Question

If $z={\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^{2009}+{\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)}^{2009}$, then

• A. $Im\left(z\right)=0$
• B. $Re\left(z\right)>0$
• C. $Im\left(z\right)>0$
• D. $Re\left(z\right)<0,Im\left(z\right)>0$

Answer 1

Answer: (A)

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

As we know that,
$\frac{\sqrt{3}}{2}+\frac{i}{2}=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$
and $\frac{\sqrt{3}}{2}-\frac{i}{2}=\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}$
$z={(\frac{\sqrt{3}}{2}+\frac{i}{2})}^{2009}+{(\frac{\sqrt{3}}{2}-\frac{i}{2})}^{2009}$
$\Rightarrow z={(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})}^{2009}+{(\cos \frac{\pi }{6}-i\sin \frac{\pi }{6})}^{2009}$ ......{ De Moivre's Theorem}

$\Rightarrow z=\cos \frac{2009\pi }{6}+i\sin \frac{2009\pi }{6}+\cos \frac{2009\pi }{6}-i\sin \frac{2009\pi }{6}$
$\Rightarrow z=2\cos \frac{2009\pi }{6}$

Therefore, $Im\left(z\right)=0$
Ans: B