Chapter 5 – Complex Numbers And Quadratic Equations

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

If ( x + iy ) 3 = a + ib, where x, y, a, b ∈ R, show that − = – 2 (a2 + b2) 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 = 3a2b – b3 x y

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) COMPLEX NUMBERS AND QUADRATIC EQUATIONS 79 From (2), we have y = 0 or x = − When y = 0, from (1), we get x2 – x = 0, i.e., x = 0 or x = 1. 1 1 1 3 3 When x = − , from (1), we get y2 = + or y2 = , i.e., y = ± . 2 4 2 4 2 Hence, the solutions of the given equation are 1 3 1 3 0 + i0, 1 + i0, − + i , − −i . 2 2 2 2 2 z +1

If the imaginary part of is – 2, then show that the locus of the point iz + 1 representing z in the argand plane is a straight line. Solution Let z = x + iy . Then 2 z +1 2( x + iy ) +1 (2 x +1) + i 2 y = = iz + 1 i ( x + iy ) +1 (1− y ) + ix {(2 x +1) + i 2 y} {(1− y ) − ix} = × {(1− y ) + ix} {(1− y ) − ix} (2 x + 1 − y ) + i (2 y − 2 y 2 − 2 x 2 − x) = 1+ y 2 − 2 y + x 2 2 z +1 2 y − 2 y 2 − 2 x2 − x Thus Im = iz +1 1+ y 2 − 2 y + x 2 2 z +1 But Im = –2 (Given) iz +1 2 y − 2 y 2 − 2x2 − x So =−2 1+ y 2 − 2 y + x 2 ⇒ 2y – 2y2 – 2x2 – x = – 2 – 2y2 + 4y – 2x2 i.e., x + 2y – 2 = 0, which is the equation of a line. 2 2

If z −1 = z + 1 , then show that z lies on imaginary axis. Solution Let z = x + iy. Then | z2 – 1 | = | z |2 + 1 ⇒ x 2 − y 2 −1 + i 2 xy = x + iy + 1 ⇒ (x2 – y2 –1)2 + 4x2y2 = (x2 + y2 + 1)2 ⇒ 4x2 = 0 i.e., x=0 Hence z lies on y-axis.

Let z1 and z2 be two complex numbers such that z1 + i z2 = 0 and arg (z1 z2) = π. Then find arg (z1). Solution Given that z1 + i z2 = 0 ⇒ z1 = i z2 , i.e., z2 = – i z1 Thus arg (z1 z2) = arg z1 + arg (– i z1) = π ⇒ arg (– i z12 ) = π ⇒ arg (– i ) + arg ( z12 ) = π ⇒ arg (– i ) + 2 arg (z1) = π −π ⇒ + 2 arg (z1) = π 3π ⇒ arg (z1) =

Let z1 and z2 be two complex numbers such that z1 + z2 = z1 + z2 . Then show that arg (z1) – arg (z2) = 0. Solution Let z1 = r1 (cosθ1 + i sin θ1) and z2 = r2 (cosθ2 + i sin θ2) where r1 = z1 , arg ( z1 ) = θ1, r2 = z2 , arg (z2) = θ2. We have, z1 + z2 = z1 + z2 = r1 (cos θ1 + cos θ2 ) + r2 (cos θ2 + sin θ2 ) = r1 + r2 = r12 + r22 + 2r1r2 cos (θ1 − θ2 ) = (r1 + r2 ) 2 ⇒ cos (θ1 – θ2 ) =1 ⇒ θ1 – θ2 i.e. arg z1 = arg z2

If z1, z2, z3 are complex numbers such that 1 1 1 z1 = z2 = z3 = + + = 1 , then find the value of z + z + z . z1 z2 z3 1 2 3 Solution z1 = z2 = z3 =1 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 81 2 2 2 ⇒ z1 = z2 = z3 =1 ⇒ z1 z1 = z2 z2 = z3 z3 = 1 1 1 1 ⇒ z1 = , z2 = , z3 = z1 z2 z3 1 1 1 Given that + + =1 z1 z2 z3 ⇒ z1 + z2 + z3 = 1 , i.e., z1 + z 2 + z 3 = 1 ⇒ z1 + z2 + z3 = 1

