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Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two points representing the complex numbers z_{1} and z_{2} respectively in the Argand plane as show in figure.
If the origin 'O' is joined to P and Q & QR is drawn through Q in such a way that it is parallel and equal to OP, then the coordinates of R will be (x_{1} + x_{2}, y_{1} + y_{2}). Therefore the point 'R' denotes the complex number z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2}). In triangle OPR, OR = | z_{1} + z_{2} |, OP = | z_{1} | and PR = OQ = | z_{2} | Using triangle inequality in Δ OPR, OR < OP + PR ⇒ | z_{1} + z_{2} | < | z_{1} | + | z_{2} | ...... (i) If O, P, Q are collinear, then OR = OP + PR ⇒ OR = OP + OQ (∵ PR = OQ) ∴ | z_{1} + z_{2} | = | z_{1} | + | z_{2} | ...... (ii) From (i) and (ii), | z_{1} + z_{2} | ≤ | z_{1} | + | z_{2} | ...... (iii) This is called the triangle inequality. Similarly we can say | z_{1} – z_{2} | ≤ | z_{1} | + | z_{2} | ...... (iv) From (iii) and (iv), we get | z_{1} + z_{2} |^{2} + | z_{1} – z_{2} |^{2} = 2(| z_{1} |^{2} + | z_{2} |^{2}) Now its geometrical interpretation is OR^{2} + PQ^{2} = OP^{2} + PR^{2} + RQ^{2} + QO^{2} (∵ OP = QR, OQ = PR) i.e., In a parallelogram the sum of the squares of the diagonals is equal to the sum of the squares of the four sides.
Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two points represents the complex numbers z_{1} and z_{2} respectively in the Argand plane as shown in figure.
Let Q_{1} be the image of Q w.r.t. the origin 'O'. Then Q_{1} represents the complex number (– z_{2}). Now completing the parallelogram OPRQ_{1}, the point R represents the complex number z_{1} – z_{2}. We know that "The absolute difference of two sides of a triangle is less than the third side" ∴ | z_{1} – z_{2} | > | |z_{1}| – |z_{2}| | ...... (i) If O, P, Q_{1} are collinear, then OR = OP + PR ⇒ OR = OP + OQ_{1} (∵ PR = OQ_{1}) ∴ | z_{1} – z_{2} | = | |z_{1}| – |z_{2}| | ...... (ii) From (i) and (ii), we get | z_{1} – z_{2} | ≥ | |z_{1}| – |z_{2}| | This inequality is also called triangle inequality.
Let P and Q be two points represents the complex numbers z_{1} and z_{2} respectively in the Argand plane as shown in figure. Join the origin with the points P and Q.
∴ OP = | z_{1} | = r_{1} and OQ = | z_{2} | = r_{2} Let A(1, 0) be a point on the real axis. Let ∠AOP = θ_{1}, ∠AOQ = θ_{2} Construct Δ OQR similar to Δ OAP, as shown in figure.
From the figure above, ∠AOR = ∠AOQ + ∠ROQ = θ_{1} + θ_{2} Hence the polar coordinates of R are (r_{1}r_{2}, θ_{1} + θ_{2})
Let P and Q be two points represents the complex numbers z_{1} and z_{2} respectively in Argand plane as shown in adjacent figure.
∴ OP = | z_{1} | = r_{1}, OQ = | z_{2} | = r_{2} Let A(1, 0) be a point on the real axis. Since ∠AOP = θ_{1}, ∠AOQ = θ_{2} Construct Δ ORP similar to Δ AOQ as shown in figure.
Let us study the applications of complex numbers in the field of Co-ordinate geometry.
The distance between two points P(z_{1}) and Q(z_{2}) is given by PQ = | z_{2} – z_{1} |
If R(z) internally divides the line segment joining the points P(z_{1}) and Q(z_{2}) in the ratio m_{1} : m_{2} (m_{1}, m_{2} > 0), then (i) z = If R(z) divides the line segment externally, then (ii) z = Note : If R(z) is the midpoint of PQ, then z =
If P(z_{1}) and Q(z_{2}) are two fixed points and R(z) is moving point such that it is always at equal distance from P(z_{1}) and Q(z_{2}), then the locus of 'z' is perpendicular bisector of PQ. ∴ PR = QR ⇒ | z – z_{1} | = | z – z_{2} | ⇒ | z – z_{1} |^{2} = | z – z_{2} |^{2} ⇒ (z – z_{1}) (z – z_{1}) = (z – z_{2}) (z – z_{2}) ⇒ zz – zz_{1} – z_{1}z + z_{1}z_{1} = zz – zz_{2} – z_{2}z + z_{2}z_{2} ⇒ | z |^{2} + | z_{1} |^{2} – | z |^{2} – | z_{2} |^{2} = zz_{1} – zz_{2} + z_{1}z– z_{2}z ⇒ | z_{1} |^{2} – | z_{2} |^{2} = z (z_{1} – z_{2}) + z (z_{1} – z_{2})
(i) General equation of a straight line The general equation of a straight line is of the form az + az + b = 0, where 'a' is complex number and 'b' is real number.
