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JEE Advanced 2025
PAPER-2 2025
Question
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 $\_\_\_\_$
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Question Tags
JEE Advance
Mathematics
Easy
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