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

Correct answer: 0 aligned & α= (-ω+ω^2-ω^3+ +(-ω)^ 2025 ) & α= ( -ω((-ω)^ 2025 -1) -ω-1 ) & α= ((-ω)/(-ω-1)(-2)) & α= ((-2 ω)/(ω+1)) & α= ((-2 ω)/(-ω^2)) & α= ((2)/(ω)) & α= (2 ω^2) & α=(-2 π)/(3) & (3 α)/(π)=-2 aligned

JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Complex Number

JEE Advanced 2025 · Paper 2 · Mathematics · Complex Number

For a non-zero complex number z , let (z) denote the principal...

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

Step-by-step 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}

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