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Notes On Topics Of Algebra

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Published in: Mathematics
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Attached files contain the basic concept of Algebra topics.

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  1. The complex number system Indian mathematician Mahavira (850 A.D.) was first to mention in his work 'Ganitasara sangraha'; 'As in nature of things a negative (quantity) is not a square (quantity), it has, therefore, no square root'. Hence, there is no real number x which satisfies the polynomial equation A symbol V(-l) , denoted by letter i was introduced by Swiss Mathematician, Leonhard Euler to provide solution of equation x2+1=0.'i'was regarded as a fictions or imaginary number which could be manipulated algebraically like an ordinary real number, except that its square was -1. The letter i was used to denote possibly because i is the first letter of the Latin word, 'imaginarius'. To permit solutions of such polynomial equations, the set of complex numbers is introduced. We can consider a complex number as having the form a+bi where a and b are real numbers. It is denoted by z, i.e., z=a+bi. 'a' is called as real part of j which is denoted by (Re z) and 'b' is called as imaginary poart of z which is denoted by (Img). Any complex number is:- (ii) (iii) Note: Purely real, if b = 0 Imaginary, if b 0. Purely imaginary, if a (a) The set R of real numbers is a proper submit of the complex numbers. number system is (b) (c) (d) NCWCICQCRCC Zero is purely real as well as purely imaginary but not imaginary. i = V(-l), is called the imaginary unit. Also, i2 VaVbVc = V(abc........) iff. At least one of a,b,c . . is non-negative. If j = a+ib, then a-ib is called complex conjugate of j and written as Real numbers satisfy order relations where as imaginary order relation, meaningless. Hence, the complete 3 + i < 2, are
  2. Alqebraic Operations on Complex Numbers Fundamental Operations with complex numbers: In performing operations with complex numbers we can proceed as in the algebra of real numbers replacing P by -1 when it occurs. 1. Addition (a +bi) + (C+dl) 2. Subtraction (a 4-bl) — • (c+di) = (a-c)i + (b-d)i In real numbers if a2 + b2=0 then a=b=0; however in complex numbers, 3. Multiplication (a+bi)(c+di) = ac+bc i+ bi2 = (ac-bd)+(ad+bc)i 4. Division (a+bi)/(c+di)=(a+bi)/(c+di)x(c-di)/(c-di) =(ac+bd)/(c2+d2 )+(bc-ad)/(c2+d2) i Inequalities in imaginary numbers are not defined. positive or negative. GC: z 4 + zi < 2+4i are meaningless. Z12 + z22 does not imply Zl = 7-2 (ac+bd+(bc-ad)i)/(c2-d2i2 ) There is no validity if we say that imaginary is
  3. Equality in complex numbers Two complex numbers Zl = al + ib, and = are equal. = z2 = Rs(Z1) = Re(zl) and Im(Z1) Representation of a complex number a2 + ib2 are equal if their real and imaginary parts Early in the 19th century, Karl Friedrich Gauss and William Hamilton independently and almost simultaneously proposed idea of redefining complex number as ordered pair of real numbers, i.e., a+ib = (a,b). To each complex number there corresponds one and are point in plane, and conversely to each point in the plane there corresponds one and only are complex number. Because of this we often refer to the complex number z as the point z. (a) Cartesian form (Geometric Representation): Every complex number z = X+iy can be represented by a point on the Cartisiar plane known as complex plane (Argand plane) by the ordered pair (x, y). Imaginary axis p(x, Y) 8 eal axis Length OP is called modules of the complex number which is denoted by Izl and 0 is the argument or amplitude. Izl = V(x2 + y2) and tano = (y/x) (angle made by positive x-axis). Note: (i) Argument of a complex number is a many values function. If 9 is the argument of a complex number then ann + 9; n E I will also be the argument of that complex number. Any two arguments of a complex number differ by 2nn . (ii) The unique value of 9 such that -11
  4. (iii) By specifying the modules and argument a complex number is defined completely. For the complex number 0+0i, the argument is not defined as this is the only complex number which is only given by its modules. r(cos9 - i sin 9). (b) Trigonometric / Polar Representation: — r, arg z = 91 i = — r(cos9 + i sin 9) where Izl — cos9 + i sine is also written as cis9 . Note: (c) Euler's Formula Z = re19, Izl=r, arg z i = re-IO Proof of this formula is beyond scope of present discussion. A heuristic proof serving as motivation for this formula is by considering expansion. = 14- X/ 1! + X 2/2! + X 3/3! Put x = i 9 = (l - Q2/2 + Q4/41 = cos 9 + sin 9 If 9 is real then cos9 = Note: - e-i0)/2i sino = (c) Vectorial Representation Q3/3! + Q5/51 (ei6 + Every complex number can be considered as the position vector of a point. If the point P represents the complex number z then OP = z and I OPI = Izl.
