184 Chapter – 8 COMPLEX NUMBERS 7.1 Introduction: The real number system had limitations that were at first accepted and later overcome by a series of improvements in both concepts and mechanics. In connection with, quadratic, equations we encountered the concept of imaginary number and the device invented for handling it, the notation i2 = - 1or i = -1 . In this chapter we continue the extension of the real number system to include imaginary' numbers. The extended system is called the complex number system. 7.2 Complex Number: A complex number is a number of the form a + bi, where a and b are real and i2 = -1 or i = -1 . The letter ‘a’ is called the real part and ‘b’ is called the imaginary part of a + bi. If a = 0, the number ib is said to be a purely imaginary number and if b = 0, the number a is real. Hence, real numbers and pure imaginary numbers are special cases of complex numbers. The complex numbers are denoted by Z , i.e., Z = a + bi. In coordinate form, Z = (a, b). Note : Every real number is a complex number with 0 as its imaginary part. 7.3 Properties of Complex Number: (i) The two complex numbers a + bi and c + di are equal if and only if a = c and b =d for example if. x – 2 + 4yi = 3 + 12 i Then x – 2 = 3 and y = 3 (ii) If any complex number vanishes then its real and imaginary parts will separately vanish. For, if a + ib = 0, then a = – ib Squaring both sides 185 Math 123 Complex Numbers a2 = – b2 a2 + b2 = 0 Which is possible only when a = 0, b = 0 7.4 Basic Algebraic Operation on Complex Numbers: There are four algebraic operations on complex numbers. (i) Addition: If Z1 = a1 + b1 i and Z2 = a2 + b2i, then Z1 +Z2 = (a1+ b1i) + (a2 + b2 i) = (a1 + a2) + i(b1 + b2) (ii) Subtraction: Z1 Z2 = (a1+ b1i) – (a2 + b2 i) = (a1 – a2) + i(b1 – b2) (iii) Multiplication: Z1 Z2 = (a1+ b1i) . (a2 + b2 i) 2 = a1 a2 + b1bi + a1b2i + b1a2i = (a1 a2 – b1b2) + i( a1b2 + b1a2) (iv) Division: z1 a1 + b1i = z2 a2 + b2i Multiply Numerator and denominator by the number a2 – b2 i in order to make the denominator real. z1 a1 + b1i a2 – b2i = z2 a2 + b2i a2 – b2i (a1a2 + b1b2) + i(b1a2 – a1 b2) = 2 2 a2 + b2 b1a2 – b1b2 = + i a 2 + b 2 2 2 Generally result will be expressed in the form a + ib. Example 1: Add and subtract the numbers 3 + 4i and 2 – 7i. 186 Math 123 Complex Numbers Solution: Addition: (3 + 4i) + (2 – 7i) = (3 + 2) + i(4 – 7) = 5 – 3i Subtraction: (3 + 4i) – (2 – 7i) = (3 – 2) + i(4 + 7) = 1 + 11i Example 2: Find the product of the complex numbers: 3 + 4i and 2 – 7i. Solution: (3 + 4i) (2 – 7i) = 6 – 21i + 8i – 28i2 = 6 + 28 – 13i = 34 – 13i Example 3: Divide 3 + 4i by 2 – 7i. 3 + 4i 3 + 4i 2 + 7i Solution: 2 – 7i = 2 – 7i 2 + 7i 6 + 28i2 + i(21 + 8) = 4 + 49 –22 + 29i = 53 –22 29 = 53 + i 53 (2 + i) (1 – i) Example 