If a complex number z lies 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 z from the centre of the circle is z − (− 4 + i 0) = z + 4 . According to given condition z + 4 ≤ 3 . Now z + 1 = z + 4 – 3 ≤ z + 4 + −3 ≤ 3 + 3 = 6 Therefore, greatest value of |z + 1| is 6. Since least value of the modulus of a complex number is zero, the least value of z +1 = 0 .

Locate the points for which 3 < z < 4 Solution z < 4 x2 + y2 < 16 which is the interior of circle with centre at origin and radius 4 units, and z > 3 x2 + y2 > 9 which is exterior of circle with centre at origin and radius 3 units. Hence 3 < z < 4 is the portion between two circles x2 + y2 = 9 and x2 + y2 = 16.

Find the value of 2x4 + 5x3 + 7x2 – x + 41, when x = – 2 – 3 i Solution x + 2 = – 3 i ⇒ x2 + 4x + 7 = 0 Therefore 2x4 + 5x3 + 7x2 – x + 41 = (x2 + 4x + 7) (2x2 – 3x + 5) + 6 = 0 × (2x2 – 3x + 5) + 6 = 6.

Find the value of P such that the difference of the roots of the equation x2 – Px + 8 = 0 is 2. Solution Let α, β be the roots of the equation x2 – Px + 8 = 0 Therefore α + β = P and α . β = 8. Now α–β=± α β αβ Therefore 2= ± − ⇒ P2 – 32 = 4, i.e., P = ± 6.

Find the value of a such that the sum of the squares of the roots of the equation x2 – (a – 2) x – (a + 1) = 0 is least. Solution Let α, β be the roots of the equation Therefore, α + β = a – 2 and αβ = – ( a + 1) Now α2 + β2 = (α + β)2 – 2αβ = (a – 2)2 + 2 (a + 1) = (a – 1)2 + 5 Therefore, α2 + β2 will be minimum if (a – 1)2 = 0, i.e., a = 1.

Find the value of k if for the complex numbers z1 and z2, 2 2 2 2 1− z1 z2 − z1 − z2 = k (1− z1 )(1− z2 ) Solution 2 2 L.H.S. = 1− z1 z2 − z1 − z2 = (1− z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 ) = (1 − z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 ) = 1 + z1 z1 z2 z2 − z1 z1 − z2 z2 2 2 2 2 = 1+ z1 ⋅ z2 − z1 − z2 2 2 = (1− z1 ) (1− z2 ) 2 2 R.H.S. = k (1 – z1 ) (1 − z2 ) ⇒ k =1 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 83 Hence, equating LHS and RHS, we get k = 1. π

If z1 and z2 both satisfy z + z = 2 z − 1 arg (z1 – z2) = , then find Im (z1 + z2). Solution Let z = x + iy, z1 = x1 + iy1 and z2 = x2 + iy2. Then z + z = 2 z −1 ⇒ (x + iy) + (x – iy) = 2 x −1 + iy ⇒ 2x = 1 + y2 ... (1) Since z1 and z2 both satisfy (1), we have 2x1 = 1 + y12 ... and 2x2 = 1 + y22 ⇒ 2 (x1 – x2) = (y1 + y2) (y1 – y2) y1 − y2 ⇒ 2 = (y1 + y2) ... (2) x1 − x2 Again z1 – z2 = (x1 – x2) + i (y1 – y2) y1 − y2 Therefore, tan θ = , where θ = arg (z1 – z2) x1 − x2 π y −y π ⇒ tan = 1 2 since θ =

Fill in the blanks:

