Chapter 5.1 Overview We know that the square of a real number is always non-negative e.g. (4)2 = 16 and (– 4)2 = 16. Therefore, square root of 16 is ± 4. What about the square root of a negative number? It is clear that a negative number can not have a real square root. So we need to extend the system of real numbers to a system in which we can find out the square roots of negative numbers. Euler (1707 - 1783) was the first mathematician to introduce the symbol i(iota) for positive square root of – 1 i.e., i= 5.1.1 Imaginary numbers Square root of a negative number is called an imaginary number., for example, 91−=− 9 = i3, −=−7 17 =i7 5.1.2 Integral powers of i i= −1, i2 = – 1, i3 = i2 i= – i, i4 = (i2)2 = (–1)2 = 1. nTo compute ifor n > 4, we divide nby 4 and write it in the form n= 4m+ r, where mis quotient and ris remainder (0 ≤ r≤ 4) n 4m+r4)m)r)rrHence i= i=(i . (i= (1)m(i= iFor example, (i)39 = i4 × 9 + 3 =(i4)9 . (i)3 = i3 = – i and (i)–435 = i– (4 × 108 + 3) =(i)– (4 × 108) . (i)– 3 11 i =. ==i4108 3 4() i ()i ()i (i) If aand bare positive real numbers, then −×−= − 1 a×− 1 =ab ×biaib =−ab (ii) a. b=abif aand bare positive or at least one of them is negative or zero. However, if aand b, both are negative.5.1.3 Complex numbers (a) A number which can be written in the form a + ib, where a, b are real numbers and i = −1 is called a complex number. (b) If z = a + ib is the complex number, then a and b are called real and imaginary parts, respectively, of the complex number and written as Re (z) = a, Im (z) = b. (c) Order relations “greater than” and “less than” are not defined for complex numbers. (d) If the imaginary part of a complex number is zero, then the complex number is known as purely real number and if real part is zero, then it is called purely imaginary number, for example, 2 is a purely real number because its imaginary part is zero and 3i is a purely imaginary number because its real part is zero. 5.1.4 Algebra of complex numbers (a) Two complex numbers z1 = a + ib and z2 = c + id are said to be equal if a = c and b = d. (b) Let z = a + ib and z = c + id be two complex numbers then12z1 + z2 = (a + c) + i (b + d). 5.1.5 Addition of complex numbers satisfies the following properties 1. As the sum of two complex numbers is again a complex number, the set of complex numbers is closed with respect to addition. 2. Addition of complex numbers is commutative, i.e., z + z = z + z3. Addition of complex numbers is associative, i.e., (z + z) + z = z + (z + z)1 221 1 2 31234. For any complex number z = x + i y, there exist 0, i.e., (0 + 0i) complex number such that z + 0 = 0 + z = z, known as identity element for addition. 5. For any complex number z = x + iy, there always exists a number – z = – a – ib such that z + (– z) = (– z) + z = 0 and is known as the additive inverse of z. 5.1.6 Multiplication of complex numbers Let z1 = a + ib and z2 = c + id, be two complex numbers. Then z1 . z2 = (a + ib) (c + id) = (ac – bd) + i (ad + bc) 1. As the product of two complex numbers is a complex number, the set of complex numbers is closed with respect to multiplication. 2. Multiplication of complex numbers is commutative, i.e., z1.z2 = z2.z1 3. Multiplication of complex numbers is associative, i.e., (z1.z2) . z3 = z1 . (z2.z3) 4. For any complex number z = x + iy, there exists a complex number 1, i.e., (1 + 0i) such that z . 1 = 1 . z = z, known as identity element for multiplication. 15. For any non zero complex number z = x + iy, there exists a complex number z 111 aib −such that z ⋅=⋅ z =1, i.e., multiplicative inverse of a + ib = = 2 2. z z aib a ++ b 6. For any three complex numbers z1, z2 and z3 , z1 . (z2 + z3) = z1 . z2 + z1 . z3 and (z + z) . z = z . z + z . z12 31323 i.e., for complex numbers multiplication is distributive over addition. 