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 ______.

$\omega^3 = 1, \quad \omega \neq 1, \quad 0 < \arg(\omega) < \pi$ $\alpha = \arg\left(\sum_{n=1}^{2025} (-\omega)^n\right)$ $\text{To find: } \dfrac{3\alpha}{\pi}$ $\text{Roots of unity lie on } |z| = 1$ $\omega^3 = 1, \ \omega \neq 1 \ \Rightarrow \ 1 + \omega + \omega^2 = 0$ $\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} = e^{2\pi i/3}$ $\omega^2 = -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} = e^{4\pi i/3}$ $r = -\omega = e^{-i\pi/3} = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$ $(-\omega)^3 = -1, \quad (-\omega)^6 = 1$ $r^6 = 1 \ \Rightarrow \ r, r^2, r^3, r^4, r^5, r^6 = \text{all 6th roots of unity}$ $r + r^2 + r^3 + r^4 + r^5 + r^6 = 0$ $2025 = 6 \cdot 337 + 3$ $\sum_{n=1}^{2025} r^n = 337 \cdot 0 + r^{2023} + r^{2024} + r^{2025}$ $\sum_{n=1}^{2025} r^n = r + r^2 + r^3$ $r + r^2 + r^3 = \left(\tfrac{1}{2} - i\tfrac{\sqrt3}{2}\right) + \left(-\tfrac{1}{2} - i\tfrac{\sqrt3}{2}\right) + (-1)$ $S = \sum_{n=1}^{2025} (-\omega)^n = -1 - i\sqrt{3}$ $|S| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{4} = 2$ $\alpha = \arg(S) = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$ $\dfrac{3\alpha}{\pi} = \dfrac{3\,(4\pi/3)}{\pi} = 4$ $\boxed{\dfrac{3\alpha}{\pi} = 4}$