Question:
If $\omega (\ne 1)$ be a cube root of unity and $(1 + \omega^2)^n = (1 + \omega^4)^n,$ then the least positive value of n is
If $\omega (\ne 1)$ be a cube root of unity and $(1 + \omega^2)^n = (1 + \omega^4)^n,$ then the least positive value of n is
Updated On: Jun 14, 2022
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The Correct Option is B
Solution and Explanation
Given, $(1+\omega^2)^n=(1+\omega^4)^n$
$\Rightarrow \, \, \, \, (-\omega)^n =(-\omega^2)^n \, [\because \, \, w^3=1 \, and \, 1+\omega+\omega^2=0]$
$\Rightarrow \, \, \, \, \, \omega^n =1$
$\Rightarrow \, \, \, \, \, $ n = 3 is the least positive value of n.
$\Rightarrow \, \, \, \, (-\omega)^n =(-\omega^2)^n \, [\because \, \, w^3=1 \, and \, 1+\omega+\omega^2=0]$
$\Rightarrow \, \, \, \, \, \omega^n =1$
$\Rightarrow \, \, \, \, \, $ n = 3 is the least positive value of n.
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