5.1.7 Let z1 = a + ib and z2( ≠ 0) = c + id. Then z + (ac+bd)(aib bc−ad)z1 ÷ z2 = z 1= ++i22 22 2 cid = c +dc +d 5.1.8 Conjugate of a complex number Let z = a + ib be a complex number. Then a complex number obtained by changing the sign of imaginary part of the complex number is called the conjugate of z and it is denoted by z , i.e., z = a – ib. Note that additive inverse of z is – a – ib but conjugate of z is a – ib. We have : 1. ()z =z 2. z + z = 2 Re (z) , z – z = 2 i Im(z) 3. z = z , if z is purely real. 4. z + z = 0 ⇔ z is purely imaginary 5. z . z = {Re (z)}2 + {Im (z)}2 . 6. (z +z ) =+ z ,( z −z ) =z –z z1 21212 12 ⎛⎞ zz ()(.) =zz ), 1 =1( z ≠0) zz ( )( 7.12 12 ⎜⎟2 z2 ()z2⎝⎠ 5.1.9 Modulus of a complex number Let z = a + ib be a complex number. Then the positive square root of the sum of square of real part and square of imaginary part is called modulus (absolute value) of z and it is denoted by z i.e., z =a2 +b2 In the set of complex numbers z1 > z2 or z1 < z2 are meaningless but z > z or z < z1 2 1 2 are meaningful because and are real numbers.z1z2 5.1.10 Properties of modulus of a complex number 1. z = 0 ⇔ z = 0 i.e., Re (z) = 0 and Im (z) = 0 2. z = z = − z 3. – z ≤ Re (z) ≤ z and – z ≤ Im (z) ≤ z 22 2z = z4. zz = z , zz 11zz = z . z , = (z2 ≠ 0) 5. 12 1 2 z z2 2 2 26. z + z = z + z 2 + 2Re ( zz )1 2 1 2 12 2 27. z1 − z2 = z + z 2 −2Re ( zz 1 2)1 2 8. z + z ≤ z + z1 2 1 2 9. z1 − z2 ≥ z − z1 2 2 2 2)2 2210. az 1 − bz2 + bz1 + az2 = (a +b )( z + z1 2 In particular: 2 2 2)zz + z + z 2 =2( z + z1 − 2 12 1 2 11. As stated earlier multiplicative inverse (reciprocal) of a complex number z = a + ib (≠ 0) is aib 1 − z = = 2 2 2z+abz 5.2 Argand Plane A complex number z = a + ib can be represented by a unique point P (a, b) in the cartesian plane referred to a pair of rectangular axes. The complex number 0 + 0i represent the origin 0 ( 0, 0). A purely real number a, i.e., (a+ 0i) is represented by the point (a, 0) onx - axis. Therefore, x-axis is called real axis. A purely imaginary number ib, i.e., (0 + ib) is represented by the point (0, b) on y-axis. Therefore, y-axis is called imaginary axis. Similarly, the representation of complex numbers as points in the plane is known as Argand diagram. The plane representing complex numbers as points is called complex plane or Argand plane or Gaussian plane. If two complex numbers z1 and z2 be represented by the points P and Q in the complex plane, then z1 − z2 = PQ 5.2.1 Polar form of a complex number Let P be a point representing a non-zero complex number z = a + ib in the Argand plane. If OP makes an angle θ with the positive direction of x-axis, then z = r (cosθ + isinθ) is called the polar form of the complex number, where b r = = a2 +b2 and tanθ = . Here θ is called argument or amplitude of z and wez a write it as arg (z) = θ. The unique value of θ such that – π ≤θ ≤π is called the principal argument. arg (z . z) =arg (z) + arg (z)12 12⎛ z1 ⎞ arg ⎜⎟ = arg (z1) – arg (z2)z⎝ 2 ⎠ 5.2.2 Solution of a quadratic equation The equations ax2 + bx + c = 0, where a, b and c are numbers (real or complex, a ≠ 0) is called the general quadratic equation in variablex. The values of the variable satisfying the given equation are called roots of the equation. The quadratic equation ax2 + bx + c = 0 with real coefficients has two roots given –b +D –b –Dand by , where D =b2 – 4ac, called the discriminant of the equation.2a 2a When D < 0, roots of the quadratic equation are non real (or complex). 