Mathematics-2.Unit-2-Complex Number

COMPLEX NUMBER SYLLABUS Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations. BASIC CONCEPTS A number in the form of x + iy, where x, y are real numbers and i = is called a complex number. (i) Clearly , for an integer n. (ii) If z = x + iy, then the real part of z is denoted by Re (z) and the imaginary part by Im(z). (iii) A complex number is said to be purely real if Im(z) = 0, and is said to be purely imaginary if Re (z) = 0. The complex number 0 = 0 + i0 is both purely real and purely imaginary. (iv) Two complex numbers are said to be equal if and only if their real parts and imaginary parts are separately equal i.e. a + ib = c + id implies a = c and b = d. However, there is no order relation between complex numbers and the expressions of the type a + ib < ( or > ) c + id are meaningless. (v) Since a real number a can be written as a + i.0, therefore every real number can be considered as a complex number whose imaginary part is zero. Thus the set R of real numbers is a proper subset of the complex numbers C. Remark: ⮚ Clearly , for an integer n. REPRESENTATION OF COMPLEX NUMBERS IN ARGAND PLANE A complex number z = x + iy written as ordered pair (x, y) can be represented by a point P whose Cartesian coordinates are (x, y) referred to axes OX and OY, usually called the real and the imaginary axes. The plane of OX and OY is called the Argand diagram or the complex plane. MODULUS OF A COMPLEX NUMBER Let z = x + iy be a complex number then its magnitude is defined by the real number and is denoted by |z|. ARGUMENT OF A COMPLEX NUMBER If z = x + iy then angle θ given by tan is said to be the argument or amplitude of the complex number z and is denoted by arg (z) or amp (z). In case of x = 0 (where y ≠ 0), arg (z) = + π/2 or –π/2 depending upon y > 0 or y < 0 and the complex number is called purely imaginary. If y = 0 (where x≠0), then arg (z) = 0 or π depending upon x > 0 or x < 0 and the complex number is called purely real. The argument of the complex number 0 is not defined. We can define the argument of a complex number also as any value of the θ which satisfies the system of equations cosθ = . PRINCIPAL ARGUMENT OF A COMPLEX NUMBER The value of θ satisfying the inequality – π < θ ≤ π is called the principal value of the argument. Method of finding the principle argument of a complex number z = x + iy Stet 1: Find tanθ = and this gives the value of θ in the first quadrant. Step 2: Find the quadrant in which z lies, with the help of sign of x and y co–ordinates. Step 3: Then argument of z will be θ, π – θ, θ – π, and –θ according as z lies in the first second, third or fourth quadrant Example -1: For z = √3 – i, find the principal value of arg(z). Solution: Here x = √3, y = -1 ⇒ tanθ = ⇒ θ = ⇒ Principal value of arg z = - . (Since z lies in the fourth quadrant) POLAR FORM OF A COMPLEX NUMBER Let OP = r, then x = r cos θ , and y = r sin θ ⇒ z = x + iy = r cos θ + ir sin θ = r ( cos θ + i sin θ ). This is known as Polar form(Trigonometric form) of a Complex Number. Here we should take the principal value of θ. For general values of the argument z = r [ cos ( 2nπ + θ) + i sin ( 2nπ + θ)] (where n is an integer) Note: sometimes cos θ + i sin θ is, in short, written as cis(θ). Euler's formula: cos θ + i sin θ = ei θ. Note: When complex numbers are multiplied their arguments get added i..e (cos θ1 + i sin θ1) (cos θ2 + i sin θ2) = cos (θ1 + θ2) + i sin (θ1 + θ2) generalizing, (cos θ1+ i sin θ1)(cos θ2+ i sin θ2)…(cos θn + i sin θn) = cos(θ1+ θ2+ …+θn) + i sin(θ1+ θ2+…+θn) when complex numbers are divided, their arguments get subtracted = cos (θ1 – θ2) + i sin (θ1 – θ2) Illustration 1: Find the modulus and the principal argument of the numbers: (i) 6(cos 310° – i sin 310°) (ii) Solution: (i) 6(cos 310° – i sin 310°) = 6[cos(360° – 50°) – i sin(360° – 50°)] = 6(cos 50° + i sin 50°) = ∴ modulus = 6 and principal value of the argument = (ii) = – i = 1 ∴ Modulus = 1 and the principal value of the argument = – Illustration 4: If z = (i)i, where , then find the value of . Solution: z = i i = Unimodular Complex Number: A complex number z such that |z| = 1 is said to be unimodular complex number. Since |z| = 1, z lies on a circle of radius 1 unit and centre (0, 0). If |z| = 1 ⇒ z = cos θ + i sin θ, ⇒ 1/z = (cos θ + i sinθ)-1 = cos θ - i sinθ Illustration 9: If z1 and z2 are two nonzero complex numbers and is a uni–modular then show that is purely real. Solution: ⇒ ⇒ lies on the right bisector of line joining A (i) and B(– i ) that mean lies on real axis. Hence is purely real Algebraic Operations with Complex Numbers: ● Addition : (a + ib) + (c+id) = (a + c) + i ( b+d) ● Subtraction : (a + ib) - (c+id) = (a - c) + i ( b - d) ● Multiplication : (a + ib) (c+id) = (ac - bd) + i ( ad + bc) ● Division : ( when at least one of c and d is non-zero) = Square root of a Complex Number: Let z1 = x1 + iy1 be the given complex number and we have to obtain its square root. Let x + iy = (x1 + iy1)½ ⇒ x2 − y2 + 2ixy = x1 + iy1 ⇒ x1 = x2 − y2 and y1 = 2xy ⇒ x2 − ⇒ x2 = , If y1 > 0 ⇒ x = ⇒ = ± If y1 < 0 ⇒ x = ⇒ = ± Example -2: Find the square root of 8 - 15i. Solution: Here y = -15 < 0 ⇒ = ± = ± (5 - 3i). CONJUGATE OF A COMPLEX NUMBER The conjugate of the complex number z = a + ib is defined to be a – ib and is denoted by . In other words is the mirror image of z in the real axis. If z = a + ib, z + = 2 a ( real), z – = 2 ib ( imaginary ) and z = ( a+ib)(a–ib) = a2 + b2 (real ) = |z|2 = . Also Re (z) = Properties of Conjugate: ● ● |z| = | | ● z + =2Re(z). z – = 2i Im(z). ● If z is purely real z = . whenever we have to show a complex number purely real we use this property. ● If z is purely imaginary z+ =0, whenever we have to show that a complex number is purely imaginary we use this property. ● = ● = 1 + 2 In general, ● = ● In general ● ● Properties of Modulus: ● |z| = 0 ⇒ z = 0 + i0 ● |z1 – z2 | denotes the distance between z1 and z2 . ● –|z| ≤ Re(z) ≤ |z| ; equality holds on right or on left side depending upon z being positive real or negative real. ● –|z| ≤ Im z ≤ |z| ; equality holds on right side or on left side depending upon z being purely imaginary and above the real axes or below the real axes. ● |z| ≤ |Re(z)| + |Im(z)| ≤ |z| ; equality holds on left side when z is purely imaginary or purely real and equality holds on right side when |Re(z)| = |Im(z)|. ● |z|2 = z ● |z1z2| = |z1| |z2| In general |z1 z2 . . . . .zn| = |z1| |z2| . . . . . |zn| ● |zn| = |z|n , n ∈ I ● ● |z1+z2| ≤ |z1| + |z2| ⇒ |z1+z2+ ... +zn| ≤ |z1| + |z2| + ... + |zn|; equality holds if origin, z1, z2, z3 …, zn are collinear and z1 , z2, z3, …,zn are on the same side of the origin. ● |z1 – z2| ≥ ||z1| – |z2|| ; equality holds when arg(z1/z2) = π i.e. origin, z1, z2 are collinear and z1 and z2 are on the opposite side of the origin. ● |z1 + z2|2 = (z1 + z2) ( 1 + 2) = |z1|2 + |z2|2 + z1 2 + z2 1 = |z1|2 + |z2|2 + 2Re(z1 2) ● |z1 – z2|2 = (z1 – z2) ( 1 – 2) = |z1|2 + |z2|2 – z1 2 – z2 1 = |z1|2 + |z2|2 – 2Re(z1 2) Properties of Argument: ● arg(z1z2) = θ1 + θ2 = arg(z1) + arg(z2) ● arg (z1/z2) = θ1 – θ2 = arg(z1) – arg(z2) ● arg (zn) = n arg(z), n ∈I Note: ● In the above result θ1 + θ2 or θ1 – θ2 are not necessarily the principle values of the argument of corresponding complex numbers. E.g arg(zn) = n arg(z) only shows that one of the argument of zn is equal to n arg(z) (if we consider arg(z) in the principle range) ● arg(z) = 0, π ⇒ z is a purely real number ⇒ z = . ● arg(z) = π/2, –π/2 ⇒ z is a purely imaginary number ⇒ z = – . Note that the property of argument is the same as the property of logarithm. Example -3: Consider two pairs of non-zero conjugate complex numbers ( z1, z2) and (z3 , z4). Find the value of arg +arg . Solution: arg +arg = arg = arg (as z2 = ) = 0 (as argument of a positive real number is zero). Example -4: If |z1| = |z2| = |z3| = 1, prove that . Solution: We know that |z| = ⇒ |z1 + z2 + z3| = = = = ( |z1|2 = |z2|2 = |z3|2 = 1) Example -5: Prove that |z1| + |z2| = . Solution: RHS = = = 2 Using parallelogram law. = |z1| + |z2| = LHS. DE MOIVRE’S THEOREM If n is any integer, then (cos θ + i sin θ)n = cos nθ + i sin nθ. This is known as De Movre’s Theorem. Remarks: ● Writing the binomial expansion of (cos  + i sin )n and equating the real part to cos n and the imaginary part to sin n, we get ? cos n = cosn  – nc2 cosn–2 sin2 + nc4 cosn–4 sin4 + ……… ? sin n = nc1 cosn–1 sin – nc3 cosn–3 sin3 + nc5 cosn–5 sin5 + ……… ? ⇒ tan n = ● If n is rational number, then one of the values of (cos  + i sin )n is cos n + i sin n. Let n = p/q, where p and q are integers (q > 0) and p, q have no common factor. Then (cos  + i sin )n has q distinct values, one of which is cos n + i sin n. ● If z = r (cos  + i sin ), and n is a positive integer, then ? z1/n = r1/n , k = 0, 1, 2, ……, n –1. ? Here if can be noted that any ‘n’ consecutive values of k will serve the purpose. ? 2.1 APPLICATIONS OF DE MOIVER’S THEOREM This is a fundamental theorem and has various applications. Here we will discuss few of these which are important from the examination point of view. nth Roots of Unity One very important application of De–Moivre’s Theorem is in solving equation of nth powers in complex number. Let x be the nth root of unity. Then xn = 1 = cos 2kπ + i sin 2kπ (where k is an integer) ⇒ x = cos + i sin k = 0, 1, 2, ……, n – 1 Let α = cos + i sin . When k = 2 = [By De–Moivre’s Theorem] = α2 when k = 3 = α3 Similarly, when k = t Then, = αt ∴ The roots are 1, α, α2, ……, αn –1 Sum of the Roots 1 + α + α2 + .... + αn – 1 = = 0 ( αn = 1) and Thus the sum of the roots of unity is zero. Product of the Roots 1.α.α2. .......... αn – 1 = = = cos{π(n – 1)} + i sin{π(n – 1)} If n is even =–1 If n is odd =1 Note : ● The points represented by n nth roots of unity are located at the vertices of a regular polygon of n sides inscribed in a unit circle having centre at the origin, one vertex being on the positive real axis. Geometrically represented as follows. 2.2 CUBE ROOTS OF UNITY For n = 3, we get the cube roots of unity and they are 1, cos + i sin and cos + i sin i.e. 1, and . They are generally denoted by 1, ω and ω2 and geometrically represented by the vertices of an equilateral triangle whose circumcentre is origin and circumradius is unity. Note: ● ω3 = 1 and 1 + ω + ω2 = 0 ● It can be easily proved that 1 + ωn + ω2n = 3 (n is a multiple of 3) 1 + ωn +ω2n = 0 (n is an integer, not a multiple of 3) Illustration 14: Given z1 + z2 + z3 = A , z1 + z2w + z3w2 = B z1 + z2w2 + z3w = C where w is cube root of unity. Express z1, z2, z3 in term of A, B , C. Solution: Adding given three condition z1 = Multiplying z1 + z2 + z3 = A, z1 + z2w + z3w2 = B, z1 + z2w2 + z2w = C by 1 , w2 ,w and adding we get also z3 = Exercise 7. If ω is a cube root of unity then find the value of Solution Ex. 7: = CONCEPT OF ROTATION If z and z′ are two complex numbers then argument of is the angle through which Oz′ must be turned in order that it may lie along Oz. = In general, let z1, z2, z3 , be the three vertices of a triangle ABC described in the counter-clock wise sense. Draw OP and OQ parallel and equal to AB and AC respectively. Then the point P is z2 – z1 and Q is z3 – z1 and Note that arg. (z3-z1)–arg(z2-z1) = α is the angle through which OP must be rotated in the anti-clockwise direction so that it becomes parallel to OQ. Here we can write also. In this case we are rotating OP in clockwise direction by an angle (2π - α). Since the rotation is in clockwise direction, we are taking negative sign with angle (2π - α) Example -10: Consider a square ABCD such that z1, z2, z3 and z4 represent its vertices A, B, C and D respectively. Express ‘z3’ and ‘z4’ in terms of z1 and z2 . Solution: Consider the rotation of AB about A through an angle π/4, we get ⇒ z3 = z1 +( z2 – z1)( 1+i) Similarly , ⇒ z4 = z1 +i( z2 – z1) Illustration 17: If |z – 3| = 3 then show that i tan (Argz) Solution: By figure | z – 3 | = 3 is a circle with centre (3, 0) by rotation eiπ / 2 = i tan θ = i tan (arg z) GEOMETRICAL APPLICATIONS 4.1 SECTION FORMULA Let z1 and z2 be any two complex numbers representing the points A and B respectively in the argand plane. Let C be the point dividing the line segment AB internally in the ratio m : n i.e , and let the complex number associated with point C be z. Let us rotate the line BC about the point C so that it becomes parallel to CA . The corresponding equation of rotation will be , = ⇒ nz1 – nz = –m z2 +mz ⇒ z = . Similarly if C(z) divides the segment AB externally in the ratio of m : n, then z = . In the specific case, if C(z) is the mid point of AB then . Example -12: If z1, z2 and z3 ( in anticlockwise sense) represents the vertices of a triangle, find the centroid, incentre, circumcentre and the orthocentre of the triangle. Solution: Let G be the centriod and let the line joining A and G meets the line BC at the point D. We have, BD = DC D ≡ G divides AD internally in ratio 2 : 1 ⇒ G ≡ ≡ Let I be the incentre and let the line connecting A and I meet the line BC at D1. We have and = ⇒ = Let ‘O’ be the circum-centre and let the liane connecting A and O meet the line BC at D2. We have and ⇒ D2 = and O = = Let ‘P’ be the orthocentre and let the line connecting the points A and P meet the line BC and D3. We have, and ⇒ D3 = and P = = Condition for Collinearity: If there are three real numbers (other than 0) l, m and n such that lz1 + mz2 + nz3 = 0 and l + m + n = 0 then complex numbers z1, z2 and z3 will be collinear. Equation of a Straight Line: Equation of Straight Line Joining the Points z1 and z2: Writing re-arranging terms, we find that the equation of the line through z1 and z2 is given by General equation of a straight line is , where a is a complex number and b is a real number. The length of the perpendicular from a point z1 to the line is given by . Example -13: Let α and β be two fixed non-zero complex numbers and z a variable complex number. If the two straight lines and are mutually perpendicular, then prove that = 0 Solution: Slope of the first line = – Slope of the second line = – Lines will be perpendicular if – + = 0 ⇒ = 0. Equation of a Circle: Consider a fixed complex number z0 and let z be any