Complex Numbers

If $\;$ $x = a + b$, $\;$ $y = a \omega + b \omega^2$ $\;$ and $\;$ $z = a \omega^2 + b \omega$, $\;$ show that $\;$ $x^3 + y^3 + z^3 = 3 \left(a^3 + b^3\right)$ $\;$ where $\omega$ is the complex cube root of unity.

$x= a + b$

$\therefore$ $\;$ $x^3 = \left(a + b\right)^3 = a^3 + b^3 + 3 a^2 b + 3 a b^2$ $\;\;\; \cdots \; (1)$

$y = a\omega + b \omega^2$

$\begin{aligned} \therefore \; y^3 & = \left(a \omega + b \omega^2\right)^3 \\\\ & = a^3 \omega^3 + b^3 \left(\omega^2\right)^3 + 3 a^2 b \omega^4 + 3 a b^2 \omega^5 \\\\ & = a^3 \omega^3 + b^3 \left(\omega^3\right)^2 + 3 a^2 b \omega^3 \times \omega + 3 a b^2 \omega^3 \times \omega^2 \\\\ & = a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \;\;\; \cdots \; (2) \;\;\; [\text{Note: } \omega^3 = 1] \end{aligned}$

$z = a \omega^2 + b \omega$

$\begin{aligned} \therefore \; z^3 & = \left(a \omega^2 + b \omega\right)^3 \\\\ & = a^3 \left(\omega^2\right)^3 + b^3 \omega^3 + 3 a^2 b \omega^5 + 3 a b^2 \omega^4 \\\\ & = a^3 \left(\omega^3\right)^2 + b^3 \omega^3 + 3 a^2 b \omega^3 \times \omega^2 + 3 a b^2 \omega^3 \times \omega \\\\ & = a^3 + b^3 + 3 a^2 b \omega^2 + 3 a b^2 \omega \;\;\; \cdots \; (3) \end{aligned}$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$\begin{aligned} x^3 + y^3 + z^3 & = a^3 + b^3 + 3 a^2 b + 3 ab^2 \\ & \hspace{1cm} a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \\ & \hspace{2cm} a^3 + b^3 + 3 a^2 b \omega^2 + 3 ab^2 \omega \\\\ & = 3 a^3 + 3 b^3 + 3 a^2 b \left(1 + \omega + \omega^2\right) + 3 ab^2 \left(1 + \omega^2 + \omega\right) \\\\ & = 3 \left(a^3 + b^3\right) \;\;\; [\because \; 1 + \omega + \omega^2 = 0] \end{aligned}$

Hence proved.

Complex Numbers

Find the value of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$

Let $\left(- \sqrt{3} - i\right) = r \left(\cos \theta + i \sin \theta\right)$

Then, $\;$ $r \cos \theta = - \sqrt{3}$; $\;\;$ $r \sin \theta = -1$

$\therefore$ $\;$ $r = \sqrt{\left(- \sqrt{3}\right)^2 + \left(-1\right)^2} = 2$

Now, $\;$ $\cos \theta = - \dfrac{\sqrt{3}}{2}$; $\;$ $\sin \theta = - \dfrac{1}{2}$ $\implies$ $\theta = -\pi + \dfrac{\pi}{6} = \dfrac{- 5 \pi}{6}$

$\begin{aligned} \therefore \; \left(- \sqrt{3} - i\right)^{\frac{2}{3}} & = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5\pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^{\frac{2}{3}} \\\\ & = 2 ^{\frac{2}{3}} \left\{\left[\cos \left(\dfrac{-5 \pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^2\right\}^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{3}\right) + i \sin \left(\dfrac{-5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left[\cos \left(2 k \pi - \dfrac{5 \pi}{3}\right) + i \sin \left(2 k \pi - \dfrac{5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left\{\cos \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] + i \sin \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] \right\} \;\;\; where \; k = 0, 1, 2 \end{aligned}$

$\therefore$ $\;$ The values of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$ $\;$ are

$2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{9}\right) + i \sin \left(\dfrac{-5 \pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{\pi}{9}\right) + i \sin \left(\dfrac{\pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{7\pi}{9}\right) + i \sin \left(\dfrac{7\pi}{9}\right)\right]$

Complex Numbers

If $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$ $\;$ and $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$, $\;$ prove that $\;$ $\dfrac{a^2 b^2 + c^2}{abc} = 2 \cos 2 \left(\alpha + \beta - \gamma\right)$

Given: $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$, $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$

$\therefore$ $\;$ $a^2 = \left(\cos 2 \alpha + i \sin 2 \alpha\right)^2 = \cos 4 \alpha + i \sin 4 \alpha$

$b^2 = \left(\cos 2 \beta + i \sin 2 \beta\right)^2 = \cos 4 \beta + i \sin 4 \beta$

$c^2 = \left(\cos 2 \gamma + i \sin 2 \gamma\right)^2 = \cos 4 \gamma + i \sin 4 \gamma$

