JEE Advanced · Mathematics · 3. Complex Numbers
For a non-zero complex number \(z\), let \(\arg (z)\) denote the principal argument of \(z\), with \(-\pi <\arg (z) \leq \pi\). Let \(\omega\) be the cube root of unity for which \(0 <\arg (\omega) <\pi\). Let \(\alpha=\arg \left(\sum_{n=1}^{2025}(-\omega)^n\right)\).
Then the value of \(\frac{3 \alpha}{\pi}\) is ___________
- A -2
- B -4
- C -6
- D -8
Answer & Solution
Correct Answer
(A) -2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \alpha=\arg \left(-\omega+\omega^2-\omega^3+\ldots \ldots \ldots+(-\omega)^{2025}\right) \\ & \alpha=\arg \left(\frac{-\omega\left((-\omega)^{2025}-1\right)}{-\omega-1}\right) \\ & \alpha=\arg \left(\frac{-\omega}{-\omega-1}(-2)\right) \\ & \alpha=\arg \left(\frac{-2 \omega}{\omega+1}\right) \\ & \alpha=\arg \left(\frac{-2 \omega}{-\omega^2}\right) \\ & \alpha=\arg \left(\frac{2}{\omega}\right) \\ & \alpha=\arg \left(2 \omega^2\right) \\ & \alpha=\frac{-2 \pi}{3} \\ & \frac{3 \alpha}{\pi}=-2\end{aligned}\)
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