4: Express 4 – 3i in the form of a + ib. (2 + i)(1 – i) (2 + 1) + i(1 – 2) 3 – i Solution: 4 – 3i = 4 – 3i = 4 – 3i 3 – i 4 + 3i (12 + 3) + i(9 – 4) = 4 – 3i 4 + 3i = 16 + 9 15 + i(5) 15 5 = 25 = 25 + 25 i 3 1 = 5 + 5 i 1 + 4i Example 5: Separate into real and imaginary parts: 3 + i . Solution : 1 + 4i 1 + 4i 3 – i 3 + i = 3 + i 3– i 187 Math 123 Complex Numbers (3 + 4) + i(12 – 1) 7 + 11i = 9 + 1 = 10 7 11 7 11 = 10 + 10 i = 10 + 10 i 7 Here, real part = a = 10 11 And imaginary part = b = 10 Extraction of square roots of a complex number: Example 6: Extract the square root of the complex numbers 21 – 20i. Solution: Let a + ib = 21 – 20i Squaring both sides (a + ib)2 = 21 – 20 i a2 – b2 + 2abi = 21 – 20 i Comparing both sides a2 – b2 = 21…………………….(1) 2ab = –20………………………(2) 10 From (2) b = – a Put b in equation (1), 2 100 a – a2 = 21 a4 – 21a2 – 100 = 0 (a2 – 25) (a2 + 4) = 0 a2 = 25 or a2 = –4 a = +5 or a = + –4 = + 2i But a is not imaginary, so the real value of a is a = 5 or a = –5 The corresponding value of b is 188 Math 123 Complex Numbers b = –2 or b = 2 Hence the square roots of 21 – 20i are: 5 – 2i and –5 + 2i Factorization of a complex numbers: Example 7: Factorise: a2 + b2 Solution: We have a2 + b2 = a2 ( b2) =a2 –( i2 b2) , i2 = - 1 = (a)2 – (ib)2 = (a + ib) (a – ib) 7.5 Additive Inverse of a Complex Number: Let Z– = a + ib be a complex number, then the number – a – ib is called the additive inverse of Z– . It is denoted by – Z– i.e., Z– = a – ib. Also Z– – Z– = 0 Example 8: Find the additive inverse of 2 – 5i Solution: Let Z– =2 – 5i Then additive inverse of Z– is: – Z– = – (2 – 5i) = – 2 + 5i. 7.6 Multiplicative inverse of a complex number: Let a + ib be a complex number, then x + iy is said to be multiplicative inverse of a + ib if (x + iy) (a + ib) = 1 1 Or x + iy = a + ib 1 a – ib = a + ib a –ib a – ib x + iy = a2 + b2 a b x + iy = a2 + b2 – i a2 + b2 189 Math 123 Complex Numbers a b So x = a2 +b2 y = - a2 + b2 a b Hence multiplicative inverse of (a, b) is 2 2 – 2 2 a + b a + b Example 9: Find the multiplicative inverse of 4 + 3i or (4, 3). Solution: 1 The multiplicative inverse of 4 + 3i is: 4 + 3i 1 1 3 – 3i Since, 4 + 3i = 4 + 3i 4 –3i 4 – 3i 4 3 4 3 = = – i = – 16 + 9 25 25 25 25 7.7 Conjugate of a complex number: Two complex numbers are called the conjugates of each other if their real parts are equal and their imaginary parts differ only in sign. If Z = a + bi, the complex number a – bi is called the conjugate _ of Z. it is denoted by Z– . _ _____ i.e., Z– = a + bi = a – bi _ Moreover Z.Z– = (a + bi) ( a – bi) = a2 + b2 _ _ Z– + Z– = 2a and Z– – Z– = 2 bi Theorem: If Z1 and Z2 are complex numbers, then _____ _ _ (i) z1 + z2 = z1 + z2 _____ _ _ (ii) z1 + z2 = z1 . z2 190 Math 123 Complex Numbers __ _ z1 