Now,

$\begin{aligned} a^2 b^2 & = \left(\cos 4 \alpha + i \sin 4 \alpha\right) \left(\cos 4 \beta + i \sin 4 \beta\right) \\\\ & = \left(\cos 4 \alpha \; \cos 4 \beta - \sin 4 \alpha \; \sin 4 \beta\right) + i \left(\sin 4 \alpha \; \cos 4 \beta + \cos 4 \alpha \; \sin 4 \beta\right) \\\\ & = \cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right) \end{aligned}$

$\begin{aligned} \therefore \; a^2 b^2 + c^2 & = \left[\cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right)\right] + \left[\cos 4 \gamma + i \sin 4 \gamma\right] \\\\ & = \left[\cos \left(4 \alpha + 4 \beta\right) + \cos 4 \gamma\right] + i \left[\sin \left(4 \alpha + 4 \beta\right) + \sin 4 \gamma\right] \\\\ & = 2 \cos \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\ & \hspace{1cm} + 2 i \sin \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\\\ & = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right] \end{aligned}$

$\begin{aligned} abc & = \left(\cos 2 \alpha + i \sin 2 \alpha\right) \left(\cos 2 \beta + i \sin 2 \beta\right) \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\left(\cos 2 \alpha \; \cos 2 \beta - \sin 2 \alpha \; \sin 2 \beta \right) + i \left(\sin 2 \alpha \; \cos 2 \beta + \cos 2 \alpha \; \sin 2 \beta \right)\right] \\ & \hspace{9cm} \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\cos \left(2 \alpha + 2 \beta\right) + i \sin \left(2 \alpha + 2 \beta\right)\right] \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\cos \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma - \sin \left(2 \alpha + 2 \beta\right) \; \sin 2 \gamma\right] \\ & \hspace{1cm} + i \left[\sin \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma + \cos \left(2 \alpha + 2 \beta \right) \; \sin 2 \gamma\right] \\\\ & = \cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right) \end{aligned}$

$\begin{aligned} \therefore \; \dfrac{a^2 b^2 + c^2}{abc} & = \dfrac{2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right]}{\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)} \\\\ & = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right) \\\\ & = 2 \cos 2 \left(\alpha + \beta - \gamma\right) \end{aligned}$

Hence proved.

Complex Numbers

If $x = \cos \alpha + i \sin \alpha$; $\;$ $y = \cos \beta + i \sin \beta$, $\;$ then prove that $\;$ $x^m y^n + \dfrac{1}{x^m y^n} = 2 \cos \left(m \alpha + n \beta\right)$

Given: $\;$ $x = \cos \alpha + i \sin \alpha$; $\;\;$ $y = \cos \beta + i \sin \beta$

$\implies$ $\dfrac{1}{x} = \cos \alpha - i \sin \alpha$; $\;$ $\dfrac{1}{y} = \cos \beta - i \sin \beta$

Now, $\;$ $x^m = \left(\cos \alpha + i \sin \alpha\right)^m = \cos \left(m \alpha\right) + i \sin \left(m \alpha \right)$

$y^n = \left(\cos \beta + i \sin \beta\right)^n = \cos \left(n \beta\right) + i \sin \left(n \beta\right)$

$\begin{aligned} \therefore \; x^m y^n & = \left[\cos \left(m \alpha\right) + i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) + i \sin \left(n \beta\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right) \end{aligned}$

$\dfrac{1}{x^m} = \left(\cos \alpha - i \sin \alpha\right)^m = \cos \left(m \alpha\right) - i \sin \left(m \alpha\right)$

$\dfrac{1}{y^n} = \left(\cos \beta - i \sin \beta\right)^n = \cos \left(n \beta\right) - i \sin \left(n \beta\right)$

$\begin{aligned} \therefore \; \dfrac{1}{x^m y^n} & = \left[\cos \left(m \alpha\right) - i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) - i \sin \left(n \beta\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right) \end{aligned}$

$\begin{aligned} \therefore \; x^m y^n + \dfrac{1}{x^m y^n} & = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right) \\ & \hspace{2cm} + \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right) \\\\ & = 2 \cos \left(m \alpha + n \beta\right) \end{aligned}$

Hence proved.

Complex Numbers

If $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$, $\;$ prove that $\;$ $x^n + \dfrac{1}{x^n} = 2 \cos \left(n \theta\right)$ $\;$ and $\;$ $x^n - \dfrac{1}{x^n} = 2 i \sin \left(n \theta\right)$

Let $x = \cos \theta + i \sin \theta$

$\begin{aligned} Then, \; \dfrac{1}{x} & = \dfrac{1}{\cos \theta + i \sin \theta} \\\\ & = \dfrac{\cos \theta - i \sin \theta}{\left(\cos \theta + i \sin \theta\right) \left(\cos \theta - i \sin \theta\right)} \\\\ & = \dfrac{\cos \theta - i \sin \theta}{\cos^2 \theta - i^2 \sin^2 \theta} \\\\ & = \cos \theta - i \sin \theta \end{aligned}$

so that $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$

Now,

$x^n = \left(\cos \theta + i \sin \theta\right)^n = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$

$\dfrac{1}{x^n} = \left(\cos \theta - i \sin \theta\right)^n = \cos \left(n \theta\right) - i \sin \left(n \theta\right)$

$\therefore \; x^n + \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) + \cos \left(n \theta\right) - i \sin \left(n \theta\right) = 2 \cos \left(n \theta\right)$

$x^n - \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right) = 2 i \sin \left(n \theta\right)$

Hence proved.