complex number which moves in such a way that it’s distance from z0 is always equals to ‘r’. This implies z would lie on a circle whose centre is z0 and radius r. And it’s equation would be |z –z0| = r . ⇒ |z –z0|2 = r2 ⇒ ( z – z0) ⇒ . Let - a = z0 and =b ⇒ , where centre = –a and radius = Example -15: Find the centre and radius of the circle . Solution: Given equation can be rewritten as So, it represents a circle with centre at –1 –i and radius = . Example -16: If z1, z2, z3 are complex numbers such that , show that the points represented by z1, z2, z3 lie on a circle passing through the origin. Solution: Since P(z1), Q(z2), R(z3) and S(z4) are concyclic points, ∠PSQ = ∠PRQ ⇒ arg. = arg ⇒ arg = 0 ⇒ = real If z4 = 0 + i0, then = real …(1) We have from which z3 = …(2) From (1) and (2), = real ⇒ = real ⇒ = real, which is true. Therefore z1, z2, z3 and the origin are concyclic. Alternative Solution: ⇒ − ⇒ ⇒ ⇒ = π + arg ⇒ α = π − β ⇒ α + β = π ⇒ points A, B, C, D are concyclic. OBJECTIVES 1. If ‘z’ be any complex number such that , then locus of ‘z’ is (A) A circle (B) An ellipse (C) A line segment (D) None of these Sol: (C) ⇒ Let P(z), A = , B = then (i) represents PA + PB = 4/3. Clearly AB = 4/3 PA + PB = AB Thus P is any point on the line segment AB. Hence (C) is the correct answer. 2. The value of the expression 2 +3 + 4 + . . . + (n + 1) , where ω is an imaginary cube root of unity, is (A) (B) (C) (D) none of these Sol: (C) tn = (n+1) = n3 + n2 = n3 + n2(ω + ω2 + 1)+ n(ω + ω2 +1) +1 = n3 +1 ∴ Sn = . Hence (C) is the correct answer. 3 If ω is an imaginary cube root of unity, then (1 + ω – ω2)7 equals (A) 128 ω (B) – 128 ω (C) 128 ω2 (D) – 128 ω2 Sol: (D) We have (1 + ω + ω2)7 = –(ω2 – ω2)7 = (–2)7 (ω2)7 = 128 ω14 = –128 ω2 Hence (D) is the correct answer. 4. Let ‘z’ be a complex number and ‘a’ be a real parameter such that z2 + az + a2 = 0, then locus of z is (A) a pair of straight lines (B) a circle (C) an ellipse (D) none of these Sol: (A) z2 + az + a2 = 0 ⇒ z = aω, aω2 ( where ‘ω’ is non real root of unity ) ⇒ locus of z is a pair of straight lines Hence (A) is the correct answer. 5. The equation | z + i | – | z – i | = k represents a hyperbola if (A) –2 < k < 2 (B) k > 2 (C) 0 < k < 2 (D) none of these Sol: (A) |z + i| –|z –i| = k represents a hyperbola if 4 – < 0 i.e. k2 < 4. Hence (A) is the correct answer. 6. The complex number z = 1 + i is rotated through an angle in anticlockwise direction about the origin and stretched by additional units, Then the new complex number is (A) – – i (B) – (C) 2 – i (D) 2 – 2i Sol: If z1 be the new complex number then |z1| = |z| + = 2 Also ⇒ z1 = z. 2 = 2( 1+i) ( 0 – i) = – 2i +2 = 2( 1– i) Hence (D) is the correct answer. 7. Let Δ = , where i = , then Δ is (A) purely +ve real number (B) purely imaginary (C) of the form a + ib with a, b both non–zero real numbers. (D) purely negative real number Sol: = – Δ ⇒ Δ is purely imaginary. Hence (B) is the correct answer. 8. If is real, where n1, n2 are positive integers then (A) n1 = n2 (B) n1 = n2 + 1 (C) n2 = n1 + 1 (D) n1n2 ∈ N Sol: = = 2 = Real number. Thus n1, n2 ∈ N Hence (D) is the correct answer. 9. If (ω ≠ 1) be a cube root of unity and (1 + ω)7 = l + mω, then (A) l = 0, m = 1 (B) l = 1, m = 1 (C) l = 1, m = 0 (D) l = –1, m = 1 Sol: We have (1 + ω)7 = l + mω ⇒ (–ω2)7 = l + mω ⇒ –ω14 = l + mω ⇒ –ω2 = l + mω ⇒ 1 + ω = l + mω ⇒ l = m = 1 Hence (B) is the correct answer. 10. If =1 and arg (z1 z2) = 0, then (A) z1 = z2 (B) |z2|2 = z1z2 (C) z1z2 = 1 (D) none of these. Sol: Let z1 = r1( cosθ1 + i sinθ1) then = 1 ⇒ |z1| = |z2| ⇒|z1| = |z2| = r1 . Now arg (z1 z2) = 0 ⇒ arg( z1) + arg(z2) = 0 ⇒ arg(z2) = – θ1 Therefore, z2 = r1 ( cos(–θ1) + i sin(–θ1)) = r1( cosθ1 – i sinθ1) = ⇒ = = z1 ⇒|z2|2 = z1 z2 . Hence (B) is the correct answer. 11. If |z| < 4, then | iz +3 – 4i| is less than (A) 4 (B) 5 (C) 6 (D) 9 Sol: | iz + ( 3 – 4i)| ≤ | iz| + |3 – 4i| = |z| +5 < 4 + 5 = 9. Hence (D) is the correct answer. 12. If z is a complex number, then z2 + = 2 represents (A) a circle (B) a straight line (C) a hyperbola (D) an ellipse Sol: Let z = x + iy, then z2 + = 2 ⇒ x2 – y2 =1, which represents a hyperbola. Hence (C) is the correct answer. 13. If = A + iB, then A2 +B2 equals to (A) 1 (B) α2 (B) –1 (D) – α2 Sol: A +iB = ⇒ A – iB = ⇒ ( A+ iB) ( A – iB) = = 1 ⇒ A2 + B2 = 1. Hence (A) is the correct answer. 14. If |z1| = |z2| and arg(z1) +arg(z2) = π/2 , then (A) z1z2 is purely real (B) z1z2 is purely imaginary (C) (z1+z2) is purely imaginary (D) none of these Sol: Let |z1| = |z2| = r ⇒ z1 = r ( cosθ + isinθ) and z2 = r ⇒ z1z2 = r2 i, which is purely imaginary Hence (B) is the correct answer. 15. If z1 and z2 are two complex numbers satisfying the equation , then is a (A) purely real (B) of unit modulus (C) purely imaginary (D) none of these Sol: (z1 + iz2) = (z1 – iz2) ⇒ ⇒ ⇒ is purely real. Hence (A) is the correct Alternatives. 16. If z = –2 + , then z2n + 22n zn + 24n, n is a multiple of 3, is equal to (A) 22n (B) 0 (C) 3. 24n (D) none of these Sol: z = –2 + 2√3i = 4w z2n + 22n zn + 24n = 42n w2n + 22n ⋅ 4n ⋅ wn + 24n = 42n [w2n + wn + 1] = 0, if n is not a multiple of 3 = 3.42n, if n is a multiple of 3. Hence (C) is the correct answer. 17. The complex numbers z = x + iy which satisfy the equation lie on (A) x–axis (B) the straight line y=5 (C) a circle passing through the origin (D) None of these. Sol: ⇒ z would lie on the right bisector of the line segment connecting the points 5i and – 5i . Thus z would lie on the x–axis. Hence (A) is the correct answer. 18. The points z1, z2, z3, z4 in the complex plane are the vertices of a parallelogram taken in order if and only if (A) z1+z4=z2+z3 (B) z1+z3=z2+z4 (C) z1+z2=z3+z4 (D) None of these. Sol: In parallelogram, diagonals bisects each other, thus mid–point of AC and BD should be same . ⇒ ⇒ z1+ z3 = z2 +z4 Hence (B) is the correct answer. 19. If z = x + iy, and then ⏐w⏐ = 1 implies that in the complex plane (A) z lies on the imaginary axis (B) z lies on the real axis (C) z lies on the unit circle (D) None of these. Sol: As |w| = 1 ⇒ | z – i| = |1 – iz| = |z + i| ( as 1/i = –i) ⇒ z lies on the right bisector of the line segment connecting the points i and –i. Thus ‘z’ lies on the real axis. Hence (B) is correct answer. 20. The complex number z =1+i is rotated through an angle 3π/2 in anticlockwise direction about the origin and stretched by additional unit, then the new complex number is (A) – – i (B) – i (C) 2– i (D) none of these Sol: If z1 be the new complex number then |z1| = |z| + = 2 Also ⇒ z1 = z. 2 = 2( 1+i) ( 0 – i) = – 2i +2 = 2( 1– i) Hence (D) is the correct answer. 21. The value of , where i = equals (A) i (B) i –1 (C) –i (D) none of these Sol: Given summation = = i = Hence (B) is correct answer. 22. If i = , then 4 + 5 equals (A) 1 –i (B) –1 + i (C) i (D) –i Sol: We have 4 + 5 = 4 + 5ω334 + 3ω365 = 4 + 5ω + 3ω2 = 4 + 5 Hence (C) is correct answer. 23. If n1, n2 are positive integers then is a real number if and only if (A) n1 = n2 + 1 (B) n1 + 1 = n2 (C) n1 = n2 (D) none of these Sol: Given expression = = = real Hence (C) is the correct answer. 