z1 (iii) = z2 _ z2 Proof: Let Z1 = a1 + bi and Z2 = a2 + b2i (i) Z1 + Z2 = (a1 + a2) + i (b1 + b2) _____ z1 + z2 = (a1 + a2) – i (b1 + b2) = (a1 – ib1) + (a2 – ib2) __ _ = z1 + z2 (ii) Z1 Z2 = (a1 + ib1) (a2 + ib2) = (a1 a2 – b1 b2) + i ((a1b2 + b1b2) _____ z1 . z2 = (a1a2 – b1b2) – i(a1b2 + b1a2) = a1a2 – ia1b2 – ib1 a2 – b1b2 2 =a1 (a2 – ib2) – ib1a2 + i b1 b2 =a1 (a1 – ib2) – ib (a2 – ib2) = a1(a2 – ib2) – ib1(a2 – ib2) = (a1 – ib1) (a2 – ib2) __ __ = z1 . z2 z a + ib a + ib a – ib (iii) 1 = 1 1 = 1 1 2 2 z2 a2 +ib2 a2 +ib2 a2 – ib2 = z1 = z = 2 __ _______________ _____ _____ z1 (a1 + ib1) (a2 - ib2) (a1 + ib 1 )(a2 – ib2 ) = 2 2 = 2 2 by (ii) z2 a2 + b2 a2 + b2 191 Math 123 Complex Numbers (a1 – ib1) (a2 + ib2) a1 – ib1 = = (a2 – ib2) (a2 + ib2) a2 – ib2 __ __ z1 z1 = z2 __ z2 _____ Example 10: Evaluate (3 + 4i) ( 3 + 4i ) Solution: _____ (3 + 4i) ( 3 + 4i ) = (3 + 4i) ( 3 – 4i) = 9 + 16 = 25 Example 11: Find the conjugate of the complex number 2i(– 3 + 8i) Solution: 2i (– 3 + 8i) = – 16 – 6i ________ 2i(-3 + 8i) = – 16 + 6i ________ __ ________ Alternately 2i(-3 + 8i) = 2i (-3 + 8i) By above theorem = 2i (– 3 – 8i) =– 16 + 6i Exercise 7.1 Q.1: Write each ordered pair (complex number) in the form: a + bi. (i) (2, 6) (ii) (5, – 2) (iii) (– 7, – 3) (iv) (4, 0) Q.2: Write each complex number as an ordered pair. (i) (2 + 3i) (ii) (- 3 + i) (iii) (4i) (iv)(0) Q.3: Find the value of x and y in each of the following: (i) x + 3i + 3 = 5 + yi (ii) x + 2yi = ix + y + 1 (iii) (x, y) (1, 2) = (-1, 8) (iv) (x, -y) (3, -4) = (3, -29) 192 Math 123 Complex Numbers (v) (2x – 3y) + i(x – y) 6 = 2 – i (2x – y + 3) Q.4: Simplify the following: (i) (2 – 3i) + (1 + 2i) (ii) (3 + 5i) – (5 – 3i) (iii) (9 + 7i) – (–9 + 7i) + (-18 + i) (iv) (2 – 3i)(3 + 5i) (v) (4 – 3i)2 (vi) (3 + 4i) (4 + 3i) (2 – 5i) -3 1 3 (vii) (–1 +i 3 )3 (viii) – + i 2 2 1 1 2 + i (ix) 8 + 3i + 8 - 3i (x) 1 - 3i (3 + 4i) (1 - 2i) 2 + –1 (xi) (xii) 1 + i 3 – –4 3 1 3 ( xiii) - - i 2 2 Q.5: Find the conjugate of each of the following: (i) – 2 + 3i (ii) (1 + i)(– 2 – i) (iii) – 3i(2 + 5i) (iv) (-5 + 3i) (2 – 3i) Q.6: Reduce of the following to the form a + bi. 3 – i (2 + 3i) (3 + 2i) (i) 3 + 2i (ii) 4– 3i _____ ____ ______ (iii) (2 – 3i)2 ( 3 +4i ) (iv) ( 4 + i ) ( – 1 +3i ) –3 (v) (1 – i)2 (1 + i) (vi) 1 – –7 (vii) (2 + – 3 ) (2 – –3 ) Q.7: Factorize the following: (i) 4m2 + 9n2 (ii) 49a2 + 625b2 a2 b2 (iii) 9 + 81 Q.8: Find the multiplicative inverse of the following: (i) (–3, 4) (ii) ( 2 ,– 5 ) 193 Math 123 Complex Numbers 2 (iii) (iv) – 6 – 3i 1 + –1 (v) 4 – – 7 Q.9: Extract the square root of the following complex numbers: (i) –3 + 4i (ii) 8 – 6i (iii) 24 + 10 i 1 Q.10: Prove that : = Cos + i Sin Cos – i Sin Q.11: Express x2 + y2 = a2 in terms of conjugate co-ordinates.