Question

If $z={\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)}^5+{\left(\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)}^5$, then Im(z) =

Answer 1

Given that
$z={\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)}^5+{\left(\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)}^5$
$\Rightarrow z={[cos\left(\frac{\pi }{6}\right)+isin\left(\frac{\pi }{6}\right)]}^5+{[cos\left(\frac{\pi }{6}\right)-isin\left(\frac{\pi }{6}\right)]}^5$
$\Rightarrow z=cos\frac{5\pi }{6}+isin\frac{5\pi }{6}+cos\frac{5\pi }{6}-isin\frac{5\pi }{6}$
Hence, $Im\left(z\right)=0$.
Ans: 0