3 Introduction We know that the equation x2 + 1 = 0 has no real solution 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. 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 numbers.
4 solution of the equation x2 + 1 = 0. Complex NumbersLet us denote −1 by the symbol i. Then, we have i2 = −1 . This means that i is asolution of the equation x2 + 1 = 0.A number of the form a + ib, where a and b are real numbers, is defined to be acomplex number. For example, 2 + i3, (– 1) + i 3 ,Is a complex numbers.For 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. For example,if z = 2 + i5, then Re z = 2 and Im z = 5.Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d.
5 Algebra of Complex Numbers In this Section, we shall develop the algebra of complex numbers.Addition of two complex numbers Let z1 = a + ib and z2 = c + id be any twocomplex numbers. Then, the sum z1 + z2 is defined as follows:z1 + z2 = (a + c) + i (b + d), which is again a complex number.For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8The addition of complex numbers satisfy the following properties:(i) The closure law The sum of two complex numbers is a complexnumber, i.e., 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
6 Difference of two complex numbers Given any two complex numbers z1 and z2, the difference z1 – z2 is defined as follows:z1 – z2 = z1 + (– z2).For example, (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4iand (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4iCOMPLEX NUMBERS AND QUADRATIC EQUATIONS 99Multiplication 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 × × 2) = – 24 + i28The multiplication of complex numbers possesses the following properties, whichwe state without proofs.
7 (ii) The commutative law For any two complex numbers z1 and z2, (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.1 = 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 number
8 The square roots of a negative real number Note that i2 = –1 and ( – i)2 = i2 = – 1Therefore, the square roots of – 1 are i, – i. However, by the symbol − , we wouldmean i only.Now, we can see that i and –i both are the solutions of the equation x2 + 1 = 0 orx2 = –1.Similarly ( ) ( ) 2 23i = 3 i2 = 3 (– 1) = – 3( )2− 3i = ( )2− 3 i2 = – 3Therefore, the square roots of –3 are 3 i and − 3i .Again, the symbol −3 is meant to represent 3 i only, i.e., −3 = 3 i .Generally, if a is a positive real number, −a = a −1 = a i ,
9 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101 We already know that a× b = 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 examine.Note thatCOMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = −1 −1= (−1) (−1) (by assuming a× b = ab for all real numbers)= 1 = 1, which is a contradiction to the fact that i = − .Therefore, a× b≠ ab if both a and b are negative real numbers.Further, if any of a and b is zero, then, clearly, a× b= ab= 0.5.3.7 Identities We prove the following identity( )2 2 2z1+z2 =z1+z2+2z1z2, for all complex numbers z1 and z2.
10 of z, denoted as z , is the complex number a – ib, i.e., z = a – ib. The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number. Then, the modulus of z, denoted by | z |, is definedto be the non-negative real number a2+b2, i.e., | z | = a2+b2 and the conjugateof z, denoted as z , is the complex number a – ib, i.e., z = a – ib.For example, 3+i =32+12=10, 2−5i = 22+(−5)2= 29 ,and 3+i=3−i, 2−5i=2+5i, −3i −5 = 3i – 5
11 SummaryA number of the form a + ib, where a and b are real numbers, is called acomplex number, a is called the real part and b is called the imaginary partof the complex number.Let z1 = a + ib and z2 = c + id. Then(i) z1 + z2 = (a + c) + i (b + d)(ii) z1 z2 = (ac – bd) + i (ad + bc)For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists thecomplex number
12 For any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = – 1, i4k + 3 = – i The conjugate of the complex number z = a + ib, denoted by z , is given byz = a – ib.The polar form of the complex number z = x + iy is r (cosè + i sinè), wherer = x2+ y2 (the modulus of z) and cosè =xr , sinè =yr . (è is known as theargument of z. The value of è, such that – ð < è ≤ ð, is called the principalargument of z.
13 SUMS Convert the given complex number in polar form: –3 Answer Discussion–3Let r cos θ = –3 and r sin θ = 0On squaring and adding, we obtain
14 On squaring and adding, we obtain Convert the given complex number in polar form: –3AnswerDiscussion–3Convert the given complex number in polar form: –3–3Let r cos θ = –3 and r sin θ = 0On squaring and adding, we obtain
15 system was recognised by the Greeks. But the credit goes to the Indian Historical NoteThe fact that square root of a negative number does not exist in the real numbersystem was recognised by the Greeks. But the credit goes to the Indianmathematician Mahavira (850 A.D.) who first stated this difficulty clearly. “Hementions in his work ‘Ganitasara Sangraha’ as in the nature of things a negative(quantity) is not a square (quantity)’, it has, therefore, no square root”. Bhaskara,another I
16 Indian mathematician, also writes in his work Bijaganita, written in 1150. A.D. “There is no square root of a negative quantity, for it is not asquare.” Cardan (1545 A.D.) considered the problem of solvingx + y = 10, xy = 40.He obtained x = 5 + −15 and y = 5 – −15 as the solution of it, whichwas discarded by him by saying that these numbers are ‘useless’. Albert Girard(about 1625 A.D.) accepted square root of negative numbers and said that this
17 Euler was the first to introduce the symbol i for −1 and W.R. Hamilton will enable us to get as many roots as the degree of the polynomial equation.Euler was the first to introduce the symbol i for −1 and W.R. Hamilton(about 1830 A.D.) regarded the complex number a + ib as an ordered pair ofreal numbers (a, b) thus giving it a purely mathematical definition and avoidinguse of the so called ‘imaginary numbers’.