Question

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

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

Answer 1

Answer: (A)

$z=\frac{\sqrt{3}}{2}+\frac{i}{2}=cos\frac{\pi }{6}+isin\frac{\pi }{6}$
$\Rightarrow z^5=cos\frac{5\pi }{6}+isin\frac{5\pi }{6}=\frac{-\sqrt{3}+i}{2}$
and $Z^8=cos\frac{4\pi }{3}+isin\frac{4\pi }{3}=-\left(\frac{1+i\sqrt{3}}{2}\right)$
$\Rightarrow {\left(1+iz+z^5+i{z}^8\right)}^9={\left(1+\frac{i\sqrt{3}}{2}-\frac{1}{2}-\frac{\sqrt{3}}{2}+\frac{i}{2}+\frac{i}{2}+\frac{\sqrt{3}}{2}\right)}^9$
$={\left(\frac{1+i\sqrt{3}}{2}\right)}^9=cos3\pi +isin3\pi =-1$