JEE Advance - Mathematics (2025 - Paper 2 Online - No. 13)
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
$$ \alpha=\arg \left(\sum\limits_{n=1}^{2025}(-\omega)^n\right) $$
Then the value of $\frac{3 \alpha}{\pi}$ is ________________.
Answer
-2
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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