(i) To find the roots of \(z^2 - z + 1 = 0\), use the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = 1\).

Calculate the discriminant: \(b^2 - 4ac = (-1)^2 - 4 \times 1 \times 1 = 1 - 4 = -3\).

Thus, \(z = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}\).

The roots are \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\) and \(\frac{1}{2} - i\frac{\sqrt{3}}{2}\).

(ii) The modulus of each root \(z = x + iy\) is \(\sqrt{x^2 + y^2}\).

For \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\), the modulus is \(\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\).

The argument is \(\arctan\left(\frac{\sqrt{3}/2}{1/2}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3}\).

For \(\frac{1}{2} - i\frac{\sqrt{3}}{2}\), the modulus is also 1, and the argument is \(-\frac{\pi}{3}\).

(iii) To show that each root satisfies \(z^3 = -1\), calculate \(z^3\) for each root.

For \(z = \frac{1}{2} + i\frac{\sqrt{3}}{2}\), \(z^3 = \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^3 = -1\).

Similarly, for \(z = \frac{1}{2} - i\frac{\sqrt{3}}{2}\), \(z^3 = \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)^3 = -1\).