x=y+h. The transformed equation is y3+(3h-15)y2+(3h2-30h-33)y+(h3-15h2-33h+847)=0. Equating coefficient of y2 to zero, h=5. Thus the transformed equation is y3-108y+432=0 (*)

Step 2 Cardan’s Method

Let a=u+v be a solution of (*). Then a3-108a+432=0 .also a3=u3+v3+3uv(u+v)= u3+v3+3uva; so a3-3uva-(u3+v3)=0. Comparing, uv=36 and u3+v3=-432. Hence u3

and v3 are the roots of t2+432t+363=0. Hence u3=-216=v3. The three values of u are -6,-6w and -6w2, where w is an imaginarycube roots of unite. Since uv=36, the corresponding values of v are -6,-6w2,-6w. Thus the roots of (*) are -12,6,6 and thus the roots of the given equation are (using x=y+5) -7,11,11.

Example1.60 Solve the equation:x3+3x+1=0.

FERRARI’S METHOD OF SOLVING A BIQUADRATIC EQUATION

We try to write down the given biquadratic equation ax4+4bx3+6cx2+4dx+e=0 in the form (ax2+2bx+p)2-(mx+n)2=0; equating coefficients of like powers of x, we get p,m,n; in the process we express the given biquadratic expression as product of two quadratic expressions and hence solve the given equation.

Example1.61 Solve the equation x4-10x3+35x2-50 x+24=0.

»We try to find p,m,n such that the given equation can be written in the form (x2-5x+p)2-(mx+n)2=0. Equating coefficients of like powers of x, we get 35=25+2p-m2,-50=-10p-2mn and 24=p2-n2. Eliminating m and n,

(p2-24)(2p-10)=m2n2=(5p-25)2 (*)

p=5 is a solution of (*), hence m=0,n=±1. Thus the given equation is (x2-5x+5)2-1=0, that is, (x2-5x+6)(x2-5x+4)=0. Hence the roots of given equation are 1,2,3,4.

Example1.62 Solve the equation x4+12x-5=0.

»We try to find p,m,n such that the given equation can be written in the form (x2+p)2-(mx+n)2=0; equating coefficients of like powers of x, we get p2-n2=-5, 2mn=-12,2p-m2=0. Eliminating m,n, we get 36=m2n2=(p2+5)2p; hence p=2.

26

Thus m=±2, n=±3. Since mn=-6<0, we take m=2, n=-3. Thus the given equation can be written in the form (x2+2)2-(2x-3)2=0, that is, x2+2x-5=0 and x2-2x+5=0. Hence the roots of given equation are -1±√2, 1±2i.

CHAPTER III

INEQUALITIES

I1(Weierstrass) If a1,…,an are all positive real numbers less than 1 and

sn=a1+…+an, then 1-sn<(1-a1)…(1-an)<1

1+sn and 1+sn<(1+a1)…(1+an)<

11−sn

.

I2(Cauchy-Schwarz) If a1,…,an and b1,…,bn be real numbers , then (a12+

…+an2)(b1

2+…+bn2)≥(a1b1+…+anbn)2, equality occurs when ai=kbi for all i

for some non-zero real k or when ai=0 for all I or bi=0 for all i. I3If a1,…,an be n positive real numbers, not all equal, and p,q are rational

numbers, then a1

p+q+a2p+q+…+an

p+q

n> or <

a1p+a2

p+…+anp

n.a1

q+a2q+…+an

q

naccording

as p and q have the same or opposite signs. I4(A.M.,G.M.,H.M.)If a1,…,an be n positive real numbers, then

a1+…+an

n≥ n√a1 …an≥

n1a1

+…+ 1an

.

I5(Weighted A.M.,G.M.,H.M.)If a1,…,an be n positive real numbers and p1,…,pn be n positive rational numbers, then p1a1+…+ pnan

p1+…+ pn

≥ (a1p1 a2

p2…anpn )

1p1+…+pn ≥

p1+…+ pn

p1

a1

+…+pn

an

.

I6If a>0,a≠1, and x,y are positive rational numbers, then ax−1x

> a y−1y

if

x>y. I7If a >0,a≠1 and m be a rational number, then am-1> or <m(a-1)

according as m does not or does lie between 0 and 1. I8If a1,…,an be n positive real numbers, not all equal, and m be a rational

number, then a1

m+…+anm

n>or<( a1+…+an

n )m

according as m does not or does

lie between 0 and 1.

27

I9(Holder) If a1,…,an; b1,…,bn;….;l1,…,ln be m sets of positive real numbers each containing n members and if α 1,…,αmbe positive rational numbers such that α 1+…+αm=1, then a1

α1 b1α 2 …l1

αm+…+anα1 bn

α2 …lnαm≤ (a1+…+an )α 1… (l1+…+ln )αn.

I10(Jensen) Let a1,…,an be n positive real numbers and r,s are positive rational numbers. Then (a1

r+…+anr)1/r>(a1

s+…+ans)1/s, if r<s.

