Show that for any complex number $z = e^{i\theta}$, $z^{n} + \dfrac{1}{z^{n}} = 2\cos n\theta .$ By taking $n = 1$, show that $\cos^{3}\theta = \tfrac{1}{8}\left(z^{3} + 3z + \tfrac{3}{z} + \tfrac{1}{z^{3}}\right).$ Deduce that $\cos^{3}\theta = \tfrac{1}{4}\cos 3\theta + \tfrac{3}{4}\cos\theta.$ Find $\displaystyle \int \cos^{3}3\theta\, d\theta$.

1. Express $z^n$ and $\frac{1}{z^n}$ in trigonometric form using Euler's formula

Given $z = e^{i\theta}$, we have:
$z^n = (e^{i\theta})^n = e^{in\theta} = \cos n\theta + i\sin n\theta$

For $\frac{1}{z^n}$, we use negative indices:
$\frac{1}{z^n} = z^{-n} = e^{-in\theta} = \cos(-n\theta) + i\sin(-n\theta)$

Using the even/odd properties of cosine and sine ($\cos(-\alpha) = \cos\alpha$ and $\sin(-\alpha) = -\sin\alpha$):
$\frac{1}{z^n} = \cos n\theta - i\sin n\theta$

2. Add the two expressions to obtain the final result

$\begin{aligned} z^n + \frac{1}{z^n} &= (\cos n\theta + i\sin n\theta) + (\cos n\theta - i\sin n\theta) \\ &= 2\cos n\theta \end{aligned}$

1. Substitute $n = 1$ into the identity from part (i).

From part (i), we have $z^{n} + \frac{1}{z^{n}} = 2\cos n\theta$.
Taking $n = 1$:
$z + \frac{1}{z} = 2\cos\theta$

2. Cube both sides of the equation.

$(2\cos\theta)^3 = \left(z + \frac{1}{z}\right)^3$
$8\cos^3\theta = z^3 + 3z^2\left(\frac{1}{z}\right) + 3z\left(\frac{1}{z^2}\right) + \frac{1}{z^3}$

3. Simplify to show the first target identity.

$8\cos^3\theta = z^3 + 3z + \frac{3}{z} + \frac{1}{z^3}$
Dividing by 8 gives the required result:
$\cos^3\theta = \frac{1}{8}\left(z^3 + 3z + \frac{3}{z} + \frac{1}{z^3}\right)$

4. Group the terms and use the identity from part (i) to deduce the second identity.

Regrouping the terms inside the bracket:
$\cos^3\theta = \frac{1}{8}\left[ \left(z^3 + \frac{1}{z^3}\right) + 3\left(z + \frac{1}{z}\right) \right]$
Using the identity from part (i) with $n=3$ ($z^3 + \frac{1}{z^3} = 2\cos 3\theta$) and $n=1$ ($z + \frac{1}{z} = 2\cos\theta$):
$\cos^3\theta = \frac{1}{8}\left(2\cos 3\theta + 3(2\cos\theta)\right)$
$\cos^3\theta = \frac{1}{4}\cos 3\theta + \frac{3}{4}\cos\theta$

1. Express $\cos^3 3\theta$ in terms of multiple angles using the identity from part (ii).

From (ii), substituting $3\theta$ for $\theta$ into the derived identity $\cos^3\theta = \frac{1}{4}\cos 3\theta + \frac{3}{4}\cos\theta$, we get:
$\cos^3 3\theta = \frac{1}{4}\cos 9\theta + \frac{3}{4}\cos 3\theta$

2. Integrate the linearised trigonometric expression with respect to $\theta$.

$\begin{aligned} \int \cos^3 3\theta \, d\theta &= \int \left( \frac{1}{4}\cos 9\theta + \frac{3}{4}\cos 3\theta \right) d\theta \\ &= \frac{1}{4} \left( \frac{1}{9}\sin 9\theta \right) + \frac{3}{4} \left( \frac{1}{3}\sin 3\theta \right) + C \\ &= \frac{1}{36}\sin 9\theta + \frac{1}{4}\sin 3\theta + C \end{aligned}$