24. The smallest positive integral value of n for which is purely imaginary with positive part, is (A) 1 (B) 3 (C) 5 (D) none of these Sol: = –i ∴ (–i)n = imaginary ⇒ n = 1, 3, 5, …… Hence (A) is the correct answer. 25. If (a + ib)5 = α + iβ then (b + ia)5 is equal to (A) β + iα (B) α – iβ (C) β – iα (D) – α – iβ Sol: (b + ia)5 = i5(a –ib)5 = i(α –iβ) Hence (C) is the correct answer. 26. If the area of the triangle on the complex plane formed by the points z, iz and z + iz is 50 square units, then |z| is (A) 5 (B) 10 (C) 15 (D) none of these Sol: |z|2 = 50 ⇒ |z|2 = 100 ⇒ |z| = 10 Hence (B) is the correct answer. 27. If z is a complex number satisfying the relation |z + 1| = z + 2(1 + I) then z is (A) 1/2 (1 + 4i) (B) 1/2 (3 + 4i) (C) 1/2 (1 – 4i) (D) 1/2 (3 – 4i) Sol: Given that |z + 1| = z + 2(1 + i) Let z = x + iy ∴ = x + 2 and 0 = y + 2 Hence (C) is the correct answer. 28. If z = x + iy such that |z + 1| = |z –1| and amp then (A) x = + 1, y = 0 (B) x = 0, y = + 1 (C) x = 0, y = 1 (D) none of these Sol: = 1 ⇒ = 1 ∴ z = Hence (D) is the correct answer. 29. Let z = . Then arg z is (A) 2θ (B) 2θ –π (C) π + 2θ (D) none of these Sol: z = (cos θ + i sin θ)2 = cos 2θ + i sin 2θ, < 2θ < π ⇒ z is a complex number in the second quadrant. ⇒ < arg (z) < π ⇒ arg (z) = tan–1 (tan 2θ) = 2θ Hence (A) is the correct answer. 30. If the cube root of unity are 1, ω, ω2, then the roots of the equation (x – 1)3 + 8 = 0 are (A) –1, 1 + 2ω, 1 + 2ω2 (B) –1, 1 – 2ω, 1 – 2ω2 (C) –1, – 1, –1 (D) none of these Sol: (x – 1)3 + 8 = 0  (x – 1)3 = – 8 . Hence (B) is the correct answer. 31. If eiθ = cos θ + i sin θ then for the Δ ABC, eiA . eiB . eiC is (A) –i (B) 1 (C) –1 (D) none of these Sol: eiA.eiB.eiC = ei(A + B + C) = cos (A + B + C) + i sin (A + B + C) = –1 Hence (C) is the correct answer. 32. If z = reiθ, then ⏐eiz⏐ is equal to (A) e–rcosθ (B) e–rsinθ (C) ersinθ (D) none of these Sol: |eiz| = |ei(rcos θ + ri sin θ)| = |e(–rsin θ + ri sin θ)| = |e–r sin θ||eir cos θ| = e–r sin θ Hence (B) is the correct answer. 33. If z lies on the circle ⏐z⏐=1, then lies on (A) circle (B) straight line (C) parabola (D) none of these Sol: ⇒ straight line Hence (B) is the correct answer. 34. If |z – i| < 1, then the value of |z + 12 – 6i| is less than (A) 14 (B) 2 (C) 28 (D) none of these Sol: |z + 12 –6i| = |(z –i) + (12 –5i)| ≤ |z –i| + |12 –5i| < 1 + 13 = 14 Hence (A) is the correct answer. 35. The value of amp (iω) – amp (iω2), where i = and ω = = non–real, is (A) 0 (B) (C) π (D) none of these Sol: amp (iω) + amp (iω2) = amp (i2ω3) = amp (–1) = π Hence (C) is the correct answer. 36. For a complex number z, the minimum value of | z | + | z –2 | is (A) 1 (B) 2 (C) 3 (D) none of these Sol: From the triangle |z| + |z –2| ≥ 2. Hence (B) is the correct answer. 37. If | z | = 1 then is equal to (A) z (B) (C) z + (D) none of these Sol: Hence (A) is the correct answer. 38. If | z1 –1 | < 1, | z2 –2 | < 2, | z3 –3 | < 3 then | z1 + z2 + z3 | is (A) less than 6 (B) more than 3 (C) less than 12 (D) lies between 6 and 12 Sol: |z1 + z2 + z3| = |(z1 –1) +(z2 –2) + (z3 –3) + 6| < |z1 –1| + |z2 –2| + |z3 –3| + 6 < 12. Hence (C) is the correct answer. 39. The roots of the equation 1 + z + z3 + z4 = 0 are represented by the vertices of (A) a square (B) an equilateral triangle (C) a rhombus (D) none of these Sol: z4 + z3 + z + 1 = 0 ⇒ z3(z + 1) + 1(z + 1) = 0 ⇒ (z + 1)(z3 + 1) = 0 ⇒ z = –1, –ω, –ω2 Clearly it represents vertices of an equilateral triangle. Hence (B) is the correct answer. 40. The equation | z + i | – | z – i | = k represents a hyperbola if (A) –2 < k < 2 (B) k > 2 (C) 0 < k < 2 (D) none of these Sol: |z + i| –|z –i| = k represents a hyperbola if 4 – < 0 i.e. k2 < 4. Hence (A) is the correct answer. 41. If amp (z1z2) = 0 and | z1 | = | z2| = 1 then (A) z1 + z2 = 0 (B) z1z2 = 1 (C) z1 –z2 = 0 (D) none of these Sol: amp(z1z2) = 0 ⇒ amp (z1) = amp |z1| = |z2| ⇒ |z1| = | |, so z1 = also z1z2 = z2 = |z2|2 = 1 (since |z2| = 1). Hence (B) is the correct answer. 42. If i = and n is a positive integer, then in + in + 1 + in + 2 + in + 3 = (A) 1 (B) i (C) in (D) 0 Sol: in + in + 1 + in + 2 + in + 3 = in(i + i + i2 + i3) = 0 Hence (A) is the correct answer. 43. (sin θ + i cos θ)4 equals (A) sin 4θ + cos 4θ (B) sin 4θ –i cos 4θ (C) sin 4θ + i sin 4θ (D) cos 4θ –i sin 4θ Sol: (sin θ – i cos θ)4 = i4(cos θ – i sin θ)4 = cos 4θ – i sin 4θ. Hence (D) is the correct answer. 44. The complex number lies in (A) Ist quadrant (B) IInd quadrant (C) IIIrd quadrant (D) IVth quadrant Sol: ⇒ Given complex number lies in the IInd quadrant. Hence (B) is the correct answer. 45. The origin and the roots of the equation z2 + pz + q=0 form an equilateral triangle, if (A) p2 = q (B) p2 = 3q (C) q2 = 2p (D) q2 = p Sol: z1 + z2 = –p, z1z2 = q, also z3 = 0 + 0i For equilateral triangle = z1z2 + z1z3 + z2z3 ⇒ (z1 + z2 + z3)2 –2(z1z2 + z1z3 + z2z3) = z1z2 + z1z3 + z2z3 ⇒ p2 = 3q. Hence (B) is the correct answer. 46. If z = (λ+3) + i , then locus of z is (A) circle (B) parabola (C) line (D) none of these Sol: Let z = x + iy x + iy = (λ 3) + i ⇒ λ + 3 = x and y = λ = x –3 ∴ y = ⇒ x2 + y2 –6x + 6 = 0, which is a circle. Hence (A) is the correct answer. 47. is equal to (A) i (B) 2i (C) 1 – i (D) 1 – 2i. Sol: Hence (B) is the correct answer. 48. If α is a complex number such that , then is equal to (A) α (B) α2 (C) 0 (D) 1 Sol: . Hence (A) is the correct answer. 49. If ω is a complex cube root of unity, then the value of is (A) 1 (B) 0 (C) 2 (D) – 1 Sol: . Hence (D) is the correct answer. 50. is equal to (A) 1 (B) – 1/2 (C) (D) – 1 Sol: . Hence (A) is the correct answer. 51. is equal to (A) 32 (B) 64 (C) – 64 (D) None of these. Sol: Similarly,  Required sum = – 32 + (– 32) = – 64. Hence (C) is the correct answer. 52. If ω is an imaginary cube root of unity, then the value of is (A) – 2 (B) – 1 (C) 1 (D) 0 Sol: { } Hence (D) is the correct answer. 53. is equal to (A) 1 + i (B) 1 – i (C) 1 (D) – 1 Sol: Given exp. . Hence (D) is the correct answer. 54. If z is any complex number such that , then the value of is (A) 1 (B) – 1 (C) 2 (D) – 2 Sol: . Case I. z = – ω . Case II z = – ω2 Hence (D) is the correct answer. 55. The value of is (A) 262 (B) 264 (C) – 262 (D) 0 Sol: . Hence (C) is the correct answer. 56. If , then b is equal to (A) (B) (C) 1 (D) None of these Sol: . Hence (A) is the correct answer. 57. The value of is equal to (A) 0 (B) 2 ω (C) 2 ω2 (D) – 3 ω2 Sol: (Operating C1 → C1 + C2) . Hence (D) is the correct answer. 58. If ω is a complex cube root of unity, then the value of is (A) x3 (B) 2x3 (C) 3x3 (D) None of these Sol: (Operating C1 → C1 + C2 + C3) . Hence (A) is the correct answer. 