  5. Arqument of a Complex Number Argument of a non-zero complex number p(z) is denoted and defined by arg (z)= angle which OP makes with the positive direction of real axis. If OP=lzl and arg (z)= 9, then obviously z=r (cos 9 + i sin 9), called the polar form of z. 'Argument of z' would mean principal argument of z (i.e., argument lying in (-TT,TT )) unless the context requires otherwise. Thus argument of a complex number z=a+ib = r (cos 9 + i sin 9) is the value of 9 satisfying r cos 9 = a and r sin 9 = b. Let 9 = tan-I lb/al P.V. arg z= 9 P.V. arg z = 11/2 (iii) a0
  6. P(a i b) P.V. arg (iv) a
  7. (vi) P.V. arg z = -n/ 2 (vii) a >0, b0, b=0 P.V. arg z L-
  8. Geometrical Representation of fundamental operations (i) Geometrical representation of addition - Q(Z2) (PZI) o If two points P and Q represents complex Zl and z2 respectively in the Argard Plane, then the sum Zl+Z2 is represented by the extremely R of the diagonal OR of parallelogram OPRQ having OP and OQ as two adjacent sides. Geometrical representation of subtraction - o (iii) Modules and argument of multiplication of two complex numbers - Theorem: For any two complex numbers Zl, z2 We have, IZI, Z21 = IZII IZ21 and arg(zl, z2) = arg(zl) + arg(z2)
  9. Proof: Zl = = rl r2 ei(91+02) ZIZ2 = > IZI arg (ZIZ2) = arg(zl) + arg(z2) i.e., to multiply two complex numbers, we multiply their absolute values and add their arguments. Note: (i) P.V. arg(z1Q) * P.V. arg(zl) + P. V. arg(z2) (ii) IZIZ2 . (iii) arg(z1.z2 . znl = IZII IZ21 .... ..... Iznl argz1+argz2+ . + argzn Geometrical representation of multiplication of complex numbers - Let P, Q be represented by zl=rl e191, zl=rl e162 respectively. To find point R representing complex number Zl z2, we take a point L on real axis such that OL=I and draw triangle OQR similar to triangle OLP. Therefore, R(ZI Z2) Q(Z2) 62—61 P(ZI) OR/OQ = OP/OL = OR rl a and (QOR = R(z1z2) Q(Z2) P(ZI) 9, LOR = Z LOP + /_ POQ + (QOR
  10. = QI + Q2 - = QI + Q2 QI + QI Hence, R is represented by Zl z2 = rl r2 ei(Q1+Q2) Modules and argument of division of two complex numbers - Theorem - If Zl and z2 ( *0 ) are two complex numbers, then IZ1/Z21 = Iz11/lz21 and arg(z1/z2) = arg Zl - arg z2 Note: P.V. arg(z1/z2) * P.V. arg (Zl) - P.V. arg (Q) (vi) Geometrical representation of the division of complex numbers- r, eiQ and z2 — Let P, Q be represented by r2 respectively. To find point R representing complex number Zl/Z2 , we tale a point L on real axis such that OL=I and draw a triangle OPR similar to OQL. Therefore, OP/OQ = OR/OL = > OR = rl/r2 L LOP - L ROP = QI-Q2 p(Z1) R(Z1/Z2) 02 and L LOR= Hence, R is represented by Zl/Z2 = rl/r2 ei(Ql-Q2)
  11. Coniuqate of a complex numbers Conjugate of a complex number z = a + ib is denoted and defined by = a-ib. In a complex number if we replace i by -i, we get conjugate of the complex number. is the mirror image of z about real axis on Argand's Plane. Geometrical representation of conjugate of complex number - lip = zi arg (i) = - arg (z) General value of arg Properties: 2nn P.V. arg(z) O p(z) If z = x+y, then x= z+ 2/2, y = z+i/2 z is purely real = 0 = j is purely imaginary (iii) (iv) (v)
  12. (vii) (viii) (ix) (x) 2 = --2 (32 0) Imaginary roots of polynomial equations with real coefficient occur is conjugate pairs. If w=f(z), then v,
