Value of Principal Argument Sum with Cube Root of Unity

To find the value of $\frac{3\alpha}{\pi}$, we begin by determining the complex number $\omega$.

Since $\omega$ is a non-real cube root of unity with argument $0 < \text{arg}(\omega) < \pi$, we have: $\omega = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$

Let $r = -\omega$. We can write $r$ in polar form as: $r = -\omega = -e^{i\frac{2\pi}{3}} = e^{i\left(\frac{2\pi}{3} - \pi\right)} = e^{-i\frac{\pi}{3}} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$

Notice that $r^6 = \left(e^{-i\frac{\pi}{3}}\right)^6 = e^{-i2\pi} = 1$. Therefore, the terms $r^n = (-\omega)^n$ are periodic with period $6$, and the sum of any $6$ consecutive powers of $r$ is zero: $\sum_{k=1}^{6} r^k = r + r^2 + r^3 + r^4 + r^5 + r^6 = r \left(\frac{1 - r^6}{1 - r}\right) = 0$

Since $2025 = 6 \times 337 + 3$, the given sum simplifies to the sum of the first $3$ terms: $S = \sum_{n=1}^{2025} (-\omega)^n = \sum_{n=1}^{3} r^n = r + r^2 + r^3$

Now, calculating each term:

1. $r = -\omega = \frac{1}{2} - i\frac{\sqrt{3}}{2}$
2. $r^2 = (-\omega)^2 = \omega^2 = e^{i\frac{4\pi}{3}} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$
3. $r^3 = (-\omega)^3 = -\omega^3 = -1$

Summing these up: $S = \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) + (-1) = -1 - i\sqrt{3}$

The principal argument $\alpha = \text{arg}(S)$ of $S = -1 - i\sqrt{3}$ lies in the third quadrant ($-\pi < \text{arg}(z) \le \pi$): $\alpha = \text{arg}\left(2 e^{-i\frac{2\pi}{3}}\right) = -\frac{2\pi}{3}$

Finally, we substitute $\alpha$ into the required expression: $\frac{3\alpha}{\pi} = \frac{3}{\pi} \left( -\frac{2\pi}{3} \right) = -2$