(i) Multiply the numerator and denominator by \(\sqrt{3} + i\) to rationalize the denominator:

\(z = \frac{(9\sqrt{3} + 9i)(\sqrt{3} + i)}{(\sqrt{3} - i)(\sqrt{3} + i)}\)

\(= \frac{18 + 18\sqrt{3}i}{4}\)

\(= \frac{9}{2} + \frac{9}{2}\sqrt{3}i\)

Find the modulus \(r\) and argument \(\theta\):

\(r = \sqrt{\left(\frac{9}{2}\right)^2 + \left(\frac{9}{2}\sqrt{3}\right)^2} = 9\)

\(\theta = \arctan\left(\frac{\frac{9}{2}\sqrt{3}}{\frac{9}{2}}\right) = \frac{\pi}{3}\)

Thus, \(z = 9e^{i\frac{\pi}{3}}\).

(ii) The square roots of \(z\) are given by \(\sqrt{9}e^{i\frac{\pi}{6}}\) and \(\sqrt{9}e^{i\left(\frac{\pi}{3} - \pi\right)}\).

\(= 3e^{i\frac{\pi}{6}}\) and \(3e^{-i\frac{5\pi}{6}}\).