Question 2 - A-Level Maths

You are given the complex numbers $w = 3 \mathrm { e } ^ { - \frac { 1 } { 12 } \pi \mathrm { j } }$ and $z = 1 - \sqrt { 3 } \mathrm { j }$.
1. Find the modulus and argument of each of the complex numbers $w , z$ and $\frac { w } { z }$.
2. Hence write $\frac { w } { z }$ in the form $a + b \mathrm { j }$, giving the exact values of $a$ and $b$.
In this part of the question, $n$ is a positive integer and $\theta$ is a real number with $0 < \theta < \frac { \pi } { n }$.
1. Express $\mathrm { e } ^ { - \frac { 1 } { 2 } \mathrm { j } \theta } + \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { j } \theta }$ in simplified trigonometric form, and hence, or otherwise, show that $$1 + \mathrm { e } ^ { \mathrm { j } \theta } = 2 \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { j } \theta } \cos \frac { 1 } { 2 } \theta$$ Series $C$ and $S$ are defined by $$\begin{aligned} & C = 1 + \binom { n } { 1 } \cos \theta + \binom { n } { 2 } \cos 2 \theta + \binom { n } { 3 } \cos 3 \theta + \ldots + \binom { n } { n } \cos n \theta
  & S = \binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta + \binom { n } { 3 } \sin 3 \theta + \ldots + \binom { n } { n } \sin n \theta \end{aligned}$$
2. Find $C$ and $S$, and show that $\frac { S } { C } = \tan \frac { 1 } { 2 } n \theta$.