2. Let α, β be the roots of the quadratic equation ax2 + bx + c = 0, then sum of the roots −b c (α + β) = and the product of the roots ( α . β) = . a a 3. Let S and P be the sum of roots and product of roots, respectively, of a quadratic equation. Then the quadratic equation is given by x2 – Sx + P = 0. 5.2 Solved Exmaples Short Answer Type Example 1 Evaluate : (1 + i)6 + (1 – i)3 Solution (1 + i)6 = {(1 + i)2}3 = (1 + i2 + 2i)3 = (1 – 1 + 2i)3 = 8 i3 = – 8i and (1 – i)3 = 1 – i3 – 3i + 3i2=1 + i – 3i – 3 = – 2 – 2i Therefore, (1 + i)6 + (1 – i)3 =–8i – 2 – 2i = – 2 – 10i 1 xyExample 2 If (x +iy)3 = a + ib, where x, y, a, b ∈ R, show that − = – 2 (a2 + b2)ab1 Solution (x +iy)3 = a + ib ⇒ x + iy = (a + ib)3 i.e., x + iy = a3 + i3 b3 + 3iab (a + ib) = a3 – ib3 + i3a2b – 3ab2 = a3 – 3ab2 + i (3a2b – b3) ⇒ x = a3 – 3ab2 and y = 3a2 b – b3 x yThus a= a2 – 3b2 and b = 3a2 – b2 x ySo, − = a2 – 3b2 – 3a2 + b2 = – 2 a2 – 2b2 = – 2 (a2 + b2).abExample 3 Solve the equation z2 = z , where z = x + iy Solution z2= z ⇒ x2 – y2 + i2xy = x – iy Therefore, x2 – y2 = x ... (1) and 2xy = – y ... (2) 1From (2), we have y = 0 or x = − 2 When y = 0, from (1), we get x2 – x = 0, i.e., x = 0 or x = 1. 1 113 When x = − , from (1), we get y2 = + or y2 = 4, i.e., y = ± .242 2 Hence, the solutions of the given equation are 1 3130 + i0, 1 + i0, −+ i , −−i .2 222 21Example 4 If the imaginary part of z +is – 2, then show that the locus of the pointiz+1 representing z in the argand plane is a straight line. Solution Let z = x + iy . Then 21 2( x++ (2 x++ i2 yz + iy)1 1) == +1 ix( ++ )1 (1 y)iziy −+ ix {(2 x 1) iy {(1 −− ix }++2} y)×= {(1 )} {(1 −−y) ix }−+yix (2x+− + 1 yi (2y −2 y2 −22) x −x) = 221 y 2 yx+ −+ ⎛21z +⎞ 2 y−2 y2 −2x2 −x Thus Im ⎜ ⎟= 22⎝ ⎠ +−+ xiz +11 y 2y ⎛21z +⎞But Im ⎜⎟= – 2 (Given)iz 1⎝ +⎠ 2y −2 y2 −2x2 −x =−2So 221 y 2yx+−+ ⇒ 2y – 2y2 – 2x2 – x = – 2 –2y2 + 4y – 2x2 i.e., x + 2y – 2 = 0, which is the equation of a line. Example 5 If z21−=z 2 +1 , then show that z lies on imaginary axis. Solution Let z = x + iy. Then | z2 – 1 | = | z |2+ 1 x2 −−y212+i xy =x⇒ ⇒ (x2 – y2 –1)2 + 4x2 y2 ⇒ 4x2 = 0 i.e., Hence z lies on y-axis. +iy 2 +1 = (x2 + y2 + 1)2 x = 0 Example 6 Let z and zbe two complex numbers such that12 arg (z1 z2) = π. Then find arg (z1). Solution Given that z1 +iz2 =0 ⇒ z1 = i z2 , i.e., z2 = – i z1 Thus arg (zz) = arg z + arg (– i z) = π1211⇒ arg (– iz12) = π ⇒ arg (– i ) + arg z12 = π()⇒ arg (– i ) + 2 arg (z) = π1 −π ⇒ + 2 arg (z1) = π2 3π⇒ arg (z) =14 Example 7 Let z1 and z2 be two complex numbers such that z1 +iz2 =0 and .+zz =z +z12 1 2 Then show that arg (z1) – arg (z2) = 0. Solution Let z = r (cosθ + i sin θ) and z = r (cosθ + i sin θ)11112222where r1 = , arg ()= 1, r2 =z1θ, arg (z2) = θ2.z1 z2 +zzWe have, = z +z12 1 2 = r1 2= r1 ⇒θ1(cos cos ) r (cos θ+ θ)θ+ θ+ sin =r +r1 222 2 12 ++22rr cos( θ−θ=+ )( r )2 ⇒ cos (θ – θ ) =1rr212 1212 12 – θ2 i.e. arg z1 = arg z2 Example 8 If z1, z2, z3 are complex numbers such that 111 z =z =z = ++ =12 3 , then find the value ofzzz12 3 Solution z =z =z =11 2 3 ++zzz .123 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 81 ⇒ ⇒ ⇒ 22 2 1 2 3 1= = =z z z 1 1 22 33 1= = =zz zz zz 1 2 3 1 2 3 1 1 1 , ,= = =z z z z z z Given that 1 2 3 1 1 1+ + z z z 1= ⇒ ++zzz =1 , i.e., z++ zz =1123 123 ⇒ ++zzz =1123 Example 9 If a complex number zlies in the interior or on the boundary of a circle of radius 3 units and centre (– 4, 0), find the greatest and least values of z+1. Solution Distance of the point representing zfrom the centre of the circle is (4 i0) =z−−+ z+4 . According to given condition z+≤43. Now z 1+=z+4–3 ≤z+4 +−3 ≤+=633Therefore, greatest value of |z+ 1| is 6. Since least value of the modulus of a complex number is zero, the least value of z+=10. Example 10 Locate the points for which 3