  13. Distance, Trianqle inequality If Zl = + iY1, z2 = + iY2, then distance between points Zl, z2 is argard plane is Izvz21= + (Yi-Y2)2) B(Z2) C(Z1+Z2) A(ZI) o In triangle OAC, OC OA + AC OA AC + OC AC OA + OC Using these inequalities we have IIZII - IZ211 Iz1+z21 IZII + IZ21 Similarly from triangle OAB, we have IIZII - IZ211 IZ1-Z21 IZII + IZ21 Note: (a) Il = Iz1+z21 , I Zl-Z21 between Zl and z2. (b) IZI + Z21 = Iz11+lz21 , the same side of origin. = IZII + IZ21 iff origin, Zl, and z2 are collinear and origin lies Il = IZ1-Z21 iff origin, Zl and z2 are collinear and Zl and z2 lies on
  14. Important results in context with rotation: = 0 represents points (non-zero) on a ray emanating from origin making an angle 9 arg z with positive direction of real axis. z O Re(z) (ii) arg (z-zl)= 0 represents points ( *3,) on ray emanating from z, positive direction of real axis. z 6 Rotation theorem: making an angle 9 with If P(ZI) and Q(Z2) are two complex numbers such that IZII where 9 = (POQ. = IZ21, then z2=z1 eld If P(ZI), Q(Z2) and R(Z3) are three complex numbers and /_PQR=Q, then
  15. 23 — p (21) If P(ZI), Q(Z2), R(Z3) and s(z4) are three complex numbers and ZSTQ= 9, then (iii) 2. 3. 4. 5. p(Z1) 6 R (23) S(Z4) Q(Z3) amp (z) = 9 is a ray emanating from the origin inclined at an angle 9 to the x-axis. Iz-al=lz-bl is the perpendicular bisectors of the line joining a to b. The equation of a line joining Zl and z2 is given by z = Zl + t(u-z2) where t is a real parameter. = Zl (1+it) where t is a real parameter is a line through the point z, and perpendicular to the z line joining z, to the origin. 6. The equation of a line passing through Zl and z2 can be expressed in the determinant form as
  16. z 1 1 1 This is also the condition for three complex numbers z, Zl, z2 to b collinear. 7. The equation of a circle having centre zo and radius p is Iz-zol = porzZ -zoi- a -z + Zozo -p 2 = 0 which is of the form zi+ a z + a: + k = 0, kis real. Centre is -a and radius - k) circle will be real if a - k 2 0. 8. The equation of the circle described on the line segment joining Zl and z2 as diameter is arg )= ±11/2 or (z-ZI) 9. Condition for four given points Zl, u, z3 and z4 to be concyclic is the number should be Hence, the equation of a circle through 3 non-collinear points Zl, z2 and z3 can be taken as (z-ZI = > (z-Z2 is 10. (iii) 11. Arg((z-z1)/(z-z2)) = 0 represent a line segment if 9 pair of ray if 9 = 0 a part of circle, if Area of triangle formed by the points Zl, z2 and z3 is 1 1 1 1
  17. 12. 13. 14. Perpendicular distance of a point zo from the line a z + ai+r=0 is + a: + r)/alall (i) Complex slope of a line a z + ai + r = 0 w = - a/a (ii) Complex slope of a line joining the points Zl and z2 is (iii) Complex slope of a line making Q angle with real axis w=12i6 Dot and Cross product Let Zl = Xl + i and z2 = + i be two complex numbers (vectors). The dot product (also called the scalar product) of and z2 is defined by Zl.Z2 = IZII IZ21 cos 9 Re { äIZ2} - + YIY= = = 1/2{ilZ2 + Where 9 is the angle between zl and z2 which lies between 0 and n . If vectors Zl, z2 are perpendicular then Zl.Z2 = 0 = > Zl/Z2 + z2/z2 = 0 i.e. sum of complex slopes = 0 The cross product of Zl and z2 is defined by = IZII IZ21 sin 91 XiY2 - YiX2 = Im{i1Q } = l/zi (iZ2 - Zli2) If vectors Zl, z2 are parallel then Zl x z2 = 0 Zl/Z2 = z2/z2 i.e., complex slopes are equal. Note: WI and w2 are complex slopes of two lines.
