JEE Advanced 2025 Paper 2 · Q13 · Complex Numbers

Question

For a non-zero complex number $z$, let $\arg(z)$ denote the principal argument of $z$, with $-\pi < \arg(z) \le \pi$. 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 $\dfrac{3\alpha}{\pi}$ is __________.

SolveKar AI · Accepted Answer

Take $\omega = e^{i 2\pi/3}$. Let $w = -\omega = e^{i(2\pi/3 + \pi)} = e^{i 5\pi/3} = e^{-i\pi/3}$. Then $w$ is a primitive $6$th root of unity, so $w^6 = 1$ and $1 + w + w^2 + \ldots + w^5 = 0$. The geometric sum $S = w + w^2 + \ldots + w^{2025}$. Since $2025 = 6 \cdot 337 + 3$, the first $2022$ terms form $337$ full periods (each summing to $0$). The remaining three terms are $w^{2023} + w^{2024} + w^{2025} = w^{2023}(1 + w + w^2)$. $2023 \bmod 6 = 1$, so $w^{2023} = w$. Compute $1 + w + w^2$ with $w = e^{-i\pi/3}$: real part $1 + \cos(-\pi/3) + \cos(-2\pi/3) = 1 + 1/2 - 1/2 = 1$; imaginary part $\sin(-\pi/3) + \sin(-2\pi/3) = -\sqrt{3}/2 - \sqrt{3}/2 = -\sqrt{3}$. So $1 + w + w^2 = 1 - i\sqrt{3} = 2 e^{-i\pi/3}$. Hence $S = w \cdot 2 e^{-i\pi/3} = 2 e^{-i 2\pi/3}$. $\arg(S) = -2\pi/3$, so $\frac{3\alpha}{\pi} = -2$.

JEE Advanced 2025 Paper 2 · Q13 · Complex Numbers

Solution