For a non-zero complex number z , let arg (z) denote the principal argument of z , with -π<arg (z) leq π . Let ω be the cube root of unity for which 0<arg (ω)<π . Let α=arg (sum n=1 2025 (-ω) n ) . Then the value of 3 α/π is

Mathematics · Complex Numbers

JEE Advanced 2025 — Paper 2 — Question 29

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 $\qquad$

Answer: -2

Numerical answer — enter this value.

Step-by-step solution

$\alpha=\arg \left(-\omega+\omega^{2}-\omega^{3}+\ldots \ldots \ldots+(-\omega)^{2025}\right)$

$\alpha=\arg \left(\frac{-\omega\left((-\omega)^{2025}-1\right)}{-\omega-1}\right)$

$\alpha=\arg \left(\frac{-\omega}{-\omega-1}(-2)\right)$

$\alpha=\arg \left(\frac{-2 \omega}{\omega+1}\right)$

$\alpha=\arg \left(\frac{-2 \omega}{-\omega^{2}}\right)$

$\alpha=\arg \left(\frac{2}{\omega}\right)$

$\alpha=\arg \left(2 \omega^{2}\right)$

$\alpha=\frac{-2 \pi}{3}$

$\frac{3 \alpha}{\pi}=-2$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Complex Numbers
Topic: Demoivre's Theorem and Roots of Unity