Question

If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}(i=\sqrt{-1})$, then $(1+iz+z^5+i{z}^8{)}^9$ is equal to

• A. $-1$
• B. $1$
• C. $0$
• D. $(-1+2i{)}^9$

Answer 1

The correct option is A $-1$

$z=\frac{\sqrt{3}}{2}+\frac{i}{2}=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$
$\Rightarrow z^5=\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}=\frac{-\sqrt{3}+i}{2}$
and $z^8=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=-\left(\frac{1+i\sqrt{3}}{2}\right)$
$\Rightarrow (1+iz+z^5+i{z}^8{)}^9={(1+\frac{i\sqrt{3}}{2}-\frac{1}{2}-\frac{\sqrt{3}}{2}+\frac{1}{2}-\frac{i}{2}+\frac{\sqrt{3}}{2})}^9$
$={\left(\frac{1+i\sqrt{3}}{2}\right)}^9=\cos 3\pi +\sin 3\pi =-1$