COMPLEX NUMBERS AND QUADRATIC EQUATIONS

1 ChapterCOMPLEX NUMBERS ANDQUADRATIC EQUATIONSW. R. Hamilton(1805-1865) Mathematics is the Queen of Sciences and Arithmetic is the Queen ofMathematics. GAUSS IntroductionIn earlier classes, we have studied linear EQUATIONS in oneand two variables and QUADRATIC EQUATIONS in one have seen that the equation x2 + 1 = 0 has no realsolution as x2 + 1 = 0 gives x2 = 1 and square of everyreal number is non-negative. So, we need to extend thereal number system to a larger system so that we canfind the solution of the equation x2 = 1.

2 In fact, the mainobjective is to solve the equation ax2 + bx + c = 0, whereD = b2 4ac < 0, which is not possible in the system ofreal COMPLEX NumbersLet us denote 1 by the symbol i. Then, we have 21i= . This means that i is asolution of the equation x2 + 1 = number of the form a + ib, where a and b are real NUMBERS , is defined to be acomplex number . For example, 2 + i3, ( 1) + 3i, 1411i + are COMPLEX the COMPLEX number z = a + ib, a is called the real part, denoted by Re z andb is called the imaginary part denoted by Im z of the COMPLEX number z.

3 For example,if z = 2 + i5, then Re z = 2 and Im z = COMPLEX NUMBERS z1 = a + ib and z2 = c + id are equal if a = c and b = NCERTnot to be republished98 MATHEMATICSE xample 1 If 4x + i(3x y) = 3 + i ( 6), where x and y are real NUMBERS , then findthe values of x and We have4x + i (3x y) = 3 + i ( 6).. (1)Equating the real and the imaginary parts of (1), we get4x = 3, 3x y = 6,which, on solving simultaneously, give 34x= and 334y=. Algebra of COMPLEX NumbersIn this Section, we shall develop the algebra of COMPLEX Addition of two COMPLEX NUMBERS Let z1 = a + ib and z2 = c + id be any twocomplex NUMBERS .

4 Then, the sum z1 + z2 is defined as follows:z1 + z2 = (a + c) + i (b + d), which is again a COMPLEX example, (2 + i3) + ( 6 +i5) = (2 6) + i (3 + 5) = 4 + i 8 The addition of COMPLEX NUMBERS satisfy the following properties:(i)The closure law The sum of two COMPLEX NUMBERS is a complexnumber, , z1 + z2 is a COMPLEX number for all COMPLEX numbersz1 and z2.(ii)The commutative law For any two COMPLEX NUMBERS z1 and z2,z1 + z2 = z2 + z1(iii)The associative law For any three COMPLEX NUMBERS z1, z2, z3,(z1 + z2) + z3 = z1 + (z2 + z3).

5 (iv)The existence of additive identity There exists the COMPLEX number0 + i 0 (denoted as 0), called the additive identity or the zero complexnumber, such that, for every COMPLEX number z, z + 0 = z.(v)The existence of additive inverse To every COMPLEX numberz = a + ib, we have the COMPLEX number a + i( b) (denoted as z),called the additive inverse or negative of z. We observe that z + ( z) = 0(the additive identity). Difference of two COMPLEX NUMBERS Given any two COMPLEX NUMBERS z1 andz2, the difference z1 z2 is defined as follows:z1 z2 = z1 + ( z2).

6 For example,(6 + 3i) (2 i) = (6 + 3i) + ( 2 + i ) = 4 + 4iand(2 i) (6 + 3i) = (2 i) + ( 6 3i) = 4 4i NCERTnot to be republishedCOMPLEX NUMBERS AND QUADRATIC EQUATIONS Multiplication of two COMPLEX NUMBERS Let z1 = a + ib and z2 = c + id be anytwo COMPLEX NUMBERS . Then, the product z1 z2 is defined as follows:z1 z2 = (ac bd) + i(ad + bc)For example, (3 + i5) (2 + i6) = (3 2 5 6) + i(3 6 + 5 2) = 24 + i28 The multiplication of COMPLEX NUMBERS possesses the following properties, whichwe state without proofs.

