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. $(-1+2i{)}^9$
• D. $0$

Answer 1

The correct option is B $-1$

$z=\frac{\sqrt{3}}{2}+\frac{i}{2}=e^{i\pi /6}$
$iz=-\frac{1}{2}+\frac{\sqrt{3}i}{2}$
$z^5=e^{i5\pi /6}=\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}=-\frac{\sqrt{3}}{2}+\frac{i}{2}$
$z^8=e^{i4\pi /3}=\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=-\frac{1}{2}-\frac{\sqrt{3}i}{2}$
So, $i{z}^8=\frac{\sqrt{3}}{2}-\frac{i}{2}$

$\therefore (1+iz+z^5+i{z}^8{)}^9$
$={(1-\frac{1}{2}+\frac{\sqrt{3}i}{2}-\frac{\sqrt{3}}{2}+\frac{i}{2}+\frac{\sqrt{3}}{2}-\frac{i}{2})}^9$
$={(\frac{1}{2}+\frac{\sqrt{3}i}{2})}^9$
$=(e^{i\pi /3}{)}^9$
$=e^{i3\pi }=-1$