Question:
If $\omega$ and $\omega^2$ are complex cube roots of unity, then $(1-\omega+\omega^2)^5+(1-\omega^2+\omega)^5$ is equal to
If $\omega$ and $\omega^2$ are complex cube roots of unity, then $(1-\omega+\omega^2)^5+(1-\omega^2+\omega)^5$ is equal to
Updated On: Jul 6, 2022
- $8$
- $16$
- $32$
- $64$
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The Correct Option is C
Solution and Explanation
$\left(1-\omega+\omega^{2}\right)^{5}+\left(1-\omega^{2}+\omega\right)^{5}$
$=\left(-2\omega\right)^{5}+\left(-2\omega^{2}\right)^{5}$
$=-32\left(\omega^{2}+\omega\right)$
$=-32\left(-1\right)=32$.
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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.
