TS EAMCET · Maths · Complex Number
If \(\omega_0, \omega_1, \ldots, \omega_{n-1}\) are the \(n\)th roots of unity, then \(\left(1+2 \omega_0\right)\left(1+2 \omega_1\right)\left(1+2 \omega_2\right) \ldots\left(1+2 \omega_{n-1}\right)=\)
- A \(1+(-1)^n 2^n\)
- B \(1+2^n\)
- C \((-1)^n+2^n\)
- D \(1+(-1)^{n-1} 2^n\)
Answer & Solution
Correct Answer
(D) \(1+(-1)^{n-1} 2^n\)
Step-by-step Solution
Detailed explanation
If \(\omega_0, \omega_1, \omega_2 \ldots \omega_{n-1}\) are nth root of unity, then \[ x^n-1=\left(x-\omega_0\right)\left(x-\omega_1\right)\left(x-\omega_2\right) \ldots\left(x-\omega_{n-1}\right) \] put \(x=\frac{-1}{2}\), we get…
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