Complex number-(E)-03-Theory-III

(3) Vector representation : If P is the point (a, b) on the argand plane corresponding to the complex number . Then ,  and arg z = direction of the vector Therefore, complex number z can also be represented by . (4) Eulerian representation (Exponential form) : Since we have = and thus z can be expressed as , where and arg (z) Note :   Example: 32 (a) (b) (c) (d) None of these Solution: (a) Let ,  and  and , Thus Alternative method: and arg  Example: 33 If then  is equal to (a) (b) (c) (d) Solution: (c) Here  Equating real and imaginary part, we get and Hence   . Example: 34 Real part of is (a) (b) (c) (d) Solution: (a)  Real part of is . Example: 35 If then is equal to (a) (b) (c) (d) Solution: (a) Let then 2.10 Logarithm of a Complex Number. Let and .....(i) .....(ii) then , clearly and From equation (ii), Obviously, the general value is Example: 36 is equal to (a) (b) (c) (d) None of these Solution: (b) Let then   and arg log A  . Example: 37 is equal to (a) (b) (c) (d) Solution: (b) Let    .....(i) ∵ Hence (By equation (i))  2.11 Geometry of Complex Numbers. (1) Geometrical representation of algebraic operations on complex numbers (i) Sum: Let the complex numbers and be represented by the points P and Q on the argand plane. Then sum of and i.e., is represented by the point R. Complex number z can be represented by In vector notation, we have (ii) Difference : We first represent by Q', so that QQ' is bisected at O. The point R represents the difference In parallelogram ORPQ, We have in vectorial notation . (iii) Product : Let  and arg and  and arg Then, =  and arg R is the point representing product of complex numbers and Important Tips  Multiplication of i : Since and then Hence, multiplication of z with i then vector for z rotates a right angle in the positive sense. i.e., To multiply a vector by –1 is to turn it through two right angles. i.e., To multiply a vector by is to turn it through the angle  in the positive sense. (iv) Division : Let  and arg and  and arg Then   Note :  The vertical angle R is i.e., .  If and are the principal values of and then and are not necessarily the principal value of and . 2.12 Use of Complex Numbers in Co-ordinate Geometry. (1) Distance formula : The distance between two points and is given by = |affix of Q – affix of P| Note :  The distance of point z from origin Thus, modulus of a complex number z represented by a point in the argand plane is its distance from the origin.  Three points and are collinear then i.e., . (2) Section formula : If R(z) divides the joining of and in the ratio (i) If R(z) divides the segment PQ internally in the ratio of then (ii) If R(z) divides the segment PQ externally in the ratio of then Note :  If R(z) is the mid point of PQ then affix of R is  If are affixes of the vertices of a triangle, then affix of its centroid is (3) Equation of the perpendicular bisector : If and are two fixed points and is moving point such that it is always at equal distance from and i.e., PR = QR or      Hence, z lies on the perpendicular bisector of and . (4) Equation of a straight line (i) Parametric form : Equation of a straight line joining the point having affixes and is (ii) Non parametric form : Equation of a straight line joining the points having affixes and is  . Note :  Three points and are collinear (iii) General equation of a straight line: The general equation of a straight line is of the form , where a is complex number and b is real number. (iv) Slope of a line : The complex slope of the line is and real slope of the line is . Note :  If and are the are the complex slopes of two lines on the argand plane, then (i) If lines are perpendicular then (ii) If lines are parallel then  If lines and are the perpendicular or parallel, then or  or where are the complex numbers and (v) Slope of the line segment joining two points : If and represent two points in the argand plane then the complex slope of AB is defined by . Note :  If three points are collinear then slope of AB = slope of BC = slope of AC (vi) Length of perpendicular : The length of perpendicular from a point to the line is given by or (5) Equation of a circle : The equation of a circle whose centre is at point having affix and radius r is Note :  If the centre of the circle is at origin and radius r, then its equation is .  represents interior of a circle and represent exterior of the circle . Similarly, is the set of all points lying outside the circle and is the set of all points lying outside and on the circle (i) General equation of a circle : The general equation of the circle is where a is complex number and .  