• (i)The real value of ‘a’ for which 3i3 – 2ai2 + (1 – a)i + 5 is real is ________. π
• (ii)If z = 2 and arg (z) = , then z = ________. π
• (iii)The locus of z satisfying arg (z) = is _______.
• (iv)The value of (− −1) 4 n –3 , where n ∈ N, is ______. 1− i (v) The conjugate of the complex number is _____. 1+ i (vi) If a complex number lies in the third quadrant, then its conjugate lies in the ______. (vii) If (2 + i) (2 + 2i) (2 + 3i) ... (2 + ni) = x + iy, then 5.8.13 ... (4 + n2) = ______. Solution (i) 3i3 – 2ai2 + (1 – a)i + 5 = –3i + 2a + 5 + (1 – a)i = 2a + 5 + (– a – 2) i, which is real if – a – 2 = 0 i.e. a = – 2. π π 1 1 (ii) z = z cos + i sin = 2 +i = 2 (1 + i ) 4 4 2 2 (iii) Let z = x + iy. Then its polar form is z = r (cos θ + i sin θ), where tan θ = y and π θ is arg (z). Given that θ = . Thus. π y tan = y = 3 x , where x > 0, y > 0. 3 x Hence, locus of z is the part of y = 3 x in the first quadrant except origin. 4 n –3 1 (iv) Here (– −1) = (−i )4 n −3 = (−i )4 n (−i ) −3 = (−i )3 1 1 i 3== = 2 = −i −i i i 1 − i 1 − i 1 − i 1 + i 2 − 2i 1 −1 − 2i (v) = × = = = −i 1+ i 1+ i 1− i 1− i 2 1+1 1− i Hence, conjugate of is i. 1+ i (vi) Conjugate of a complex number is the image of the complex number about the x-axis. Therefore, if a number lies in the third quadrant, then its image lies in the second quadrant. (vii) Given that (2 + i) (2 + 2i) (2 + 3i) ... (2 + ni) = x + iy ... (1) ⇒ ( ) (2 + i ) (2 + 2i ) (2 + 3i )...(2 + ni ) = x + iy = ( x − iy ) i.e., (2 – i) (2 – 2i) (2 – 3i) ... (2 – ni) = x – iy ... (2) COMPLEX NUMBERS AND QUADRATIC EQUATIONS 85 Multiplying (1) and (2), we get 5.8.13 ... (4 + n2) = x2 + y2.

State true or false for the following:

• (i)Multiplication of a non-zero complex number by i rotates it through a right angle in the anti- clockwise direction.
• (ii)The complex number cosθ + i sinθ can be zero for some θ.
• (iii)If a complex number coincides with its conjugate, then the number must lie on imaginary axis.
• (iv)The argument of the complex number z = (1 +i 3 ) (1 + i) (cos θ + i sin θ) is 7π +θ (v) The points representing the complex number z for which z +1 < z −1 lies in the interior of a circle. (vi) If three complex numbers z1, z2 and z3 are in A.P., then they lie on a circle in the complex plane. (vii) If n is a positive integer, then the value of in + (i)n+1 + (i)n+2 + (i)n+3 is 0. Solution (i) True. Let z = 2 + 3i be complex number represented by OP. Then iz = –3 + 2i, represented by OQ, where if OP is rotated in the anticlockwise direction through a right angle, it coincides with OQ. (ii) False. Because cosθ + isinθ = 0 ⇒ cosθ = 0 and sinθ = 0. But there is no value of θ for which cosθ and sinθ both are zero. (iii) False, because x + iy = x – iy ⇒ y = 0 ⇒ number lies on x-axis. (iv) True, arg (z) = arg (1 + i 3 ) + arg (1 + i) + arg (cosθ + isinθ) π π 7π + + θ= +θ 3 4 12 (v) False, because x + iy +1 < x + iy −1 ⇒ (x + 1)2 + y2 < (x – 1)2 + y2 which gives 4x < 0. z +z (vi) False, because if z1, z2 and z3 are in A.P., then z2 = 1 3 ⇒ z2 is the midpoint of z1 and z3, which implies that the points z1, z2, z3 are collinear. (vii) True, because in + (i)n+1 + (i)n+2 + (i)n+3 = in (1 + i + i2 + i3) = in (1 + i – 1 – i) = in (0) = 0

