⇒
2n + 2 cos
S2 =
(n − 2)π 3
3
Find S3 Multiply (ii) be ω , (iii) with ω2 and add to (i) to get 3(C2 + C5 + C8 + ...........) = 2n + 2 cos 2n + 2 cos
⇒
S3 =
(n + 2 )π 3
(n + 2)π 3
3
Example : 28 Prove that the complex number z1, z2 and the origin form an isosceles triangle with vertical angle 2π/3. If x21 + z22 + z1 z2 = 0 Solution Let A and B are the points represented by z1 and z2 respectively on the Argand plane Consider z21 + z22 + z1z2 = 0 On factoring LHS, we get : (z2 – ωz1) (z2 – ω2z1) = 0 ⇒ z2 = ω1 z1 or z2 = ω2 z1 consider z2 = ωz1 ..........(i) Take modulus of both sides |z2| = |ωz1| ⇒ |z2| = |ω| |z1| = |z1| (Q |ω| = 1) ⇒ OA = OB ⇒ ∆OAB is isosceles. Take argument on both sides, Arg (z2) = Arg (ωz1) = Arg (ω) + Arg (z1) ⇒ Arg (z2) – Arg (z1) = 2π/3 (Q Arg (ω) = 2π/3) ⇒ ∠AOB = 2π/3. Hence vertical angle = ∠AOB = 2π/3. ⇒ z2 = z1 ei2π/3 ,we can directly conclude that z2 is obtained by rotating z1 Note : As z2 = ωz1 through 2π/3 in anti-clockwise direction ⇒ ∠AOB = 2π/3 and OA = OB Consider z2 = ω2 z1 Similarly show that ∆AOB is isosceles with vertical angle 2π/3 Example : 29 For every real number c ≥ 0, find all complex numbers z which satisfy the equation : |z|2 – 2iz + 2c (1 + i) = 0. Solution Let z = x = iy ⇒ (x2 + y2 + 2y + 2c) – i (2x – 2c) = 0 Comparing the real and imaginary parts, we get : ⇒ x2 + y2 + 2y + 2c = 0 ..........(i) and x=c ..........(ii) Solving (i) and (ii), we get ⇒ y2 + 2y + c2 + 2c = 0 ⇒
y=
− 2 ± 4 − 4(c 2 + 2c ) =–1± 2
1 − c 2 − 2c
as y is real, 1 – c2 – 2c ≥ 0 ⇒ –√2 – 1 ≤ c ≤ √2 – 1 ⇒ c ≤ √2 – 1 (Q c ≥ 0) ⇒ the solution is
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