Centre and radius are – a and respectively. Note :  Rule to find the centre and radius of a circle whose equation is given: • Make the coefficient of equal to 1 and right hand side equal to zero. • The centre of circle will be = – a • Radius (ii) Equation of circle through three non-collinear points : Let are three points on the circle and be any point on the circle, then Using coni method In ACB, .....(i) In  APB, .....(ii) From (i) and (ii) we get = Real .....(iii) (iii) Equation of circle in diametric form : If end points of diameter represented by and and be any point on circle then, which is required equation of circle in diametric form. (iv) Other forms of circle : (a) Equation of all circle which are orthogonal to and . Let the circle be = r cut given circles orthogonally  …...(i) and …..(ii) on solving and let (b) = k is a circle if and a line if k = 1. (c) The equation will represent a circle if (6) Equation of parabola : Now for parabola or where (focus) Directrix is (7) Equation of ellipse : For ellipse  where (since eccentricity <1) Then point z describes an ellipse having foci at and and . (8) Equation of hyperbola : For hyperbola  where (since eccentricity >1) Then point z describes a hyperbola having foci at and and Example: 38 If in the adjoining diagram, A and B represent complex number and respectively, then C represents (a) (b) (c) (d) Solution: (a) It is a fundamental concept. Example: 39 If centre of a regular hexagon is at origin and one of the vertex on argand diagram is then its perimeter is (a) (b) (c) (d) Solution: (d) Let the vertices be centre O and    .....(i) Similarly, Hence, the perimeter of regular polygon is . Example: 40 The complex numbers and satisfying are the vertices of a triangle which is (a) Of area zero (b) Right-angled isosceles (c) Equilateral (d) Obtuse-angled isosceles Solution: (b) Taking mod of both sides of given relation . So, . Also, amp = or or  The triangle has two sides equal and the angle between the equal sides . So it is equilateral. Example: 41 Let the complex numbers and be the vertices of an equilateral triangle. Let be the circumcentre of the triangle, then (a) (b) (c) (d) Solution: (c) Let r be the circum-radius of the equilateral triangle and the cube root of unity. Let ABC be the equilateral triangle with and as its vertices A, B and C respectively with circumcentre The vectors are equal and parallel to respectively. Then the vectors    Squaring and adding, we get, since Example: 42 The points in the complex plane are the vertices of a parallelogram taken in order, if and only if (a) (b) (c) (d) None of these Solution: (b) Diagonals of a parallelogram ABCD are bisected each other at a point i.e.,  Example: 43 If the complex number and the origin form an equilateral triangle then (a) (b) (c) (d) Solution: (a) Let OA, OB be the sides of an equilateral  OAB and OA, OB represent the complex numbers or vectors respectively. From the equilateral  OAB,  and Also, since triangle is equilateral. Thus the vectors and have same modulus and same argument, which implies that the vectors are equal, that is   2.13 Rotation Theorem. Rotational theorem i.e., angle between two intersecting lines. This is also known as coni method. Let and be the affixes of three points A, B and C respectively taken on argand plane. Then we have and = and let arg arg and arg Let , = arg arg = arg arg = arg or angle between AC and AB = arg For any complex number z we have Similarly, or Note :  Here only principal values of the arguments are considered.  arg , if AB coincides with CD, then arg or , so that is real. It follows that if is real, then the points A, B, C, D are collinear.  If AB is perpendicular to CD, then arg , so is purely imaginary. It follows that if = , where k purely imaginary number, then AB and CD are perpendicular to each other. (1) Complex number as a rotating arrow in the argand plane : Let ..…(i) be a complex number representing a point P in the argand plane. Then and Now consider complex number or {from (i)} Clearly the complex number represents a point Q in the argand plane, when and . Clearly multiplication of z with rotates the vector through angle in anticlockwise sense. Similarly multiplication of z with will rotate the vector in clockwise sense. Note :  If and are the affixes of the points A,B and C such that and . Therefore, . Then will be obtained by rotating through an angle  in anticlockwise sense, and therefore, or or  If A, B and C are three points in argand plane such that and then use the rotation about A to find , but if use coni method.  