Match the statements of column A and B. Column A Column B 2 4 6 20 (a) The value of 1+i + i + i + ... i is

• (i)purely imaginary complex number (b) The value of i–1097 is
• (ii)purely real complex number (c) Conjugate of 1+i lies in
• (iii)second quadrant 1+ 2i (d) lies in
• (iv)Fourth quadrant 1− i (e) If a, b, c ∈ R and b2 – 4ac < 0, (v) may not occur in conjugate pairs then the roots of the equation ax2 + bx + c = 0 are non real (complex) and (f) If a, b, c ∈ R and b2 – 4ac > 0, (vi) may occur in conjugate pairs and b – 4ac is a perfect square, then the roots of the equation ax2 + bx + c = 0 Solution (a) ⇔ (ii), because 1 + i2 + i4 + i6 + ... + i20 = 1 – 1 + 1 – 1 + ... + 1 = 1 (which is purely a real complex number) 1 1 1 1 i (b) ⇔ (i), because i–1097 = 1097 = 4×274+1 = 4 274 = = 2 = −i (i ) i {(i ) } (i ) i i which is purely imaginary complex number. (c) ⇔ (iv), conjugate of 1 + i is 1 – i, which is represented by the point (1, –1) in the fourth quadrant. 1+ 2i 1+ 2i 1+ i −1+ 3i 1 3 (d) ⇔ (iii), because = × = = − + i , which is 1− i 1− i 1+ i 2 2 2 1 3 represented by the point − , in the second quadrant. 2 2 (e) ⇔ (vi), If b2 – 4ac < 0 = D < 0, i.e., square root of D is a imaginary −b ± Imaginary Number number, therefore, roots are x = , i.e., roots are in 2a conjugate pairs. COMPLEX NUMBERS AND QUADRATIC EQUATIONS 87 (f) ⇔ (v), Consider the equation x2 – (5 + 2)x+5 2 = 0, where a = 1, b = – (5 + 2 ), c = 5 2 , clearly a, b, c ∈ R. Now D = b2 – 4ac = {– (5 + 2 2 )} – 4.1.5 2 = (5 – 2).

What is the value of ? i 4 n +1 − i 4 n −1 i 4 n i − i 4 n i − i Solution i, because = 2 2 2 i− i = i −1 = −2 = i = 2 2i 2i

What is the smallest positive integer n, for which (1 + i)2n = (1 – i)2n? 2n 1+ i Solution n = 2, because (1 + i)2n = (1 – i)2n = =1 1− i ⇒ (i)2n = 1 which is possible if n = 2 (∴ i4 = 1)

What is the reciprocal of 3 + 7i Solution Reciprocal of z = 2 3− 7 i 3 7i Therefore, reciprocal of 3 + 7 i= = – 16 16 16

If z1 = 3 + i 3 and z2 = 3 + i , then find the quadrant in which z1 lies. z2 z1 3+i 3 3+ 3 3− 3 Solution = = + i z2 3+i 4 4 which is represented by a point in first quadrant. 5 +12i + 5 −12i

What is the conjugate of ? 5 +12i − 5 −12i Solution Let 5 +12i + 5 −12i 5 +12i + 5 −12i z= × 5 +12i − 5 −12i 5 +12i + 5 −12i 5 +12i + 5 −12i + 2 25 +144 = 5 +12i − 5 +12i 3 3i 3 = = = 0− i 2i −2 2 Therefore, the conjugate of z = 0 + i

