Question

If $(\sqrt{3}+i)^8-(\sqrt{3}-i)^8=\alpha+i \beta$, then $\alpha-\frac{\sqrt{3}}{2} \beta=$

• A. $256$
• B. $384 \sqrt{3}$
• C. $384$
• D. $256 \sqrt{3}$

Answer 1

Answer: (C)

$ (\sqrt{3}+i)^8-(\sqrt{3}-i)^8=\alpha+i \beta $
$ \frac{\sqrt{3}}{2}+\frac{i}{2}=\cos \left(\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{6}\right)=e^{i \pi / 6}$
$ \frac{\sqrt{3}}{2}-\frac{i}{2}=\cos \left(2 \pi-\frac{\pi}{6}\right)+i \sin \left(2 \pi-\frac{\pi}{6}\right)=e^{i\left(2 \pi-\frac{\pi}{6}\right)}$
From Eq. (i),
$ \left(2 e^{i \frac{\pi}{6}}\right)^8-\left(2 e^{i \frac{11 \pi}{6}}\right)^8=\alpha+i \beta $
$ \Rightarrow 2^8\left(e^{i \frac{8 \pi}{6}}-e^{i \frac{88 \pi}{6}}\right)=\alpha+i \beta $
$ \Rightarrow 2^8\left[e^{i\left(\pi+\frac{\pi}{3}\right)}-e^{i\left(15 \pi-\frac{\pi}{3}\right)}\right]=\alpha+i \beta$
$ \Rightarrow 2^8\left[\left(-\frac{1}{2}-\frac{\sqrt{3} i}{2}\right)-\left(-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\right]=\alpha+i \beta$
$ \Rightarrow 2^8(-\sqrt{3} i)=\alpha+i \beta \Rightarrow \alpha=0, \beta=-\sqrt{3} \cdot 2^8$
$\alpha-\frac{\sqrt{3}}{2} \beta \Rightarrow 0-\frac{\sqrt{3}}{2}\left(-\sqrt{3} \cdot 2^8\right)=3 \cdot 2^7=384$