Let and be two complex numbers represented by point P and Q in the argand plane such that Then, is a vector of magnitude along and is a unit vector along Consequently, is a vector of magnitude along OQ i.e., . (2) Condition for four points to be concyclic : If points A,B,C and D are concyclic Using rotation theorem In .....(i) In .....(ii) From (i) and (ii) =Real So if and are such that is real, then these four points are concyclic. Example: 44 If complex numbers and represent the vertices A, B and C respectively of an isosceles triangle ABC of which is right angle, then correct statement is (a) (b) (c) (d) Solution: (d) and By rotation about C in anticlockwise sense       Example: 45 In the argand diagram, if O,P and Q represents respectively the origin, the complex numbers z and then the angle is (a) (b) (c) (d) Solution: (c) It is a fundamental concept. Example: 46 The centre of a regular polygon of n sides is located at the point and one of its vertex is known. If be the vertex adjacent to then is equal to (a) (b) (c) (d) None of these Solution: (a) Let A be the vertex with affix There are two possibilities of i.e., can be obtained by rotating through either in clockwise or in anticlockwise direction.    Example: 47 Let be three vertices of an equilateral triangle circumscribing the circle If and are in anticlockwise sense then is (a) (b) (c) 1 (d) – 1 Solution: (d) 2.14 Triangle Inequalities. In any triangle, sum of any two sides is greater than the third side and difference of any two side is less than the third side. By applying this basic concept to the set of complex numbers we are having the following results. (1) (2) (3) (4) Note :  In a complex plane is the distance between the points and .  The equality holds only when arg = arg i.e., and are parallel.  The equality holds only when arg – arg =  i.e., and are antiparallel.  In any parallelogram sum of the squares of its sides is equal to the sum of the squares of its diagonals i.e.  Law of polygon i.e., Important Tips  The area of the triangle whose vertices are z, iz and z + iz is .  If be the vertices of a triangle then the area of the triangle is .  Area of the triangle with vertices and is .  If be the vertices of an equilateral triangle and be the circumcentre, then .  If be the vertices of a regular polygon of n sides and be its centroid, then .  If be the vertices of a triangle, then the triangle is equilateral iff or or .  If are the vertices of an isosceles triangle, right angled at then .  If are the vertices of right-angled isosceles triangle, then .  If one of the vertices of the triangle is at the origin i.e., then the triangle is equilateral iff .  If and are the vertices of a similar triangle, then .  If be the affixes of the vertices A, B, C respectively of a triangle ABC, then its orthocentre is . Example: 48 The points and in the complex plane are (a) Vertices of a right angled triangle (b) Collinear (c) Vertices of an obtuse angled triangle (d) Vertices of an equilateral triangle Solution: (b) Let and . Then area of triangle , Hence and are collinear. Example: 49 If , then area of the triangle whose vertices are points z, iz and is (a) (b) (c) (d) Solution: (b) Let , and If A denotes the area of the triangle formed by and , then (Applying transformation ) We get = . Example: 50 is possible if (a) (b) (c) (d) Solution: (c) Squaring both sides, we get     Hence arg arg Trick: Let z1 and z2 are the two sides of a triangle. By applying triangle inequality is the third side. Equality holds only when i.e., are parallel. 2.15 Standard Loci in the Argand Plane. (1) If z is a variable point in the argand plane such that arg , then locus of z is a straight line (excluding origin) through the origin inclined at an angle with x–axis. (2) If z is a variable point and is a fixed point in the argand plane such that arg , then locus of z is a straight line passing through the point representing and inclined at an angle with x-axis. Note that the point is excluded from the locus. (3) If z is a variable point and are two fixed points in the argand plane, then (i)  Locus of z is the perpendicular bisector of the line segment joining and (ii) = constant  Locus of z is an ellipse (iii)  Locus of z is the line segment joining and (iv)  Locus of z is a straight line