(i) The modulus of \(u = 1 - (\sqrt{3})i\) is given by \(\sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2\).

The argument is \(\arctan\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}\) or \(-60^\circ\).

(ii) To show \(u^3 + 8 = 0\), calculate \(u^3 = (1 - (\sqrt{3})i)^3\).

Using the binomial theorem or De Moivre's theorem, \(u^3 = 2^3 \left(\cos(-\pi) + i\sin(-\pi)\right) = -8\).

Thus, \(u^3 + 8 = -8 + 8 = 0\).

(iii) On the Argand diagram, draw a circle centered at \(1 - (\sqrt{3})i\) with radius 2.

Draw the vertical line \(\text{Re } z = 2\).

Shade the region to the right of the line and within the circle.