Q29 / 48 · JEE Advanced 2025 (Paper 2) · Mathematics
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 ___________
Correct answer: -2
Solution
$\begin{aligned} & \alpha=\arg \left(-\omega+\omega^2-\omega^3+\ldots \ldots \ldots+(-\omega)^{2025}\right) \\ & \alpha=\arg \left(\frac{-\omega\left((-\omega)^{2025}-1\right)}{-\omega-1}\right) \\ & \alpha=\arg \left(\frac{-\omega}{-\omega-1}(-2)\right) \\ & \alpha=\arg \left(\frac{-2 \omega}{\omega+1}\right) \\ & \alpha=\arg \left(\frac{-2 \omega}{-\omega^2}\right) \\ & \alpha=\arg \left(\frac{2}{\omega}\right) \\ & \alpha=\arg \left(2 \omega^2\right) \\ & \alpha=\frac{-2 \pi}{3} \\ & \frac{3 \alpha}{\pi}=-2\end{aligned}$
Hence, the correct answer is -2.