I11(Minkowski)Let a1,…,an and b1,…,bn be all positive real numbers and r is a positive rational number. Then (a1

r+…+anr)1/r+(b1

r+…+bnr)1/r≥

[(a1+b1)r+…+(an+bn)r]1/r, when r>1.(a1

r+…+anr)1/r+(b1

r+…+bnr)1/r≤[(a1+b1)r+…+(an+bn)r]1/r, when 0<r<1.

EXAMPLES

1. If a,b,c be all real numbers, prove that a2+b2+c2≥ ab+bc+ca»a2+b2+c2−¿ ab+bc+ca)=1/2[(a-b)2+(b-c)2+(c-a)2]≥0.

2. If a,b,c be positive real numbers, prove that a2+b2

a+b+ b2+c2

b+c+ c2+a2

c+a≥a+b+c .

»2(a2+b2)=(a+b)2+(a-b)2≥(a+b)2. Hence a2+b2

a+b≥a+b

2. Similarly b

2+c2

b+c≥b+c

2

and a2+c2

a+c≥

a+c2

. Adding respective sides, we get the result.

3. If a,b,c be all positive real numbers and n be a positive rational number, prove that an(a-b)(a-c)+bn(b-a)(b-c)+cn(c-a)(c-b)≥0.»If a=b=c, nothing remains to prove. If a=b≠c, LHS equals cn(c-a)2>0. Next, let no two of a,b,c are equal. Since the expression on the left is symmetric in a,b,c, we can assume without loss of generality a>b>c. Then an(a-b)(a-c)+bn(b-a)(b-c)=(a-b)[an(a-c)-bn(b-c)]>0 (since a>b and n positive). Also cn(c-a)(c-b)>0. Hence an(a-b)(a-c)+bn(b-a)(b-c)+cn(c-a)(c-b)≥0.

4. If a,b,c,d be positive real numbers, not all equal, prove that (a+b+c+d)(1a+ 1b+ 1c+ 1d

¿>16.

»By I3, a+b+c+d4

.a−1+b−1+c−1+d−1

4> a0+b0+c0+d0

4=1. Hence.

5. If n be a positive integer, prove that 1.3.7 …(2n−1)

2.4 .8…2n < 2n

2n+1−1.

»By Weierstrass inequality, (1−12 )(1−

1

22 )…(1− 1

2n )< 1

1+12+…+

1

2n

=2n

2n+1−1.

Hence the result.

28

6. If n>1, prove (1+2+…+n)(1+12+…+ 1

n¿>n2 .

»By I3, 1+2+…+n

n .1+ 1

2+…+ 1

nn

>1.1+2.

12+…+n .

1n

n=1. Hence.

7. If a1,…,an be n positive real numbers in ascending order of magnitude,

prove that a1<a1

2+…+an2

a1+…+an

<an.

»a1=a1

2+a1a2+…+a1 an

a1+…+an

<a1

2+…+an2

a1+…+an

<a1an+a2 an+…+an

2

a1+…+an

=an.

8. If a,b,c be real numbers, prove that (a+b-c)2+(b+c-a)2+(c+a-b)2≥

ab+bc+ca.»LHS= 3(a2+b2+c2)-2(ab+bc+ca)≥3(ab+bc+ca)-2(ab+bc+ca)=ab+bc+ca.

9. If a,b,c,x,y,z are real numbers and a2+b2+c2=1=x2+y2+z2, prove that -1≤

ax+by+cz≤1.»By Cauchy-Schwarz inequality, (ax+by+cz)2≤(a2+b2+c2)(x2+y2+z2)=1. Hence.

10.If a,b,c be positive real numbers, prove ( a+b+c3 )

3

≥a( b+c2 )

2

.

»1.a+2.

b+c2

1+2≥ [a .( b+c

2 )2]

13. Hence the result.

11.If a,b,c be positive real numbers, prove ab+bc+ ca≥3.

»ab+ bc+ ca

3≥

3√ ab

bc

ca=1. Hence.

12.If a,b,c be positive real numbers, 9

a+b+c≤

2a+b

+ 2b+c

+ 2c+a

≤1a+ 1b+1c .

»

2a+b

+ 2b+c

+ 2c+a

3≥

3a+b

2+b+c

2+c+a

2

=3

a+b+c. Hence

9a+b+c

≤2

a+b+ 2b+c

+ 2c+a . Again,

1a+ 1b

2≥

2a+b

,

1c+ 1b

2≥

2c+b

,1a+ 1c

2≥

2a+c

. Hence

2a+b

+ 2b+c

+ 2c+a

≤1a+ 1b+ 1c

.

13.If a,b,c be positive real numbers, a

b+c+ bc+a

+ ca+b>

32 , unless a=b=c.

29

»( b+c

a )−1

+( c+ab )

−1

+( a+bc )

−1

3>( b+c

a+ c+a

b+ a+b

c3 )

−1

=[ 3−( 1a+ 1b+ 1c )

3 ]−1

=

[1−13 ( 1

a+ 1b+ 1c )]

−1

>1. Hence ab+c

+ bc+a

+ ca+b

>3>32

.

14.If a,b,c be positive real numbers and a+b+c=1, prove 1

1−a+ 1

1−b+ 1

1−c≥

92.