Complex Numbers

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2px + \left(p^2 + q^2\right) = 0$ and $\tan \theta = \dfrac{q}{y + p}$, show that $\dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} = q^{n - 1} \left(\dfrac{\sin n\theta}{\sin^n \theta}\right)$

The roots of the given quadratic equation $\;$ $x^2 - 2px + \left(p^2 + q^2\right) = 0$ $\;$ are

$\begin{aligned} x & = \dfrac{2p \pm \sqrt{4p^2 - 4p^2 - 4q^2}}{2} \\\\ & = \dfrac{2p \pm 2iq}{2} \\\\ & = p \pm iq \end{aligned}$

$\because$ $\;$ $\alpha$ and $\beta$ are the roots of the given quadratic equation, let

$\alpha = p + i q$ $\;$ and $\;$ $\beta = p - iq$

Given: $\;$ $\tan \theta = \dfrac{q}{y + p}$

$\implies$ $y = \dfrac{q}{\tan \theta} - p$

$\begin{aligned} \therefore \; \left(y + \alpha\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p + iq\right)^n \\\\ & = \left[q \left(\dfrac{1}{\tan \theta} + i\right)\right]^n \\\\ & = q^n \left[\dfrac{\cos \theta + i \sin \theta}{\sin \theta}\right]^n \\\\ & = \dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \left(y + \beta\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p - iq\right)^n \\\\ & = \left[q \left(\dfrac{1}{\tan \theta} - i\right)\right]^n \\\\ & = q^n \left[\dfrac{\cos \theta - i \sin \theta}{\sin \theta}\right]^n \\\\ & = \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (2) \end{aligned}$

$\alpha - \beta = p + iq - \left(p - iq\right) = 2 \; i \; q$ $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$\begin{aligned} \dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} & = \dfrac{\dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} - \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta}}{2 \;i \;q} \\\\ & = \dfrac{q^n \left[\cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right)\right]}{2 \;i \;q \; \sin^n \theta} \\\\ & = \dfrac{2 \; i \; q^n \; \sin \left(n \theta\right)}{2 \; i \; q \; \sin^n \theta} \\\\ & = q^{n - 1} \left[\dfrac{\sin \left(n \theta\right)}{\sin^n \theta}\right] \end{aligned}$

Hence proved.

Complex Numbers

Prove that $\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$

Let $\;$ $z = \cos \theta + i \sin \theta$

$\because$ $\;$ $\left|z\right| = 1$ $\implies$ $\overline{z} = \dfrac{1}{z}$

$\therefore$ $\;$ From equation $(1a)$, $\;$ $\overline{z} = \dfrac{1}{z} = \cos \theta - i \sin \theta$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \therefore \; \left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n & = \left(1 + z\right)^n + \left(1 + \dfrac{1}{z}\right)^n \\\\ & = \left(1 + z\right)^n + \dfrac{\left(1 + z\right)^n}{z^n} \\\\ & = \left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) \;\;\; \cdots \; (2) \end{aligned}$

From equation $(1)$,

$\begin{aligned} \dfrac{1}{z^n} & = \left(\cos \theta - i \sin \theta\right)^n \\\\ & = \cos \left(n \theta\right) - i \sin \left(n \theta\right) \;\; [\text{By De Moivre's theorem}] \end{aligned}$

$\begin{aligned} \therefore \; 1 + \dfrac{1}{z^n} & = 1 + \cos \left(n \theta\right) - i \sin \left(n \theta\right) \\\\ & = 2 \cos^2 \left(\dfrac{n \theta}{2}\right) - 2 i \sin \left(\dfrac{n \theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\\\ & = 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3a) \end{aligned}$

$\begin{aligned} Now, \; \left(1 + z\right) & = \left(1 + \cos \theta\right) + i \sin \theta \\\\ & = 2 \cos^2 \left(\dfrac{\theta}{2}\right) + 2 i \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \\\\ & = 2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right] \end{aligned}$

$\begin{aligned} \therefore \; \left(1 + z\right)^n & = \left\{2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right]\right\}^n \\\\ & = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3b) \end{aligned}$

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$,

$\begin{aligned} \left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) & = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \\ & \hspace{2.5cm} \times 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\ & \hspace{1cm} \times \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right]\left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) - i^2 \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) + \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \;\;\; \cdots \; (4) \end{aligned}$

$\therefore$ $\;$ From equations $(2)$ and $(4)$ we have,

$\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$

Hence proved.