59. If then is equal to (A) z (B) z2 (C) z3 (D) None of these Sol: Also . Hence (C) is the correct answer. COMPLEX NUMBER SYLLABUS Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations. BASIC CONCEPTS A number in the form of x + iy, where x, y are real numbers and i = is called a complex number. (i) Clearly , for an integer n. (ii) If z = x + iy, then the real part of z is denoted by Re (z) and the imaginary part by Im(z). (iii) A complex number is said to be purely real if Im(z) = 0, and is said to be purely imaginary if Re (z) = 0. The complex number 0 = 0 + i0 is both purely real and purely imaginary. (iv) Two complex numbers are said to be equal if and only if their real parts and imaginary parts are separately equal i.e. a + ib = c + id implies a = c and b = d. However, there is no order relation between complex numbers and the expressions of the type a + ib < ( or > ) c + id are meaningless. (v) Since a real number a can be written as a + i.0, therefore every real number can be considered as a complex number whose imaginary part is zero. Thus the set R of real numbers is a proper subset of the complex numbers C. Remark: ⮚ Clearly , for an integer n. REPRESENTATION OF COMPLEX NUMBERS IN ARGAND PLANE A complex number z = x + iy written as ordered pair (x, y) can be represented by a point P whose Cartesian coordinates are (x, y) referred to axes OX and OY, usually called the real and the imaginary axes. The plane of OX and OY is called the Argand diagram or the complex plane. MODULUS OF A COMPLEX NUMBER Let z = x + iy be a complex number then its magnitude is defined by the real number and is denoted by |z|. ARGUMENT OF A COMPLEX NUMBER If z = x + iy then angle θ given by tan is said to be the argument or amplitude of the complex number z and is denoted by arg (z) or amp (z). In case of x = 0 (where y ≠ 0), arg (z) = + π/2 or –π/2 depending upon y > 0 or y < 0 and the complex number is called purely imaginary. If y = 0 (where x≠0), then arg (z) = 0 or π depending upon x > 0 or x < 0 and the complex number is called purely real. The argument of the complex number 0 is not defined. We can define the argument of a complex number also as any value of the θ which satisfies the system of equations cosθ = . PRINCIPAL ARGUMENT OF A COMPLEX NUMBER The value of θ satisfying the inequality – π < θ ≤ π is called the principal value of the argument. Method of finding the principle argument of a complex number z = x + iy Stet 1: Find tanθ = and this gives the value of θ in the first quadrant. Step 2: Find the quadrant in which z lies, with the help of sign of x and y co–ordinates. Step 3: Then argument of z will be θ, π – θ, θ – π, and –θ according as z lies in the first second, third or fourth quadrant Example -1: For z = √3 – i, find the principal value of arg(z). Solution: Here x = √3, y = -1 ⇒ tanθ = ⇒ θ = ⇒ Principal value of arg z = - . (Since z lies in the fourth quadrant) POLAR FORM OF A COMPLEX NUMBER Let OP = r, then x = r cos θ , and y = r sin θ ⇒ z = x + iy = r cos θ + ir sin θ = r ( cos θ + i sin θ ). This is known as Polar form(Trigonometric form) of a Complex Number. Here we should take the principal value of θ. For general values of the argument z = r [ cos ( 2nπ + θ) + i sin ( 2nπ + θ)] (where n is an integer) Note: sometimes cos θ + i sin θ is, in short, written as cis(θ). Euler's formula: cos θ + i sin θ = ei θ. Note: When complex numbers are multiplied their arguments get added i..e (cos θ1 + i sin θ1) (cos θ2 + i sin θ2) = cos (θ1 + θ2) + i sin (θ1 + θ2) generalizing, (cos θ1+ i sin θ1)(cos θ2+ i sin θ2)…(cos θn + i sin θn) = cos(θ1+ θ2+ …+θn) + i sin(θ1+ θ2+…+θn) when complex numbers are divided, their arguments get subtracted = cos (θ1 – θ2) + i sin (θ1 – θ2) Illustration 1: Find the modulus and the principal argument of the numbers: (i) 6(cos 310° – i sin 310°) (ii) Solution: (i) 6(cos 310° – i sin 310°) = 6[cos(360° – 50°) – i sin(360° – 50°)] = 6(cos 50° + i sin 50°) = ∴ modulus = 6 and principal value of the argument = (ii) = – i = 1 ∴ Modulus = 1 and the principal value of the argument = – Illustration 4: If z = (i)i, where , then find the value of . Solution: z = i i = Unimodular Complex Number: A complex number z such that |z| = 1 is said to be unimodular complex number. Since |z| = 1, z lies on a circle of radius 1 unit and centre (0, 0). If |z| = 1 ⇒ z = cos θ + i sin θ, ⇒ 1/z = (cos θ + i sinθ)-1 = cos θ - i sinθ Illustration 9: If z1 and z2 are two nonzero complex numbers and is a uni–modular then show that is purely real. Solution: ⇒ ⇒ lies on the right bisector of line joining A (i) and B(– i ) that mean lies on real axis. Hence is purely real Algebraic Operations with Complex Numbers: ● Addition : (a + ib) + (c+id) = (a + c) + i ( b+d) ● Subtraction : (a + ib) - (c+id) = (a - c) + i ( b - d) ● Multiplication : (a + ib) (c+id) = (ac - bd) + i ( ad + bc) ● Division : ( when at least one of c and d is non-zero) = Square root of a Complex Number: Let z1 = x1 + iy1 be the given complex number and we have to obtain its square root. Let x + iy = (x1 + iy1)½ ⇒ x2 − y2 + 2ixy = x1 + iy1 ⇒ x1 = x2 − y2 and y1 = 2xy ⇒ x2 − ⇒ x2 = , If y1 > 0 ⇒ x = ⇒ = ± If y1 < 0 ⇒ x = ⇒ = ± Example -2: Find the square root of 8 - 15i. Solution: Here y = -15 < 0 ⇒ = ± = ± (5 - 3i). CONJUGATE OF A COMPLEX NUMBER The conjugate of the complex number z = a + ib is defined to be a – ib and is denoted by . In other words is the mirror image of z in the real axis. If z = a + ib, z + = 2 a ( real), z – = 2 ib ( imaginary ) and z = ( a+ib)(a–ib) = a2 + b2 (real ) = |z|2 = . Also Re (z) = Properties of Conjugate: ● ● |z| = | | ● z + =2Re(z). z – = 2i Im(z). ● If z is purely real z = . whenever we have to show a complex number purely real we use this property. ● If z is purely imaginary z+ =0, whenever we have to show that a complex number is purely imaginary we use this property. ● = ● = 1 + 2 In general, ● = ● In general ● ● Properties of Modulus: ● |z| = 0 ⇒ z = 0 + i0 ● |z1 – z2 | denotes the distance between z1 and z2 . ● –|z| ≤ Re(z) ≤ |z| ; equality holds on right or on left side depending upon z being positive real or negative real. ● –|z| ≤ Im z ≤ |z| ; equality holds on right side or on left side depending upon z being purely imaginary and above the real axes or below the real axes. ● |z| ≤ |Re(z)| + |Im(z)| ≤ |z| ; equality holds on left side when z is purely imaginary or purely real and equality holds on right side when |Re(z)| = |Im(z)|. ● |z|2 = z ● |z1z2| = |z1| |z2| In general |z1 z2 . . . . .zn| = |z1| |z2| . . . . . |zn| ● |zn| = |z|n , n ∈ I ● ● |z1+z2| ≤ |z1| + |z2| ⇒ |z1+z2+ ... +zn| ≤ |z1| + |z2| + ... + |zn|; equality holds if origin, z1, z2, z3 …, zn are collinear and z1 , z2, z3, …,zn are on the same side of the origin. ● |z1 – z2| ≥ ||z1| – |z2|| ; equality holds when arg(z1/z2) = π i.e. origin, z1, z2 are collinear and z1 and z2 are on the opposite side of the origin. ● |z1 + z2|2 = (z1 + z2) ( 1 + 2) = |z1|2 + |z2|2 + z1 2 + z2 1 = |z1|2 + |z2|2 + 2Re(z1 2) ● |z1 – z2|2 = (z1 – z2) ( 1 – 2) = |z1|2 + |z2|2 – z1 2 – z2 1 = |z1|2 + |z2|2 – 2Re(z1 2) Properties of Argument: ● arg(z1z2) = θ1 + θ2 = arg(z1) + arg(z2) ● arg (z1/z2) = θ1 – θ2 = arg(z1) – arg(z2) ● arg (zn) = n arg(z), n ∈I Note: ● In the above result θ1 + θ2 or θ1 – θ2 are not necessarily the principle values of the argument of corresponding complex numbers. E.g arg(zn) = n arg(z) only shows that one of the argument of zn is equal to n arg(z) (if we consider arg(z) in the principle range) ● arg(z) = 0, π ⇒ z is a purely real number ⇒ z = . ● arg(z) = π/2, –π/2 ⇒ z is a purely imaginary