Question

If $z={(\frac{\sqrt{3}}{2}+\frac{i}{2})}^{107}+{(\frac{\sqrt{3}}{2}-\frac{i}{2})}^{107}$, then what is the imaginary part of z equal to?

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $1$

Answer 1

The correct option is A $0$

$z={(\frac{\sqrt{3}}{2}+\frac{i}{2})}^{107}+{(\frac{\sqrt{3}}{2}-\frac{i}{2})}^{107}$

$\therefore z={(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})}^{107}+{(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6})}^{107}$

Using Demoivre's theorem ,we get

$\therefore z=(\cos \frac{107\pi }{6}+i\sin \frac{107\pi }{6})+(\cos \frac{\left(107\right)11\pi }{6}+i\sin \frac{\left(107\right)11\pi }{6})$

$=\left(\cos \right(16\pi +\frac{11\pi }{6})+i\sin (16\pi +\frac{11\pi }{6}\left)\right)+\left(\cos \right((16\pi +\frac{11\pi }{6})11)+i\sin ((16\pi +\frac{11\pi }{6})11\left)\right)$

$=\left(\cos \right(\frac{11\pi }{6})+i\sin (\frac{11\pi }{6}\left)\right)+\left(\cos \right(176\pi +\frac{121\pi }{6})+i\sin (176\pi +\frac{121\pi }{6}\left)\right)$

$=\left(\cos \right(\frac{11\pi }{6})+i\sin (\frac{11\pi }{6}\left)\right)+\left(\cos \right(\frac{121\pi }{6})+i\sin (\frac{121\pi }{6}\left)\right)$

$=\left(\cos \right(\frac{11\pi }{6})+i\sin (\frac{11\pi }{6}\left)\right)+\left(\cos \right(\frac{121\pi }{6})+i\sin (\frac{121\pi }{6}\left)\right)$

$=\left(\cos \right(\frac{11\pi }{6})+i\sin (\frac{11\pi }{6}\left)\right)+\left(\cos \right(20\pi +\frac{\pi }{6})+i\sin (20\pi +\frac{\pi }{6}\left)\right)$

$=\left(\cos \right(\frac{11\pi }{6})+i\sin (\frac{11\pi }{6}\left)\right)+\left(\cos \right(\frac{\pi }{6})+i\sin (\frac{\pi }{6}\left)\right)$

$=(\frac{\sqrt{3}}{2}-\frac{i}{2})+(\frac{\sqrt{3}}{2}+\frac{i}{2})$

$=\sqrt{3}$

$=\sqrt{3}$

$\therefore Im\left(z\right)=0$

So, the answer is option (A).