joining and but z does not lie between and . (v)  Locus of z is a hyperbola. (vi)  Locus of z is a circle with and as the extremities of diameter. (vii)  Locus of z is a circle. (viii) arg  Locus of z is a segment of circle. (ix) arg =  Locus of z is a circle with and as the vertices of diameter. (x) arg = 0 or  Locus z is a straight line passing through and . (xi) The equation of the line joining complex numbers and is given by or Example: 51 The locus of the points z which satisfy the condition arg is (a) A straight line (b) A circle (c) A parabola (d) None of these Solution:(c) We have  arg Hence   , which is obviously a circle. Example: 52 If then z lies on (a) An ellipse (b) The imaginary axis (c) A circle (d) The real axis Solution: (b)    0  z lies on imaginary axis. Example: 53 The locus of the point z satisfying arg (where k is non-zero) is (a) Circle with centre on y–axis (b) Circle with centre on x–axis (b) A straight line parallel to x–axis (d) A straight line making an angle 60° with x–axis Solution: (a)        It is an equation of circle whose centre is on y–axis. Example: 54 The locus of z satisfying the inequality is (a) (b) (c) (d) None of these Solution: (a)     Example: 55 If and is constant, the locus of 'z' is (a) (b) (c) (d) Solution: (a)  (conjugate),  is a constant and  is known to be eliminated    2.16 De' Moivre's Theorem. (1) If is any rational number, then . (2) If then , where . (3) If and n is a positive integer, then , where . (4) If and then , where . Deductions: If then (i) (ii) (iii) (iv) Applications (i) In finding the expansions of trigonometric functions i.e. (ii) In finding the roots of complex numbers. (iii) In finding the complex solution of algebraic equations. Note :  This theorem is not valid when n is not a rational number or the complex number is not in the form of Powers of complex numbers : Let  Number x + iy form Standard complex form General 1 1+i0 – 1 – 1+i0 i 0 +i(1) ¬– i 0 +i(–1) Example: 56 If , then (a) a = 2, b = –1 (b) a = 1, b = 0 (c) a = 0, b = 1 (d) a = –1, b = 2 Solution: (b) = Example: 57 If + , then is (a) –3 (b) –2 (c) –1 (d) 0 Solution: (c) upto ……………….. = = = Example: 58 If where then is equal to (a) (b) (c) (d) Solution: (c)  ; ;………………..   . Example: 59 (a) (b) (c) (d) Solution: (a) 2.17 Roots of a Complex Number. (1) nth roots of complex number (z1/n) : Let be a complex number. To find the roots of a complex number, first we express it in polar form with the general value of its amplitude and use the De' moivre's theorem. By using De'moivre's theorem nth roots having n distinct values of such a complex number are given by where Properties of the roots of z1/n : (i) All roots of z1/n are in geometrical progression with common ratio (ii) Sum of all roots of z1/n is always equal to zero. (iii) Product of all roots of (iv) Modulus of all roots of z1/n are equal and each equal to or (v) Amplitude of all the roots of z1/n are in A.P. with common difference (vi) All roots of z1/n lies on the circumference of a circle whose centre is origin and radius equal to Also these roots divides the circle into n equal parts and forms a polygon of n sides. (2) The nth roots of unity : The nth roots of unity are given by the solution set of the equation where . Properties of nth roots of unity (i) Let the nth roots of unity can be expressed in the form of a series i.e., Clearly the series is G.P. with common difference i.e., (ii) The sum of all n roots of unity is zero i.e., (iii) Product of all n roots of unity is (iv) Sum of pth power of n roots of unity (v) The n, nth roots of unity if represented on a complex plane locate their positions at the vertices of a regular plane polygon of n sides inscribed in a unit circle having centre at origin, one vertex on positive real axis. Note :   (3) Cube roots of unity : Cube roots of unity are the solution set of the equation    , where Therefore roots are or . Alternative :   If one of the complex roots is then other root will be or vice-versa. Properties of cube roots of unity (i) (ii) (iii) (iv) and and . (v) Cube roots of unity from a G.P. (vi) Imaginary cube roots of unity are square of each other i.e., and . (vii) Imaginary cube roots of unity are reciprocal to each other i.e., and . (viii) The cube roots of unity by, when represented on complex plane, lie on vertices of an equilateral triangle inscribed in a unit circle having centre at origin, one vertex being on positive real axis. (ix) A complex number for which or can always be expressed in terms of Note :  If , then or vice-versa .   