(i) To solve the equation \(z^2 + (2\sqrt{3})iz - 4 = 0\), we use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\sqrt{3}i\), and \(c = -4\).

Calculate the discriminant: \(b^2 - 4ac = (2\sqrt{3}i)^2 - 4 \times 1 \times (-4) = -12 - (-16) = 4\).

Thus, \(z = \frac{-2\sqrt{3}i \pm \sqrt{4}}{2} = \frac{-2\sqrt{3}i \pm 2}{2}\).

This gives the roots \(z = 1 - \sqrt{3}i\) and \(z = -1 - \sqrt{3}i\).

(ii) On the Argand diagram, plot the points \((1, -\sqrt{3})\) and \((-1, -\sqrt{3})\).

(iii) The modulus of each root is \(\sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2\).

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

The argument of \(-1 - \sqrt{3}i\) is \(\arctan\left(\frac{-\sqrt{3}}{-1}\right) = -120^\circ\).

(iv) The points \((0,0)\), \((1, -\sqrt{3})\), and \((-1, -\sqrt{3})\) form an equilateral triangle because the distances between each pair of points are equal, each being 2.