(ii) Slope of a line The complex slope of the line az + az + b = 0 is = and real slope of the line az + az + b = 0 is =
(iii) Length of the perpendicular The length of perpendicular from a point z_{1} to the line az + az + b = 0 is given by or
(iv) Parametric form Equation of a straight line joining the points z_{1} and z_{2} is z = t z_{1} + (1 – t) z_{2}, where t ∈ R
(v) Non-parametric form Equation of a straight line joining the points z_{1} and z_{2} is ⇒ z (z_{1} – z_{2}) – z (z_{1} – z_{2}) + z_{1} z_{2} – z_{2} z_{1} = 0
(vi) If three points A(z_{1}), B(z_{2}), C(z_{3}) are collinear, then Slope of AB = Slope of BC = Slope of AC ⇒ or
If A(z_{1}), B(z_{2}) and C(z_{3}) are the vertices of a ΔABC, then its centroid is given by
The centre of the circle is z_{1} and its radius is CP = r. ⇒ | z – z_{1} | = r ⇒ | z – z_{1} |^{2} = r^{2} ⇒ (z – z_{1}) (z – z_{1}) = r^{2} (∵ | z |^{2} = zz) ⇒ zz – zz_{1} – z_{1}z + z_{1}z_{1} = r^{2} ⇒ zz – z_{1}z – zz_{1} + z_{1}z_{1} – r^{2} = 0 Let a = – z_{1}, b = z_{1}z_{1}^{2} – r^{2} The above equation becomes zz + az + az + b = 0 This is called general equation of a circle.
Note : (i) The general equation of the circle is zz + az + az + b = 0, where 'a' is complex number and b ∈ R. ∴ The centre of the circle = – a The radius of the circle = (ii) If the centre of the circle is at origin and radius r, then the equation of the circle is | z | = r. (iii) | z – z_{1} | < r represents interior of a circle | z – z_{1} | = r (iv) | z – z_{1} | > r represents exterior of the circle | z – z_{1} | = r
If A(z_{1}), B(z_{2}) are end points of the diameter and P(z) is any point on the circle then, (z – z_{1}) (z – z_{2}) + (z – Z_{2}) (z – z_{1}) = 0
If z is a variable point and z_{1}, z_{2} are two fixed points in the Argand plane, then (i) | z – z_{1} | = | z – z_{2} | ⇒ locus of z is the perpendicular bisector of the line segment joining z_{1} and z_{2} (ii) | z – z_{1} | + | z – z_{2} | = | z_{1} – z_{2} | ⇒ locus of z is the line segment joining z_{1} and z_{2} (iii) | z – z_{1} | – | z – z_{2} | = | z_{1} – z_{2} | ⇒ locus of z is a straight line joining z_{1} and z_{2} but z does not lie between z_{1} and z_{2} (iv) | z – z_{1} | = k | z – z_{2} |, (k ≠ 1) ⇒ locus of z is a circle (v) | z – z_{1} |^{2} + | z – z_{2} |^{2} = | z_{1} – z_{2} |^{2} ⇒ locus of z is a circle with z_{1} and z_{2} as the extremities of diameter (vi) arg = α (fixed) ⇒ locus of z is a segment of circle (vii) arg = ± ⇒ locus of z is a circle with z_{1} and z_{2} as the extremities of diameter. (viii) arg = 0 (or) π ⇒ locus of z is a straight line passing through z_{1} and z_{2} (ix) | z – z_{1} | + | z – z_{2} | = constant (≠ | z_{1} – z_{2} |) ⇒ locus of z is an ellipse (x) | z – z_{1} | – | z – z_{2} | = constant (≠ | z_{1} – z_{2} |) ⇒ locus of z is a hyperbola
The angle between two intersecting lines in Argand plane. Let A(z_{1}), B(z_{2}) and C(z_{3}) be three points in Argand plane. From the adjacent figure, AC = z_{3} – z_{1} AB = z_{2} – z_{1} Let arg (AC) = arg (z_{3} – z_{1}) = θ and arg (AB) = arg (z_{2} – z_{1}) = ϕ Let ∠ CAB = α,
Note : (1) Let z_{1}, z_{2}, z_{3} and z_{4} be four complex numbers such that the line joining z_{4} and z_{3} is inclined to the line joining z_{2} and z_{1} at an angle α. Then α = arg (2) If z_{1}, z_{2}, z_{3} and z_{4} are complex numbers such that the line joining z_{1} and z_{2} is perpendicular to the line joining z_{3} and z_{4}, then arg = i.e., is purely imaginary ⇒ is purely imaginary ⇒ (z_{1} – z_{2}) (z_{3} – z_{4}) is purely imaginary.
Let A(z_{1}), B(z_{2}), C(z_{3}) and D(z_{4}) be four Concyclic points represented in the Argand plane as shown in figure.
From adjacent figure, ∠ADB = ∠ACB From rotation theorem, arg = arg ⇒ arg – = 0 ⇒ arg = 0 ⇒ is a positive real number
Note : If a complex number is a positive real number, then the points represented by z_{1}, Z_{2}, z_{3} and z_{4} are Concyclic in Argand plane.