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📐 Mathematics  ·  Complex Numbers and Quadratic Equations  ·  JEE

The smallest positive integer n for which (1+i)2n = (1-i)2n is:

Answer: 2.

  • A 1
  • B 2
  • C 3
  • D 4

Correct answer: B. 2

Explanation: Compute squares: (1+i)²=1+2i-1=2i and (1-i)²=1-2i-1=-2i. So equation becomes (2i)n=(-2i)n, i.e., 2n·in = 2n·(-i)n, giving in=(-i)n, or (i/(-i))n=(-1)n=1. Smallest positive n: n=2. Answer: n = 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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