Do not use a calculator in answering this question.

1. Find the modulus and argument of $\alpha$ to express it in polar form.

Given $\alpha = -\sqrt{3} + \mathrm{i}$.
$|\alpha| = \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2$
$\arg(\alpha) = \pi - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
So, $\alpha = 2\left(\cos\frac{5\pi}{6} + \mathrm{i}\sin\frac{5\pi}{6}\right)$.

2. Evaluate $\alpha^5$ using De Moivre's Theorem.

$|\alpha^5| = 2^5 = 32$
$\arg(\alpha^5) = 5 \times \frac{5\pi}{6} = \frac{25\pi}{6}$
To find the principal argument, we subtract $4\pi$:
$\frac{25\pi}{6} - 4\pi = \frac{\pi}{6}$
So, $\alpha^5 = 32\left(\cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6}\right)$.

3. Find the modulus and argument of the conjugate $\beta^*$.

Given $\beta = \cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}$.
By the properties of conjugates, $|\beta^*| = |\beta| = 1$.
$\arg(\beta^*) = -\arg(\beta) = -\frac{\pi}{3}$.
So, $\beta^* = 1\left(\cos\left(-\frac{\pi}{3}\right) + \mathrm{i}\sin\left(-\frac{\pi}{3}\right)\right)$.

4. Divide $\alpha^5$ by $\beta^*$ to find the final polar form.

$\left|\frac{\alpha^5}{\beta^*}\right| = \frac{|\alpha^5|}{|\beta^*|} = \frac{32}{1} = 32$
$\arg\left(\frac{\alpha^5}{\beta^*}\right) = \arg(\alpha^5) - \arg(\beta^*) = \frac{\pi}{6} - \left(-\frac{\pi}{3}\right) = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{\pi}{2}$
Therefore, $\frac{\alpha^5}{\beta^*} = 32\left(\cos\frac{\pi}{2} + \mathrm{i}\sin\frac{\pi}{2}\right)$.

1. Evaluate the denominator $1+z^2$ and simplify it using trigonometric identities.

$z^2 = (\cos\theta + \mathrm{i}\sin\theta)^2 = \cos^2\theta - \sin^2\theta + \mathrm{i}(2\sin\theta\cos\theta)$

$1 + z^2 = 1 + \cos^2\theta - \sin^2\theta + \mathrm{i}(2\sin\theta\cos\theta)$

Using the identity $1 - \sin^2\theta = \cos^2\theta$:
$1 + z^2 = 2\cos^2\theta + \mathrm{i}(2\sin\theta\cos\theta)$

Factorising out $2\cos\theta$:
$1 + z^2 = 2\cos\theta(\cos\theta + \mathrm{i}\sin\theta)$

2. Multiply by the complex conjugate to show the given relationship for $\frac{1}{1+z^2}$.

$\frac{1}{1+z^2} = \frac{1}{2\cos\theta(\cos\theta + \mathrm{i}\sin\theta)}$

Multiply the numerator and denominator by the complex conjugate $\cos\theta - \mathrm{i}\sin\theta$:
$\frac{1}{1+z^2} = \frac{\cos\theta - \mathrm{i}\sin\theta}{2\cos\theta(\cos^2\theta + \sin^2\theta)}$

Since $\cos^2\theta + \sin^2\theta = 1$ and $z^* = \cos\theta - \mathrm{i}\sin\theta$:
$\frac{1}{1+z^2} = \frac{z^*}{2\cos\theta(1)} = \frac{z^*}{2\cos\theta}$ (shown)

3. Find the modulus of $\frac{1}{1+z^2}$.

The modulus is given by $\left|\frac{1}{1+z^2}\right| = \left|\frac{z^*}{2\cos\theta}\right| = \frac{|z^*|}{|2\cos\theta|}$.

Since $z = \cos\theta + \mathrm{i}\sin\theta$, we have $|z| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$, which means $|z^*| = 1$.

Given $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, we have $\cos\theta > 0$, so $2\cos\theta$ is a positive real number and $|2\cos\theta| = 2\cos\theta$.

Thus, the modulus is $\frac{1}{2\cos\theta}$.

4. Find the argument of $\frac{1}{1+z^2}$.

The argument is given by $\arg\left(\frac{1}{1+z^2}\right) = \arg\left(\frac{z^*}{2\cos\theta}\right)$.

Using properties of arguments:
$\arg\left(\frac{z^*}{2\cos\theta}\right) = \arg(z^*) - \arg(2\cos\theta)$

Since $2\cos\theta$ is a positive real number, $\arg(2\cos\theta) = 0$.

$z^* = \cos\theta - \mathrm{i}\sin\theta = \cos(-\theta) + \mathrm{i}\sin(-\theta)$, so $\arg(z^*) = -\theta$.

Thus, the argument is $-\theta - 0 = -\theta$.
(Note: This is within the principal range $(-\pi, \pi]$ since $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ implies $-\frac{\pi}{2} < -\theta < \frac{\pi}{2}$.)