»

11−a

+ 11−b

+ 11−c

3≥

33−(a+b+c )

=32

. Hence the result.

15.If a,b,c be positive real numbers and a+b+c=1, prove that

(a+ 1a)

2

+(b+ 1b)

2

+(c+ 1c)

2

≥3313.

»a2+b2+c2

3>( a+b+c

3 )2

(by I8, since m=2 does not lie between 0 and 1)=19.

a−2+b−2+c−2

3>( a+b+c

3 )−2

=9. Hence the result.

16.If a,b,c,d are positive real numbers, prove that a6+b6+c6+d6≥

abcd(a2+b2+c2+d2).

»a6+b6+c6+d6

4≥

a4+b4+c4+d4

4.a

2+b2+c2+d2

4≥

14

4√a4 b4 c4 d4(a2+b2+c2+d2).Hence.

17.If a,b,c,d are positive real numbers, prove that (1+a4)(1+b4)(1+c4)(1+d4)≥(1+abcd)4.

»By Holder’s inequality, [1.1.1.1+(a4 )14 (b4 )

14 (c4 )

14 (d4 )

14]

≤(1+a4)14 (1+b4)

14 (1+c4)

14 (1+d4)

14.Hence.

18.If a,b,c,d are four positive real numbers, 4(a4+b4+c4+d4)≥(a+b+c+d)(a3+b3+c3+d3)≥16abcd.

»∑ a4

4≥∑ a3

4∑ a

4. Thus 4∑ a4 ≥∑ a3∑ a≥4 (abcd )

34 .4 (abcd )

14 =16abcd .

19.If a,b,c,d be positive and s=a+b+c+d, then 81abcd≤(s-a)(s-b)(s-c)(s-d)

≤81

256s4.

» [(s−a)(s−b)(s−c )(s−d )]14 ≤

4 s−(a+b+c+d )4

=3 s4

. Hence (s-a)(s-b)(s-c)(s-d)

≤81

256s4.

30

Again, s−a

3=b+c+d

3≥ 3√bcd. Similarly others. Hence 81abcd≤(s-a)(s-b)(s-

c)(s-d).20.If a,b,c,d be positive and s=a+b+c+d, then

16s

≤3

s−a+ 3s−b

+ 3s−c

+ 3s−d

≤1a+ 1b+ 1c+ 1

d.

»

1a+ 1b+ 1c

3≥

3a+b+c

=3

s−d. Adding similar terms,

1a+ 1b+ 1c+ 1d

≥3

s−a+ 3

s−b+ 3

s−c+ 3

s−d. Again,

1s−a

+…+ 1s−d

4≥

44 s−(a+b+c+d )

=4

3 s. Hence 16

s≤

3s−a

+ 3s−b

+ 3s−c

+ 3s−d

.

21.If n be a positive integer >1, then nn>1.3.5…(2n-1)

»n=1+(2n−1)

2>√1. (2n−1), n=

3+(2n−3)2

≥√3.(2n−3),…,n=

(2n−3 )+32

≥√(2n−3 ) .3,n=(2n−1 )+1

2>√ (2n−1 ) .1. Multiplying respective

sides, we get the result.

22.If n be a positive integer >1, prove that 1.3.5….(2n−1)

2.4 .6….(2n) >1

2√n .

»n+(n−1)

2≥√n(n−1). Thus 2n−1

2n≥√ n−1

n. Hence 1

2≥

12,34≥√ 1

2,56≥√ 2

3,…,

2n−12n

≥√ n−1n

. Multiplying respective sides, we get the result.

23.If n be a positive integer, prove 1

n+1+ 1n+2

+…+ 13n+1

>1.

»

1n+1

+…+ 13n+1

2n+1>

2n+1(n+1 )+…+(3n+1 )

=1

2n+1. Hence the result.

24.If n be a positive integer, prove (1− 1n+1 )

n+1

>(1−1n )

n

.

»Let a=1-1

n+1 , m=n+1n

. from am-1¿m(a-1), we get

(1− 1n+1 )

m

> n+1n ( −1

n+1 )+1=1−1n.Hence .

25.If a1,…,an be positive and s=a1+…+an, prove that s

s−a1+

ss−a1

+…+ ss−a1

≥n2

n−1.

»∑ s

s−a1

n≥

n

∑ s−a1

s

=n

n−ss= nn−1

. Hence.

31

26.If a,b,c be positive rational, prove ( a2+b2+c2

a+b+c )a+b+c

≥aabb cc≥( a+b+c3 )

a+b+ c

.

»a .a+b .b+c . c

a+b+c≥(aabb cc )

1a+b+c ≥

a+b+caa+ bb+ cc

=a+b+c3

. Hence.

27.If a,b,c are positive rationals,(a+b)c(b+c)a(c+a)b<{23(a+b+c)}

a+b+c

, unless

a=b=c.

»[(a+b)c (b+c )a(c+a)¿¿b ]1

a+b+c <c (a+b )+a (b+c )+b(c+a)

a+b+c=2

ab+bc+caa+b+c

=23(a+b+c)¿

32