Let $\alpha = \cos\theta + i \sin\theta$, where $0 < \theta < 2\pi$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2017 - Paper 1

To find $C$, rewrite it as:

$C = 1 + \sum_{k=1}^{n} \alpha^k = 1 + (\alpha^n + \alpha^{n-1} + \cdots + \alpha)$

This is a geometric series with first term 1 and ratio $\alpha$, which sums to:

$C = \frac{1 - \alpha^{n + 1}}{1 - \alpha}$

Now substitute from part (i) and simplify to reach:

$C = \frac{\alpha^{n+1} - (1 + \alpha)(\alpha^n)}{(1 - \alpha)(1 - \alpha)}$

From part (i), we have:

$1 + 2(\cos 0 + \cos 2\theta + \cdots + \cos n\theta) = \cos n\theta(\frac{\cos(n + 1)\theta}{1 - \cos \theta})$

Since $\cos 0 = 1$, the series can be simplified to the deduced result with correct substitutions from previous results.

To show independence from $n$, observe that:

Using double angle formulas, we get: $S = \sum_{k=1}^{m} \cos\left(\frac{2\pi k}{n}\right) = \frac{1}{2} \left[\frac{\sin((m+1)\frac{\pi}{n})}{\sin(\frac{\pi}{n})} \right],$ which is constant for different integer $n$. Therefore, it demonstrates independence with respect to $n$.

Using the formula for eccentricity $e = \frac{c}{a}$ for a hyperbola, where $c = \sqrt{a^2 + b^2}$:

Given $e = 2$, we have: $2 = \frac{\sqrt{a^2 + b^2}}{a} \Rightarrow 4a^2 = a^2 + b^2 \Rightarrow 3a^2 = b^2 \Rightarrow b = \sqrt{3}a.$

Also, from the given vertex distance, $a = 1$, thus giving possible values: $b = \sqrt{3}$.