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