  18. If lines are parallel then WI = w2 If lines are perpendicular then WI + w2 15. 16. line 17. 18. If Iz-Z11 + Iz-Z21 = K > IZ1-Z21 then locus of z is an ellipse focii are Zl and z2. If Iz-zol = I a z + ai + r/ 2K Il then locus of z is parabola whose focus is zoand directrix is the a z + ai + r = 0 (provided äzo+ aio + 0) If = k < IZ1-Z21 then locus of z is a circle If then locus of z is a hyperbola, whose focii are Zl and Q.
  19. Demoivre's theorem: Case I Statement: If n is any integer, then (i) (cos 9+ i sin = cos no + i sin no (ii) (cos 91 + i sin 91) = (cos 92 + i sin 92) ..... .... (COS On + i sin On) = cos 91 + 92 + 93 • Case 11 Statement: If p, q E and q * 0, where k = 0,1,2,3,.....,q-1 Note: + On) + i sin (91 + 92 + • then (cos 9 + i sin 9)P/q 93) Continued product of the roots of a complex quantity should be determined using theory of equations.
  20. Cube Root of Unity The cube roots of unity are 1, (-1+iV3)/2, (-1-iV3)/2 . If w is one of the imaginary cube roots of unity then 1+w+w2=0. In general 1 + wn+w2n=0• where n E I but is not the multiple of 3. (iii) In polar form the cube roots of unity are: cos O + i sin O; cos2n/3 + isin2n/3; cos4n/3 + isin4n/3. (iv) The three cube roots of unity when plotted on the argand plane constitutes the vertices of an equilateral triangle. (v) The following factorization should be remembered : (a, b, c E R and w is the cube root of unity) a3 - (a-b) (a-wb) (a-x2b); x2 +x+1=(x-w)(x-w2); a3 + (a+b) (a+wb) (a+x2b); a2 +ab+b2=(a-bw)(a-bw2) a3 + b3 + c3 - 3abc = (a+b+c) (a+wb+w2c) (a+w2b+wc); nth Roots of unity if 1, al, 02,......an-1 are the n, nth roots of unity then: 1m (Z) a 2 a Re (z) They are in G.P. with common ratio i(211/n) • an-IP = 0 if p is not an integral multiple of n. IP + a IP + + .
  21. (iii) (iv) (1 + a + 02) 1. al • 02.03. . an-I - an-I) = n and, .......(1 + an-I) = 0 if n is even and 1 if n is odd. = 1 or -1 According as n is odd or even. The sum of the following series should be remembered: cos 9 + cos 29 + cos 39 + . = (sin (n6/2)/sin9/2) sin 9 + sin 29 + . . + cos no . + sin no Note: = sin((n+1)/2)9 2 11/2 then the sum of the above series vanishes. Logarithm of a complex quantity - (i) Loge(a + iß) = 1/2 loge(a2 + ß2) + i(2nn + tan-I(ß/a)) where n E I. (ii) ii represents a set of positive real numbers given by e-(2nn + (1/2) n E I.
  22. Reflection points for a straiqht line Two given points P and Q are the reflection points for a given straight line if the given line is the perpendicular bisector of the segment PQ. Note that the two points denoted by the complex numbers Zl and z2 will be the reflection points for the straight line a z + ai + r = 0 iff + aZ2+ r where r is real and a is non-zero complex constant. Inverse points w.r.t. a circle - Two points P and Q are said to be inverse w.r.t. a circle with centre 'O' and radius P if the points O, P, Q are collinear and P, Q are on the same side of O. OP.OQ=p2 Note: Two points Zl and z2 will be the inverse points w.r.t. the circle ai + a z + ai+r= 0 iff ai2+ a Zl + aZ2+r=0 Ptolemy's theorem It states that the product of the lengths of the diagonals of a converse quadrilateral inscribed in a circle is equal to the sum of the products of lengths of the two pairs of its opposite sides, i.e., IZ1-Z31 IZ2-Z41 = IZ1-Z21 Iz3-z41+lz1-z41 IZ2-Z31