Prove that $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$, $\;$ $n \in N$
Let $\;$ $\left(1 + i\right) = r \left(\cos \theta + i \sin \theta\right)$ $\;\;\; \cdots \; (1)$
Equating the real and imaginary parts separately, we have
$r \; \cos \theta = 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $r \; \sin \theta = 1$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ $r = \sqrt{\left(1\right)^2 + \left(1\right)^2} = \sqrt{2}$
Substituting the value of $r$ in equations $(2a)$ and $(2b)$, we have
$\cos \theta = \dfrac{1}{\sqrt{2}}$, $\;\;$ $\sin \theta = \dfrac{1}{\sqrt{2}}$ $\implies$ $\theta = \dfrac{\pi}{4}$
Substituting the values of $r$ and $\theta$ in equation $(1)$ we have,
$\left(1 + i\right) = \sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]$
$\begin{aligned}
\therefore \; \left(1 + i\right)^n & = \left\{\sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]\right\}^{n} \\\\
& = \left(\sqrt{2}\right)^n \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]^n \\\\
& = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) + i \sin \left(\dfrac{n \pi}{4}\right)\right] \;\;\; \cdots \; (3a)
\end{aligned}$
Replacing $+i$ with $-i$ in equation $(3a)$, we have
$\left(1 - i\right)^n = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) - i \sin \left(\dfrac{n \pi}{4}\right)\right]$ $\;\;\; \cdots \; (3b)$
Adding equations $(3a)$ and $(3b)$, we have
$\left(1 + i\right)^n + \left(1 - i\right)^n = 2^{\frac{n}{2}} \left[2 \cos \left(\dfrac{n \pi}{4}\right)\right]$
i.e. $\;$ $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$
Hence proved.