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📐 Mathematics  ·  Complex Numbers and Quadratic Equations  ·  JEE

If z = cos(2π/5) + i sin(2π/5), find 1 + z + z² + z³ + z⁴.

Answer: 0.

  • A 1
  • B 5
  • C z⁵
  • D 0

Correct answer: D. 0

Explanation: z=e<sup>2πi/5</sup> is a primitive 5th root of unity, so z⁵=1. The sum 1+z+z²+z³+z⁴ is a geometric series: (z⁵-1)/(z-1)=(1-1)/(z-1)=0. Alternatively, it equals the sum of ALL 5th roots of unity, which always equals 0. Answer: sum = 0

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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