$\text{For a non-zero complex number}z,\text{let}\arg \left(z\right)\text{denote the principal argument of}\mathrm{ z},\text{with}-\pi <\arg \left(z\right)\le \pi .$

$\text{Let}\mathrm{ \omega }\text{ be the cube root of unity for which}0<\arg \left(\omega \right)<\pi .\text{Let}$

$\alpha =\arg \left(\sum _{n=1}^{2025}{(-\omega )}^n\right).$

$\text{Then the value of }\frac{3\alpha }{\pi }\text{ is \_\_\_\_\_\_}$

The correct answer is -2.

We are given that $\omega$ is the cube root of unity with $0<\arg \left(\omega \right)<\pi$.
Since $\omega$ is a non-real complex cube root of unity, we have:
$\omega =e^{i\frac{2\pi }{3}}$
and it satisfies:
${\omega }^3=1\text{and}1+\omega +{\omega }^2=0$

Let the sum be:
$S=\sum _{n=1}^{2025}{(-\omega )}^n$
Observe the terms of this geometric series. The common ratio is $-\omega$.
Since ${(-\omega )}^6={(-1)}^6{\omega }^6=1\cdot {\left({\omega }^3\right)}^2=1$, the terms repeat in cycles of 6, and the sum of any 6 consecutive terms in the geometric progression is zero:
$\sum _{k=1}^6{(-\omega )}^k=0$

We can divide the total number of terms, 2025, by the period of 6:
$2025=6\times 337+3$
Therefore, the sum simplifies to the sum of the first 3 terms of the series:
$S=\sum _{n=1}^3{(-\omega )}^n=(-\omega )+{(-\omega )}^2+{(-\omega )}^3$

Evaluating these terms:
$S=-\omega +{\omega }^2-1$
Using the relation $1+\omega +{\omega }^2=0$, we can substitute ${\omega }^2=-1-\omega$ into the expression:
$S=-\omega +(-1-\omega )-1=-2\omega -2=-2(\omega +1)$
Since $\omega +1=-{\omega }^2$, we have:
$S=-2(-{\omega }^2)=2{\omega }^2$

Now, we find the argument of $S$:
$\alpha =\arg \left(S\right)=\arg \left(2{\omega }^2\right)=\arg \left({\omega }^2\right)$
Since $\omega =e^{i\frac{2\pi }{3}}$, we have:
${\omega }^2=e^{i\frac{4\pi }{3}}=e^{-i\frac{2\pi }{3}}$
Since the principal argument must satisfy $-\pi <\arg \left(z\right)\le \pi$, we write:
$\alpha =-\frac{2\pi }{3}$

Finally, we calculate the requested value of $\frac{3\alpha }{\pi }$:
$\frac{3\alpha }{\pi }=\frac{3}{\pi }(-\frac{2\pi }{3})=-2$