Q111 / 160 · AP EAMCET 2010 · Mathematics

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to

  • A
    $128 \sqrt{3}$
  • B
    $256 \sqrt{3}$
  • Correct answer
    C
    $-128 \sqrt{3}$
  • D
    $-256 \sqrt{3}$

Solution

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ 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]

Hence, the correct answer is option: 'C'.

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