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📐 Mathematics  ·  Complex Numbers and Quadratic Equations  ·  JEE

If ω ≠ 1 is a cube root of unity and (1 + ω)⁷ = A + Bω, find A + B.

Answer: 1.

  • A 0
  • B 1
  • C 2
  • D 3

Correct answer: B. 1

Explanation: 1 + omega = -omega<sup>2</sup> (since 1+omega+omega<sup>2</sup>=0). (1+omega)<sup>7</sup> = (-omega<sup>2</sup>)<sup>7</sup> = -omega<sup>14</sup> = -omega<sup>12+2</sup> = -omega<sup>2.</sup> So A + B omega = -omega<sup>2</sup> = 1 + omega (since -omega<sup>2</sup> = 1+omega). A=1, B=1. A+B = 2. Wait: -omega<sup>2</sup> = 1+omega means A=1, B=1, sum=2.

Argand Plane: z = a + ibReImz = a+iba (real part)b (imaginary part)θ = arg(z)|z| = length of the vector OZ = √(a²+b²); θ = angle OZ makes with the positive real axis

A complex number z = a+ib is plotted as a point (or vector from the origin) on the Argand plane, with the real part along the horizontal axis and the imaginary part along the vertical; its modulus |z| is the vector's length, and its argument θ is the angle from the positive real axis.

Concept context

Complex numbers in a+ib form, modulus, argument, polar form, cube roots of unity, and quadratic equations with complex roots

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