$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$
Let $\quad \sqrt{3}+i=z$
$z=(\sqrt{3}+i)=r(\cos \theta+i \sin \theta)$
Then, $\quad r \cos \theta=\sqrt{3}$ $\ldots$ (i)
$r \sin \theta=1$ $\ldots$ (ii)
Eq. (i) + Eq. (ii), we get
$r^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=3+1$
$\Rightarrow \quad r^2=4$
$\Rightarrow \quad r=2$
Eq. (ii) $\div$ Eq. (i), we get
$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6}$
$\Rightarrow \quad \theta=\frac{\pi}{6}$
Then, $z=(\sqrt{3}+i)=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$
$\Rightarrow \quad \bar{z}=(\sqrt{3}-i)=2\left(\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right)$
Then, $(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$
$=\left[2\left\{\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right\}\right]^7$ $+\left[2\left\{\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right\}\right]^7$
$=2^7\left\{\cos \frac{7 \pi}{6}+i \sin \frac{7 \pi}{6}\right\}+2^7$ $\left\{\cos \frac{7 \pi}{6}-i \sin \frac{7 \pi}{6}\right\}$
(By De-moiver theorem)
$=2^7\left\{\cos \frac{7 \pi}{6}+i \sin \frac{7 \pi}{6}+\cos \frac{7 \pi}{6}-i \sin \frac{7 \pi}{6}\right\}$
$=2^7 \cdot 2 \cos \frac{7 \pi}{6}$
$=2^8 \cdot \cos (\pi+\pi / 6)$
$=-2^8 \cdot \cos \frac{\pi}{6}$
$=-2^8 \cdot \frac{\sqrt{3}}{2}$
$=-2^7 \cdot \sqrt{3}=-128 \sqrt{3}$
OBJECT END]