7 (i)The closure law The product of two COMPLEX NUMBERS is a COMPLEX number ,the product z1 z2 is a COMPLEX number for all COMPLEX NUMBERS z1 and z2.(ii)The commutative law For any two COMPLEX NUMBERS z1 and z2,z1 z2 = z2 z1.(iii)The associative law For any three COMPLEX NUMBERS z1, z2, z3,(z1 z2) z3 = z1 (z2 z3).(iv)The existence of multiplicative identity There exists the COMPLEX number1 + i 0 (denoted as 1), called the multiplicative identity such that = z,for every COMPLEX number z.(v)The existence of multiplicative inverse For every non-zero complexnumber z = a + ib or a + bi(a 0, b 0), we have the COMPLEX number22 22a biab ab+++ (denoted by 1z or z 1 ), called the multiplicative inverseof z such (the multiplicative identity).

8 (vi)The distributive law For any three COMPLEX NUMBERS z1, z2, z3,(a) z1 (z2 + z3) = z1 z2 + z1 z3(b) (z1 + z2) z3 = z1 z3 + z2 Division of two COMPLEX NUMBERS Given any two COMPLEX NUMBERS z1 and z2,where 20z , the quotient 12zz is defined by11221zzzz=For example, letz1 = 6 + 3i and z2 = 2 iThen121(6 3 )2zizi =+ = ()63i+ ()()()2222122121i + + + NCERTnot to be republished100 MATHEMATICS= ()2635ii+ + = ()()1112 36 69 1255ii + + = + Power of i we know that()321iiii i== = ,()()224211ii== =()()22521iiiii== =, ()()336211ii== = , , we have 122111,1,11iiiiiiii = == === 34341111,111iiiiiiiii == == === In general, for any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = 1, i4k + 3 = The square roots of a negative real numberNote that i2 = 1 and ( i)2 = i2 = 1 Therefore, the square roots of 1 are i, i.

9 However, by the symbol 1 , we wouldmean i , we can see that i and i both are the solutions of the equation x2 + 1 = 0 orx2 = ()()2233i= i2 = 3 ( 1) = 3()23i = ()23 i2 = 3 Therefore, the square roots of 3 are 3i and 3i .Again, the symbol 3 is meant to represent 3i only, , 3 = , if a is a positive real number , a = 1a = ai,We already know that ab = ab for all positive real number a and b. Thisresult also holds true when either a > 0, b < 0 or a < 0, b > 0. What if a < 0, b < 0?Let us that NCERTnot to be republishedCOMPLEX NUMBERS AND QUADRATIC EQUATIONS 101()()21111i= = (by assuming ab = ab for all real NUMBERS )= 1 = 1, which is a contradiction to the fact that = , ab ab if both a and b are negative real , if any of a and b is zero, then, clearly, ab ab == Identities We prove the following identity()222121 2 122zz z z zz+=++, for all COMPLEX NUMBERS z1 and We have,(z1 + z2)2 = (z1 + z2) (z1 + z2),= (z1 + z2) z1 + (z1 + z2) z2(Distributive law)

10 = 22121122zzzzzz+++(Distributive law) = 22112122zzzzzz+++(Commutative law of multiplication) = 2211222zzzz++Similarly, we can prove the following identities:(i)()22212112 22zzz zz z = +(ii)()33223121 12 12 233zz z zz zz z+=+ + +(iii)()33223121 12 12 233zz zzzzzz = + (iv)()()221 2 1212z zz z z z=+In fact, many other identities which are true for all real NUMBERS , can be provedto be true for all COMPLEX 2 Express the following in the form of a + bi:(i)()158ii (ii)()()2ii 318i Solution(i)()158ii = 258i = ()518 = 58= 508i+(ii)()()3128ii i = 512888i = ()221256i 1256ii=.