方程 $x^3 - 1 = 0$ 的三个不同的根记为 $1, \omega$ 和 $\omega^2$。

1. Factorise the difference of cubes for the given equation.

$x^3 - 1 = 0$
$(x - 1)(x^2 + x + 1) = 0$

2. Substitute the root $\omega$ and use the fact that it is distinct from $1$ to complete the proof.

Since $\omega$ is a root of the equation, it must satisfy:
$(\omega - 1)(\omega^2 + \omega + 1) = 0$

We are given that $1, \omega$, and $\omega^2$ are distinct roots. Thus, $\omega \neq 1$, which implies $\omega - 1 \neq 0$.

Dividing both sides by $(\omega - 1)$ yields:
$\omega^2 + \omega + 1 = 0$
$1 + \omega + \omega^2 = 0 \text{ (shown)}$

1. Determine the exact value of $\omega$ from its argument.

The equation $x^3 - 1 = 0$ has complex roots $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{2\pi}{3}}$ in addition to $1$.
Given the condition $0 < \arg(\omega) < \pi$, we must have $\omega = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.
We also note its modulus is $|\omega| = 1$.

2. Identify and describe the geometric locus.

The equation $|z - \omega| = |\omega|$ represents the locus of points $z$ whose distance from $\omega$ is equal to a constant radius $|\omega|$.
This forms a circle with centre $\omega \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and radius $1$.
Since the distance from the centre to the origin is $|\omega| = 1$, the circle passes exactly through the origin $(0,0)$.

1. Determine the value of $\arg(\omega^2)$ from the given roots of $x^3 - 1 = 0$.

The roots of $x^3 - 1 = 0$ are $x = 1, e^{i\frac{2\pi}{3}}, e^{-i\frac{2\pi}{3}}$.
Given the condition $0 < \arg(\omega) < \pi$, we must have $\omega = e^{i\frac{2\pi}{3}}$.
Then $\omega^2 = e^{-i\frac{2\pi}{3}}$, which gives its principal argument as:
$\arg(\omega^2) = -\frac{2\pi}{3}$

2. Calculate the angle of the locus and describe its geometric shape.

Substitute $\arg(\omega^2)$ into the locus equation:
$\arg(z + 1) = \pi + \left(-\frac{2\pi}{3}\right) = \frac{\pi}{3}$
This equation describes a half-line originating from the point $(-1, 0)$ (or $z = -1$), excluding the point itself, and making an angle of $\frac{\pi}{3}$ radians above the positive real axis.

1. Use the result from part (a) to express the half-line locus parametrically in terms of $\omega$.

From locus (ii), we are given $\arg(z + 1) = \pi + \arg(\omega^2)$.
Multiplying a complex number by $-1$ adds $\pi$ to its argument, so $\pi + \arg(\omega^2) = \arg(-\omega^2)$.
From part (a), we know $1 + \omega + \omega^2 = 0$, which rearranges to $-\omega^2 = 1 + \omega$.
Thus, $\arg(z + 1) = \arg(1 + \omega)$.
This indicates that the vector $z + 1$ points in the same direction as $1 + \omega$, allowing us to write the half-line parametrically as:
$z + 1 = r(1 + \omega) \implies z = r(1 + \omega) - 1 \quad \text{for some real } r > 0.$

2. Substitute into the circle locus to find the exact value of $z$.

Substitute $z = r(1 + \omega) - 1$ into locus (i), which is $|z - \omega| = |\omega|$:
$|r(1 + \omega) - 1 - \omega| = |\omega|$
$|r(1 + \omega) - (1 + \omega)| = 1 \quad \text{(since } |\omega| = 1 \text{ for roots of unity)}$
$|(r - 1)(1 + \omega)| = 1$
$|r - 1| |1 + \omega| = 1$
Since $|1 + \omega| = |-\omega^2| = |\omega|^2 = 1$, we have $|r - 1| = 1$.
Since $r > 0$, we must have $r - 1 = 1 \implies r = 2$.
Substitute $r = 2$ back to find $z$:
$z = 2(1 + \omega) - 1 = 1 + 2\omega$
Given the root $\omega = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ (since $0 < \arg(\omega) < \pi$):
$z = 1 + 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 0 + i\sqrt{3}$