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 ______.

The non-real cube roots of unity are $$\omega$$ and $$\omega^{2}$$, where $$\omega^{3}=1$$ and $$1+\omega+\omega^{2}=0$$.
Because $$0 \lt \arg(\omega) \lt \pi$$, we take $$\omega = e^{\,i\frac{2\pi}{3}} = -\frac12 + i\frac{\sqrt3}{2}$$, so $$\arg(\omega)=\frac{2\pi}{3}$$.

Define $$r=-\omega$$. Then
$$r = -\omega = e^{\,i\pi}\,e^{\,i\frac{2\pi}{3}} = e^{\,i\frac{5\pi}{3}} = e^{-\,i\frac{\pi}{3}}$$
and hence $$\arg(r)=-\frac{\pi}{3}$$.

Since $$r^{3}=(-\omega)^{3}=-(\omega^{3})=-1$$, we get $$r^{6}=1$$. Thus the powers of $$r$$ repeat every six terms.

The required sum is $$S=\sum_{n=1}^{2025} r^{\,n}.$$ Because one full cycle of six consecutive terms sums to zero (geometric-series property for a 6th root of unity different from 1), we split 2025 terms into cycles:

$$2025 = 6 \times 337 + 3.$$ The first $$6\times337=2022$$ terms contribute zero, so $$S = r^{2023}+r^{2024}+r^{2025}.$$

Now reduce exponents modulo 6: $$2023 \equiv 1,\; 2024 \equiv 2,\; 2025 \equiv 3 \pmod{6}.$$ Hence $$S = r^{1}+r^{2}+r^{3}.$$

Compute each term:
$$r^{1} = -\omega,$$ $$r^{2} = (-\omega)^{2} = \omega^{2},$$ $$r^{3} = -1.$$ Therefore $$S = -\omega + \omega^{2} - 1.$$

Using $$1+\omega+\omega^{2}=0 \; \Rightarrow \; \omega^{2}= -1-\omega,$$ $$S = -\omega + (-1-\omega) -1 = -2 - 2\omega = -2(1+\omega).$$

But $$1+\omega = -\omega^{2},$$ so $$S = -2(-\omega^{2}) = 2\,\omega^{2}.$$

Multiplying by the positive real number 2 does not change the argument, hence $$\alpha = \arg(S) = \arg(\omega^{2}).$$ For our chosen branch, $$\omega^{2}=e^{\,i\frac{4\pi}{3}}$$ has principal argument $$\frac{4\pi}{3}-2\pi = -\frac{2\pi}{3}.$$ Thus $$\alpha = -\frac{2\pi}{3}.$$

Finally, $$\frac{3\alpha}{\pi} = \frac{3\left(-\tfrac{2\pi}{3}\right)}{\pi} = -2.$$

Answer: -2