(i) To find the modulus of \(z\), we calculate \(|z| = \sqrt{(1 + \cos 2\theta)^2 + (\sin 2\theta)^2}\).

Using the identity \(\cos 2\theta = 2\cos^2 \theta - 1\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we have:

\(|z| = \sqrt{(2\cos^2 \theta)^2 + (2\sin \theta \cos \theta)^2} = \sqrt{4\cos^4 \theta + 4\sin^2 \theta \cos^2 \theta}\).

Factor out \(4\cos^2 \theta\):

\(|z| = \sqrt{4\cos^2 \theta (\cos^2 \theta + \sin^2 \theta)} = \sqrt{4\cos^2 \theta} = 2\cos \theta\).

For the argument, use \(\tan \theta = \frac{\sin 2\theta}{1 + \cos 2\theta} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \tan \theta\).

Thus, the argument is \(\theta\).

(ii) To find the real part of \(\frac{1}{z}\), multiply numerator and denominator by the conjugate:

\(\frac{1}{z} = \frac{1}{1 + \cos 2\theta + i \sin 2\theta} \times \frac{1 - \cos 2\theta - i \sin 2\theta}{1 - \cos 2\theta - i \sin 2\theta}\).

This gives:

\(\frac{1 - \cos 2\theta - i \sin 2\theta}{(1 + \cos 2\theta)^2 + (\sin 2\theta)^2} = \frac{1 - \cos 2\theta - i \sin 2\theta}{4\cos^2 \theta}\).

The real part is \(\frac{1 - \cos 2\theta}{4\cos^2 \theta} = \frac{1}{2}\).