number ⇒ z = – . Note that the property of argument is the same as the property of logarithm. Example -3: Consider two pairs of non-zero conjugate complex numbers ( z1, z2) and (z3 , z4). Find the value of arg +arg . Solution: arg +arg = arg = arg (as z2 = ) = 0 (as argument of a positive real number is zero). Example -4: If |z1| = |z2| = |z3| = 1, prove that . Solution: We know that |z| = ⇒ |z1 + z2 + z3| = = = = ( |z1|2 = |z2|2 = |z3|2 = 1) Example -5: Prove that |z1| + |z2| = . Solution: RHS = = = 2 Using parallelogram law. = |z1| + |z2| = LHS. DE MOIVRE’S THEOREM If n is any integer, then (cos θ + i sin θ)n = cos nθ + i sin nθ. This is known as De Movre’s Theorem. Remarks: ● Writing the binomial expansion of (cos  + i sin )n and equating the real part to cos n and the imaginary part to sin n, we get ? cos n = cosn  – nc2 cosn–2 sin2 + nc4 cosn–4 sin4 + ……… ? sin n = nc1 cosn–1 sin – nc3 cosn–3 sin3 + nc5 cosn–5 sin5 + ……… ? ⇒ tan n = ● If n is rational number, then one of the values of (cos  + i sin )n is cos n + i sin n. Let n = p/q, where p and q are integers (q > 0) and p, q have no common factor. Then (cos  + i sin )n has q distinct values, one of which is cos n + i sin n. ● If z = r (cos  + i sin ), and n is a positive integer, then ? z1/n = r1/n , k = 0, 1, 2, ……, n –1. ? Here if can be noted that any ‘n’ consecutive values of k will serve the purpose. ? 2.1 APPLICATIONS OF DE MOIVER’S THEOREM This is a fundamental theorem and has various applications. Here we will discuss few of these which are important from the examination point of view. nth Roots of Unity One very important application of De–Moivre’s Theorem is in solving equation of nth powers in complex number. Let x be the nth root of unity. Then xn = 1 = cos 2kπ + i sin 2kπ (where k is an integer) ⇒ x = cos + i sin k = 0, 1, 2, ……, n – 1 Let α = cos + i sin . When k = 2 = [By De–Moivre’s Theorem] = α2 when k = 3 = α3 Similarly, when k = t Then, = αt ∴ The roots are 1, α, α2, ……, αn –1 Sum of the Roots 1 + α + α2 + .... + αn – 1 = = 0 ( αn = 1) and Thus the sum of the roots of unity is zero. Product of the Roots 1.α.α2. .......... αn – 1 = = = cos{π(n – 1)} + i sin{π(n – 1)} If n is even =–1 If n is odd =1 Note : ● The points represented by n nth roots of unity are located at the vertices of a regular polygon of n sides inscribed in a unit circle having centre at the origin, one vertex being on the positive real axis. Geometrically represented as follows. 2.2 CUBE ROOTS OF UNITY For n = 3, we get the cube roots of unity and they are 1, cos + i sin and cos + i sin i.e. 1, and . They are generally denoted by 1, ω and ω2 and geometrically represented by the vertices of an equilateral triangle whose circumcentre is origin and circumradius is unity. Note: ● ω3 = 1 and 1 + ω + ω2 = 0 ● It can be easily proved that 1 + ωn + ω2n = 3 (n is a multiple of 3) 1 + ωn +ω2n = 0 (n is an integer, not a multiple of 3) Illustration 14: Given z1 + z2 + z3 = A , z1 + z2w + z3w2 = B z1 + z2w2 + z3w = C where w is cube root of unity. Express z1, z2, z3 in term of A, B , C. Solution: Adding given three condition z1 = Multiplying z1 + z2 + z3 = A, z1 + z2w + z3w2 = B, z1 + z2w2 + z2w = C by 1 , w2 ,w and adding we get also z3 = Exercise 7. If ω is a cube root of unity then find the value of Solution Ex. 7: = CONCEPT OF ROTATION If z and z′ are two complex numbers then argument of is the angle through which Oz′ must be turned in order that it may lie along Oz. = In general, let z1, z2, z3 , be the three vertices of a triangle ABC described in the counter-clock wise sense. Draw OP and OQ parallel and equal to AB and AC respectively. Then the point P is z2 – z1 and Q is z3 – z1 and Note that arg. (z3-z1)–arg(z2-z1) = α is the angle through which OP must be rotated in the anti-clockwise direction so that it becomes parallel to OQ. Here we can write also. In this case we are rotating OP in clockwise direction by an angle (2π - α). Since the rotation is in clockwise direction, we are taking negative sign with angle (2π - α) Example -10: Consider a square ABCD such that z1, z2, z3 and z4 represent its vertices A, B, C and D respectively. Express ‘z3’ and ‘z4’ in terms of z1 and z2 . Solution: Consider the rotation of AB about A through an angle π/4, we get ⇒ z3 = z1 +( z2 – z1)( 1+i) Similarly , ⇒ z4 = z1 +i( z2 – z1) Illustration 17: If |z – 3| = 3 then show that i tan (Argz) Solution: By figure | z – 3 | = 3 is a circle with centre (3, 0) by rotation eiπ / 2 = i tan θ = i tan (arg z) GEOMETRICAL APPLICATIONS 4.1 SECTION FORMULA Let z1 and z2 be any two complex numbers representing the points A and B respectively in the argand plane. Let C be the point dividing the line segment AB internally in the ratio m : n i.e , and let the complex number associated with point C be z. Let us rotate the line BC about the point C so that it becomes parallel to CA . The corresponding equation of rotation will be , = ⇒ nz1 – nz = –m z2 +mz ⇒ z = . Similarly if C(z) divides the segment AB externally in the ratio of m : n, then z = . In the specific case, if C(z) is the mid point of AB then . Example -12: If z1, z2 and z3 ( in anticlockwise sense) represents the vertices of a triangle, find the centroid, incentre, circumcentre and the orthocentre of the triangle. Solution: Let G be the centriod and let the line joining A and G meets the line BC at the point D. We have, BD = DC D ≡ G divides AD internally in ratio 2 : 1 ⇒ G ≡ ≡ Let I be the incentre and let the line connecting A and I meet the line BC at D1. We have and = ⇒ = Let ‘O’ be the circum-centre and let the liane connecting A and O meet the line BC at D2. We have and ⇒ D2 = and O = = Let ‘P’ be the orthocentre and let the line connecting the points A and P meet the line BC and D3. We have, and ⇒ D3 = and P = = Condition for Collinearity: If there are three real numbers (other than 0) l, m and n such that lz1 + mz2 + nz3 = 0 and l + m + n = 0 then complex numbers z1, z2 and z3 will be collinear. Equation of a Straight Line: Equation of Straight Line Joining the Points z1 and z2: Writing re-arranging terms, we find that the equation of the line through z1 and z2 is given by General equation of a straight line is , where a is a complex number and b is a real number. The length of the perpendicular from a point z1 to the line is given by . Example -13: Let α and β be two fixed non-zero complex numbers and z a variable complex number. If the two straight lines and are mutually perpendicular, then prove that = 0 Solution: Slope of the first line = – Slope of the second line = – Lines will be perpendicular if – + = 0 ⇒ = 0. Equation of a Circle: Consider a fixed complex number z0 and let z be any complex number which moves in such a way that it’s distance from z0 is always equals to ‘r’. This implies z would lie on a circle whose centre is z0 and radius r. And it’s equation would be |z –z0| = r . ⇒ |z –z0|2 = r2 ⇒ ( z – z0) ⇒ . Let - a = z0 and =b ⇒ , where centre = –a and radius = Example -15: Find the centre and radius of the circle . Solution: Given equation can be rewritten as So, it represents a circle with centre at –1 –i and radius = . Example -16: If z1, z2, z3 are complex numbers such that , show that the points represented by z1, z2, z3 lie on a circle passing through the origin. Solution: Since P(z1), Q(z2), R(z3) and S(z4) are concyclic points, ∠PSQ = ∠PRQ ⇒ arg. = arg ⇒ arg = 0 ⇒ = real If z4 = 0 + i0, then = real …(1) We have from which z3 = …(2) From (1) and (2), = real ⇒ = real ⇒ = real, which is true. Therefore z1, z2, z3 and the origin are concyclic. Alternative Solution: ⇒ − ⇒ ⇒ ⇒ = π + arg ⇒ α = π − β ⇒ α + β = π ⇒ points A, B, C, D are concyclic. OBJECTIVES 1. If ‘z’ be any complex number such that , then locus of ‘z’ is (A) A circle (B) An ellipse (C) A line segment (D) None of these Sol: (C) ⇒ Let P(z), A = , B = then (i) represents PA + PB = 4/3. Clearly AB = 4/3 PA + PB = AB Thus P is any point on the line segment AB. Hence (C) is the correct answer. 