What is the principal value of amplitude of 1 – i ? Solution Let θ be the principle value of amplitude of 1 – i. Since π π tan θ = – 1 ⇒ tan θ = tan −θ=− 4 4 25 3

What is the polar form of the complex number (i ) ? Solution z = (i25)3 = (i)75 = i4×18+3 = (i4)18 (i)3 = i3 = – i = 0 – i Polar form of z = r (cos θ + i sinθ) π π = 1 cos − + i sin − 2 2 π π = cos – i sin 2 2 π

What is the locus of z, if amplitude of z – 2 – 3i is ? Solution Let z = x + iy. Then z – 2 – 3i = (x – 2) + i (y – 3) y −3 Let θ be the amplitude of z – 2 – 3i. Then tan θ = x−2 π y −3 π ⇒ tan = since θ = 4 x−2 4 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 89 y −3 ⇒ 1= i.e. x – y + 1 = 0 x−2 Hence, the locus of z is a straight line.

If 1 – i, is a root of the equation x2 + ax + b = 0, where a, b ∈ R, then find the values of a and b. −a Solution Sum of roots = (1 – i) + (1 + i) ⇒ a = – 2. (since non real complex roots occur in conjugate pairs) Product of roots, = (1 − i ) (1 + i ) ⇒ b = 2 Choose the correct options out of given four options in each of the Examples from 28 to 33 (M.C.Q.).

1 + i2 + i4 + i6 + ... + i2n is

• (A)positive
• (B)negative
• (C)0
• (D)can not be evaluated Solution (D), 1 + i2 + i4 + i6 + ... + i2n = 1 – 1 + 1 – 1 + ... (–1)n which can not be evaluated unless n is known.

If the complex number z = x + iy satisfies the condition z +1 = 1 , then z lies on

• (A)x-axis
• (B)circle with centre (1, 0) and radius 1
• (C)circle with centre (–1, 0) and radius 1
• (D)y-axis Solution (C), z +1 =1 ⇒ ( x +1) + iy = 1 ⇒ (x +1)2 + y2 = 1 which is a circle with centre (–1, 0) and radius 1.

The area of the triangle on the complex plane formed by the complex numbers z, – iz and z + iz is: 2 2

• (A)z
• (B)z
• (C)
• (D)none of these Solution (C), Let z = x + iy. Then – iz = y – ix. Therefore, z + iz = (x – y) + i (x + y) 1 2 z Required area of the triangle = (x + y2 ) = 2 2

The equation z +1 − i = z −1+ i represents a

• (A)straight line
• (B)circle
• (C)parabola
• (D)hyperbola Solution (A), z +1 − i = z −1+ i ⇒ z − (−1 + i ) = z − (1 − i ) ⇒ PA = PB, where A denotes the point (–1, 1), B denotes the point (1, –1) and P denotes the point (x, y) ⇒ z lies on the perpendicular bisector of the line joining A and B and perpendicular bisector is a straight line.

Number of solutions of the equation z2 + z = 0 is

• (A)1
• (B)2
• (C)3
• (D)infinitely many Solution (D), z2 + z = 0, z ≠ 0 ⇒ x2 – y2 + i2xy + x2 + y2 = 0 ⇒ 2x2 + i2xy = 0 2x (x + iy) = 0 ⇒ x = 0 or x + iy = 0 (not possible) Therefore, x = 0 and z ≠ 0 So y can have any real value. Hence infinitely many solutions. π π

The amplitude of sin + i (1− cos ) is 5 5 2π π π π

• (A)
• (B)
• (C)
• (D)5 5 15 10 π π Solution (D), Here r cos θ = sin and r sin θ = 1 – cos 5 5 COMPLEX NUMBERS AND QUADRATIC EQUATIONS 91 π π 1 − cos 2sin 2 5 = 10 Therefore, tan θ = π π π sin 2sin .cos 5 10 10 π π ⇒ tan θ = tan i.e., θ= 10 10