if a, b, c are real.  Cube root of – 1 are . Important Tips          Fourth roots of unity : The four, fourth roots of unity are given by the solution set of the equation   Note :  Sum of roots = 0 and product of roots =–1.  Fourth roots of unity are vertices of a square which lies on coordinate axes. Continued product of the roots If i.e., and then continued product of roots of is , where . Thus continued product of roots of Similarly, the continued product of values of is Important Tips  If or then  If n be a positive integer then , .  If z is a complex number, then is periodic.  nth root of ¬–1 are the solution of the equation where root of unity if n is even.  If and given, then (i) (ii) (iii) (iv) then, putting, values if in these results     the summation consists 3 terms gives similarly  If the condition given be then etc. Example: 60 If the cube roots of unity be , then the roots of the equation are (a) (b) (c) (d) None of these Solution: (c)   Example: 61 is an imaginary cube root of unity. If then least positive integral value of m is (a) 6 (b) 5 (c) 4 (d) 3 Solution: (d) The given equation reduces to   . Example: 62 If is the cube root of unity, then = (a) 4 (b) 0 (c) – 4 (d) None of these Solution: (c) = Example: 63 If , then is equal to (a) (b) (c) (d) Solution: (c) Given equation is Example: 64 Let is an imaginary cube root of unity then the value of is (a) (b) (c) (d) None of these Solution: (a) Example: 65 The roots of the equation , are (a) (b) (c) (d) None of these Solution: (b) Given equation and  2.18 Shifting the Origin in Case of Complex Numbers. Let O be the origin and P be a point with affix Let a point Q has affix z with respect to the co-ordinate system passing through O. When origin is shifted to the point then the new affix Z of the point Q with respect to new origin P is given by i.e., to shift the origin at we should replace z by Example: 66 If are the vertices of an equilateral triangle with as its circumcentre then changing origin to , then (where are new complex numbers of the vertices) (a) (b) (c) Both (a) and (b) (d) None of these Solution: (a) In an equilateral triangle the circumcentre and the centroid are the same point. So,  ..... (i) To shift the origin at we have to replace and by and then equation (i) becomes  On squaring ..... (ii) But triangle with vertices and is equilateral, then .....(iii) From (ii) and (iii) we get, . Therefore, 2.19 Inverse Points. (1) Inverse points with respect to a line : Two points P and Q are said to be the inverse points with respect to the line RS. If Q is the image of P in RS, i.e., if the line RS is the right bisector of PQ. (2) Inverse points with respect to a circle : If C is the centre of the circle and P,Q are the inverse points with respect to the circle then three points C,P,Q are collinear, and also CP . CQ where r is the radius of the circle. Example: 67 are the inverse points with respect to the line if (a) (b) (c) (d) None of these Solution: (b) Let RS be the line represented by the equation .....(i) Let P and Q are the inverse points with respect to the line RS. The point Q is the reflection (inverse) of the point P in the line RS if the line RS is the right bisector of PQ. Take any point z in the line RS, then lines joining z to P and z to Q are equal. i.e., or i.e.,  .....(ii) Hence, equations (i) and (ii) are identical, therefore comparing coefficients, we get So that, (By ratio and proportion rule) Hence, = 0 or Example: 68 Inverse of a point a with respect to the circle (a and c are complex numbers, centre C and radius R) is the point (a) (b) (c) (d) None of these Solution: (a) Let a' be the inverse point of a with respect to the circle then by definition the points c, a, a' are collinear. We have,    is purely real and positive. By definition    is purely real and positive}  . Therefore, the inverse point a' of a point a, 2.20 Dot and Cross Product. Let and be two complex numbers. If then from coni method      .....(i) and .....(ii) The dot product and is defined by (From(i)) Cross product of and is defined by (From(ii)) Hence, and Important Tips  If and are perpendicular then  If and are parallel then  Projection of on  Projection of on  Area of triangle if two sides represented by and is  Area of a parallelogram having sides and is  Area of parallelogram if diagonals represents by and is Example: 69 If then projection of on is (a) 1/10 (b) (c) (d) None of these Solution: (b) Projection of on ***