2. The value of the expression 2 +3 + 4 + . . . + (n + 1) , where ω is an imaginary cube root of unity, is (A) (B) (C) (D) none of these Sol: (C) tn = (n+1) = n3 + n2 = n3 + n2(ω + ω2 + 1)+ n(ω + ω2 +1) +1 = n3 +1 ∴ Sn = . Hence (C) is the correct answer. 3 If ω is an imaginary cube root of unity, then (1 + ω – ω2)7 equals (A) 128 ω (B) – 128 ω (C) 128 ω2 (D) – 128 ω2 Sol: (D) We have (1 + ω + ω2)7 = –(ω2 – ω2)7 = (–2)7 (ω2)7 = 128 ω14 = –128 ω2 Hence (D) is the correct answer. 4. Let ‘z’ be a complex number and ‘a’ be a real parameter such that z2 + az + a2 = 0, then locus of z is (A) a pair of straight lines (B) a circle (C) an ellipse (D) none of these Sol: (A) z2 + az + a2 = 0 ⇒ z = aω, aω2 ( where ‘ω’ is non real root of unity ) ⇒ locus of z is a pair of straight lines Hence (A) is the correct answer. 5. The equation | z + i | – | z – i | = k represents a hyperbola if (A) –2 < k < 2 (B) k > 2 (C) 0 < k < 2 (D) none of these Sol: (A) |z + i| –|z –i| = k represents a hyperbola if 4 – < 0 i.e. k2 < 4. Hence (A) is the correct answer. 6. The complex number z = 1 + i is rotated through an angle in anticlockwise direction about the origin and stretched by additional units, Then the new complex number is (A) – – i (B) – (C) 2 – i (D) 2 – 2i Sol: If z1 be the new complex number then |z1| = |z| + = 2 Also ⇒ z1 = z. 2 = 2( 1+i) ( 0 – i) = – 2i +2 = 2( 1– i) Hence (D) is the correct answer. 7. Let Δ = , where i = , then Δ is (A) purely +ve real number (B) purely imaginary (C) of the form a + ib with a, b both non–zero real numbers. (D) purely negative real number Sol: = – Δ ⇒ Δ is purely imaginary. Hence (B) is the correct answer. 8. If is real, where n1, n2 are positive integers then (A) n1 = n2 (B) n1 = n2 + 1 (C) n2 = n1 + 1 (D) n1n2 ∈ N Sol: = = 2 = Real number. Thus n1, n2 ∈ N Hence (D) is the correct answer. 9. If (ω ≠ 1) be a cube root of unity and (1 + ω)7 = l + mω, then (A) l = 0, m = 1 (B) l = 1, m = 1 (C) l = 1, m = 0 (D) l = –1, m = 1 Sol: We have (1 + ω)7 = l + mω ⇒ (–ω2)7 = l + mω ⇒ –ω14 = l + mω ⇒ –ω2 = l + mω ⇒ 1 + ω = l + mω ⇒ l = m = 1 Hence (B) is the correct answer. 10. If =1 and arg (z1 z2) = 0, then (A) z1 = z2 (B) |z2|2 = z1z2 (C) z1z2 = 1 (D) none of these. Sol: Let z1 = r1( cosθ1 + i sinθ1) then = 1 ⇒ |z1| = |z2| ⇒|z1| = |z2| = r1 . Now arg (z1 z2) = 0 ⇒ arg( z1) + arg(z2) = 0 ⇒ arg(z2) = – θ1 Therefore, z2 = r1 ( cos(–θ1) + i sin(–θ1)) = r1( cosθ1 – i sinθ1) = ⇒ = = z1 ⇒|z2|2 = z1 z2 . Hence (B) is the correct answer. 11. If |z| < 4, then | iz +3 – 4i| is less than (A) 4 (B) 5 (C) 6 (D) 9 Sol: | iz + ( 3 – 4i)| ≤ | iz| + |3 – 4i| = |z| +5 < 4 + 5 = 9. Hence (D) is the correct answer. 12. If z is a complex number, then z2 + = 2 represents (A) a circle (B) a straight line (C) a hyperbola (D) an ellipse Sol: Let z = x + iy, then z2 + = 2 ⇒ x2 – y2 =1, which represents a hyperbola. Hence (C) is the correct answer. 13. If = A + iB, then A2 +B2 equals to (A) 1 (B) α2 (B) –1 (D) – α2 Sol: A +iB = ⇒ A – iB = ⇒ ( A+ iB) ( A – iB) = = 1 ⇒ A2 + B2 = 1. Hence (A) is the correct answer. 14. If |z1| = |z2| and arg(z1) +arg(z2) = π/2 , then (A) z1z2 is purely real (B) z1z2 is purely imaginary (C) (z1+z2) is purely imaginary (D) none of these Sol: Let |z1| = |z2| = r ⇒ z1 = r ( cosθ + isinθ) and z2 = r ⇒ z1z2 = r2 i, which is purely imaginary Hence (B) is the correct answer. 15. If z1 and z2 are two complex numbers satisfying the equation , then is a (A) purely real (B) of unit modulus (C) purely imaginary (D) none of these Sol: (z1 + iz2) = (z1 – iz2) ⇒ ⇒ ⇒ is purely real. Hence (A) is the correct Alternatives. 16. If z = –2 + , then z2n + 22n zn + 24n, n is a multiple of 3, is equal to (A) 22n (B) 0 (C) 3. 24n (D) none of these Sol: z = –2 + 2√3i = 4w z2n + 22n zn + 24n = 42n w2n + 22n ⋅ 4n ⋅ wn + 24n = 42n [w2n + wn + 1] = 0, if n is not a multiple of 3 = 3.42n, if n is a multiple of 3. Hence (C) is the correct answer. 17. The complex numbers z = x + iy which satisfy the equation lie on (A) x–axis (B) the straight line y=5 (C) a circle passing through the origin (D) None of these. Sol: ⇒ z would lie on the right bisector of the line segment connecting the points 5i and – 5i . Thus z would lie on the x–axis. Hence (A) is the correct answer. 18. The points z1, z2, z3, z4 in the complex plane are the vertices of a parallelogram taken in order if and only if (A) z1+z4=z2+z3 (B) z1+z3=z2+z4 (C) z1+z2=z3+z4 (D) None of these. Sol: In parallelogram, diagonals bisects each other, thus mid–point of AC and BD should be same . ⇒ ⇒ z1+ z3 = z2 +z4 Hence (B) is the correct answer. 19. If z = x + iy, and then ⏐w⏐ = 1 implies that in the complex plane (A) z lies on the imaginary axis (B) z lies on the real axis (C) z lies on the unit circle (D) None of these. Sol: As |w| = 1 ⇒ | z – i| = |1 – iz| = |z + i| ( as 1/i = –i) ⇒ z lies on the right bisector of the line segment connecting the points i and –i. Thus ‘z’ lies on the real axis. Hence (B) is correct answer. 20. The complex number z =1+i is rotated through an angle 3π/2 in anticlockwise direction about the origin and stretched by additional unit, then the new complex number is (A) – – i (B) – i (C) 2– i (D) none of these Sol: If z1 be the new complex number then |z1| = |z| + = 2 Also ⇒ z1 = z. 2 = 2( 1+i) ( 0 – i) = – 2i +2 = 2( 1– i) Hence (D) is the correct answer. 21. The value of , where i = equals (A) i (B) i –1 (C) –i (D) none of these Sol: Given summation = = i = Hence (B) is correct answer. 22. If i = , then 4 + 5 equals (A) 1 –i (B) –1 + i (C) i (D) –i Sol: We have 4 + 5 = 4 + 5ω334 + 3ω365 = 4 + 5ω + 3ω2 = 4 + 5 Hence (C) is correct answer. 23. If n1, n2 are positive integers then is a real number if and only if (A) n1 = n2 + 1 (B) n1 + 1 = n2 (C) n1 = n2 (D) none of these Sol: Given expression = = = real Hence (C) is the correct answer. 24. The smallest positive integral value of n for which is purely imaginary with positive part, is (A) 1 (B) 3 (C) 5 (D) none of these Sol: = –i ∴ (–i)n = imaginary ⇒ n = 1, 3, 5, …… Hence (A) is the correct answer. 25. If (a + ib)5 = α + iβ then (b + ia)5 is equal to (A) β + iα (B) α – iβ (C) β – iα (D) – α – iβ Sol: (b + ia)5 = i5(a –ib)5 = i(α –iβ) Hence (C) is the correct answer. 