, 2013

Solution:

Note that $z = e^{\frac{2\pi}{3} i} = \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}$, so that $z^{3} = 1$ and $z^{2} + z + 1 = 0$. Also, $s_{2} = s_{1} z$ and $s_{3} = s_{1} z^{2}$.

Then, the sum of the coefficients of $g(x)$ is $g(1) = (1 - s_{1})(1 - s_{2})(1 - s_{3}) = (1 - s_{1})(1 - s_{1} z)(1 - s_{1} z^{2}) = 1 - (1 + z + z^{2}) s_{1} + (z + z^{2} + z^{3}) s_{1}^{2} - z^{3} s_{1}^{3} = 1 - s_{1}^{3}$.

Meanwhile,
$$s_{1}^{3} = (r_{1} + r_{2} z + r_{3} z^{2})^{3} = r_{1}^{3} + r_{2}^{3} + r_{3}^{3} + 3 r_{1}^{2} r_{2} z + 3 r_{1}^{2} r_{3} z^{2} + 3 r_{2}^{2} r_{3} z + 3 r_{2}^{2} r_{1} z^{2} + 3 r_{3}^{2} r_{1} z + 3 r_{3}^{2} r_{2} z^{2} + 6 r_{1} r_{2} r_{3}.$$
Since the real parts of both $z$ and $z^{2}$ are $-\frac{1}{2}$, and since all of $r_{1}, r_{2}, r_{3}$ are real, the real part of $s_{1}^{3}$ is
$$r_{1}^{3} + r_{2}^{3} + r_{3}^{3} - \frac{3}{2}(r_{1}^{2} r_{2} + r_{1}^{2} r_{3} + r_{2}^{2} r_{3} + r_{2}^{2} r_{1} + r_{3}^{2} r_{1} + r_{3}^{2} r_{2}) + 6 r_{1} r_{2} r_{3}.$$
This can be rewritten as
$$(r_{1} + r_{2} + r_{3})^{3} - \frac{9}{2}(r_{1} + r_{2} + r_{3})(r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}) + \frac{27}{2} r_{1} r_{2} r_{3}.$$
From $f(x)$, the sum $r_{1} + r_{2} + r_{3} = 3$, $r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = -4$, $r_{1} r_{2} r_{3} = -4$.

Plugging in,
$$(r_{1} + r_{2} + r_{3})^{3} = 3^{3} = 27$$
$$-\frac{9}{2} \cdot 3 \cdot (-4) = 54$$
$$\frac{27}{2} \cdot (-4) = -54$$
So the real part is $27 + 54 - 54 = 27$.

Therefore, the answer is $1 - 27 = -26$.