26. If the area of the triangle on the complex plane formed by the points z, iz and z + iz is 50 square units, then |z| is (A) 5 (B) 10 (C) 15 (D) none of these Sol: |z|2 = 50 ⇒ |z|2 = 100 ⇒ |z| = 10 Hence (B) is the correct answer. 27. If z is a complex number satisfying the relation |z + 1| = z + 2(1 + I) then z is (A) 1/2 (1 + 4i) (B) 1/2 (3 + 4i) (C) 1/2 (1 – 4i) (D) 1/2 (3 – 4i) Sol: Given that |z + 1| = z + 2(1 + i) Let z = x + iy ∴ = x + 2 and 0 = y + 2 Hence (C) is the correct answer. 28. If z = x + iy such that |z + 1| = |z –1| and amp then (A) x = + 1, y = 0 (B) x = 0, y = + 1 (C) x = 0, y = 1 (D) none of these Sol: = 1 ⇒ = 1 ∴ z = Hence (D) is the correct answer. 29. Let z = . Then arg z is (A) 2θ (B) 2θ –π (C) π + 2θ (D) none of these Sol: z = (cos θ + i sin θ)2 = cos 2θ + i sin 2θ, < 2θ < π ⇒ z is a complex number in the second quadrant. ⇒ < arg (z) < π ⇒ arg (z) = tan–1 (tan 2θ) = 2θ Hence (A) is the correct answer. 30. If the cube root of unity are 1, ω, ω2, then the roots of the equation (x – 1)3 + 8 = 0 are (A) –1, 1 + 2ω, 1 + 2ω2 (B) –1, 1 – 2ω, 1 – 2ω2 (C) –1, – 1, –1 (D) none of these Sol: (x – 1)3 + 8 = 0  (x – 1)3 = – 8 . Hence (B) is the correct answer. 31. If eiθ = cos θ + i sin θ then for the Δ ABC, eiA . eiB . eiC is (A) –i (B) 1 (C) –1 (D) none of these Sol: eiA.eiB.eiC = ei(A + B + C) = cos (A + B + C) + i sin (A + B + C) = –1 Hence (C) is the correct answer. 32. If z = reiθ, then ⏐eiz⏐ is equal to (A) e–rcosθ (B) e–rsinθ (C) ersinθ (D) none of these Sol: |eiz| = |ei(rcos θ + ri sin θ)| = |e(–rsin θ + ri sin θ)| = |e–r sin θ||eir cos θ| = e–r sin θ Hence (B) is the correct answer. 33. If z lies on the circle ⏐z⏐=1, then lies on (A) circle (B) straight line (C) parabola (D) none of these Sol: ⇒ straight line Hence (B) is the correct answer. 34. If |z – i| < 1, then the value of |z + 12 – 6i| is less than (A) 14 (B) 2 (C) 28 (D) none of these Sol: |z + 12 –6i| = |(z –i) + (12 –5i)| ≤ |z –i| + |12 –5i| < 1 + 13 = 14 Hence (A) is the correct answer. 35. The value of amp (iω) – amp (iω2), where i = and ω = = non–real, is (A) 0 (B) (C) π (D) none of these Sol: amp (iω) + amp (iω2) = amp (i2ω3) = amp (–1) = π Hence (C) is the correct answer. 36. For a complex number z, the minimum value of | z | + | z –2 | is (A) 1 (B) 2 (C) 3 (D) none of these Sol: From the triangle |z| + |z –2| ≥ 2. Hence (B) is the correct answer. 37. If | z | = 1 then is equal to (A) z (B) (C) z + (D) none of these Sol: Hence (A) is the correct answer. 38. If | z1 –1 | < 1, | z2 –2 | < 2, | z3 –3 | < 3 then | z1 + z2 + z3 | is (A) less than 6 (B) more than 3 (C) less than 12 (D) lies between 6 and 12 Sol: |z1 + z2 + z3| = |(z1 –1) +(z2 –2) + (z3 –3) + 6| < |z1 –1| + |z2 –2| + |z3 –3| + 6 < 12. Hence (C) is the correct answer. 39. The roots of the equation 1 + z + z3 + z4 = 0 are represented by the vertices of (A) a square (B) an equilateral triangle (C) a rhombus (D) none of these Sol: z4 + z3 + z + 1 = 0 ⇒ z3(z + 1) + 1(z + 1) = 0 ⇒ (z + 1)(z3 + 1) = 0 ⇒ z = –1, –ω, –ω2 Clearly it represents vertices of an equilateral triangle. Hence (B) is the correct answer. 40. The equation | z + i | – | z – i | = k represents a hyperbola if (A) –2 < k < 2 (B) k > 2 (C) 0 < k < 2 (D) none of these Sol: |z + i| –|z –i| = k represents a hyperbola if 4 – < 0 i.e. k2 < 4. Hence (A) is the correct answer. 41. If amp (z1z2) = 0 and | z1 | = | z2| = 1 then (A) z1 + z2 = 0 (B) z1z2 = 1 (C) z1 –z2 = 0 (D) none of these Sol: amp(z1z2) = 0 ⇒ amp (z1) = amp |z1| = |z2| ⇒ |z1| = | |, so z1 = also z1z2 = z2 = |z2|2 = 1 (since |z2| = 1). Hence (B) is the correct answer. 42. If i = and n is a positive integer, then in + in + 1 + in + 2 + in + 3 = (A) 1 (B) i (C) in (D) 0 Sol: in + in + 1 + in + 2 + in + 3 = in(i + i + i2 + i3) = 0 Hence (A) is the correct answer. 43. (sin θ + i cos θ)4 equals (A) sin 4θ + cos 4θ (B) sin 4θ –i cos 4θ (C) sin 4θ + i sin 4θ (D) cos 4θ –i sin 4θ Sol: (sin θ – i cos θ)4 = i4(cos θ – i sin θ)4 = cos 4θ – i sin 4θ. Hence (D) is the correct answer. 44. The complex number lies in (A) Ist quadrant (B) IInd quadrant (C) IIIrd quadrant (D) IVth quadrant Sol: ⇒ Given complex number lies in the IInd quadrant. Hence (B) is the correct answer. 45. The origin and the roots of the equation z2 + pz + q=0 form an equilateral triangle, if (A) p2 = q (B) p2 = 3q (C) q2 = 2p (D) q2 = p Sol: z1 + z2 = –p, z1z2 = q, also z3 = 0 + 0i For equilateral triangle = z1z2 + z1z3 + z2z3 ⇒ (z1 + z2 + z3)2 –2(z1z2 + z1z3 + z2z3) = z1z2 + z1z3 + z2z3 ⇒ p2 = 3q. Hence (B) is the correct answer. 46. If z = (λ+3) + i , then locus of z is (A) circle (B) parabola (C) line (D) none of these Sol: Let z = x + iy x + iy = (λ 3) + i ⇒ λ + 3 = x and y = λ = x –3 ∴ y = ⇒ x2 + y2 –6x + 6 = 0, which is a circle. Hence (A) is the correct answer. 47. is equal to (A) i (B) 2i (C) 1 – i (D) 1 – 2i. Sol: Hence (B) is the correct answer. 48. If α is a complex number such that , then is equal to (A) α (B) α2 (C) 0 (D) 1 Sol: . Hence (A) is the correct answer. 49. If ω is a complex cube root of unity, then the value of is (A) 1 (B) 0 (C) 2 (D) – 1 Sol: . Hence (D) is the correct answer. 50. is equal to (A) 1 (B) – 1/2 (C) (D) – 1 Sol: . Hence (A) is the correct answer. 51. is equal to (A) 32 (B) 64 (C) – 64 (D) None of these. Sol: Similarly,  Required sum = – 32 + (– 32) = – 64. Hence (C) is the correct answer. 52. If ω is an imaginary cube root of unity, then the value of is (A) – 2 (B) – 1 (C) 1 (D) 0 Sol: { } Hence (D) is the correct answer. 53. is equal to (A) 1 + i (B) 1 – i (C) 1 (D) – 1 Sol: Given exp. . Hence (D) is the correct answer. 54. If z is any complex number such that , then the value of is (A) 1 (B) – 1 (C) 2 (D) – 2 Sol: . Case I. z = – ω . Case II z = – ω2 Hence (D) is the correct answer. 55. The value of is (A) 262 (B) 264 (C) – 262 (D) 0 Sol: . Hence (C) is the correct answer. 56. If , then b is equal to (A) (B) (C) 1 (D) None of these Sol: . Hence (A) is the correct answer. 57. The value of is equal to (A) 0 (B) 2 ω (C) 2 ω2 (D) – 3 ω2 Sol: (Operating C1 → C1 + C2) . Hence (D) is the correct answer. 58. If ω is a complex cube root of unity, then the value of is (A) x3 (B) 2x3 (C) 3x3 (D) None of these Sol: (Operating C1 → C1 + C2 + C3) . Hence (A) is the correct answer. 59. If then is equal to (A) z (B) z2 (C) z3 (D) None of these Sol: Also . Hence (C) is the correct answer. 60. The value of ..... 2n factors is equal to (A) 23n (B) 22n (C) 2n (D) None of these. Sol: First factor = – ω – ω = – 2 ω Second factor 3rd factor, 5th factor, ....... are equal to the first factor, which is – 2ω. 4th factor, 6th factor, ....... are equal to the second factor, which is – 2ω2.  Required product factors factors = 4.4 ...... n factors = 4n = (22)n = 22n. Hence (B) is the correct answer. 60. The value of ..... 2n factors is equal to (A) 23n (B) 22n (C) 2n (D) None of these. Sol: First factor = – ω – ω = – 2 ω Second factor 3rd factor, 5th factor, ....... are equal to the first factor, which is – 2ω. 4th factor, 6th factor, ....... are equal to the second factor, which is – 2ω2.  Required product factors factors = 4.4 ...... n factors = 4n = (22)n = 22n. Hence (B) is the correct answer.