If
Question:

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

to:

1. (1) 0

2. (2) 1

3. (3) $(-1+2 i)^{9}$

4. (4) $-1$

Correct Option: , 4

Solution:

$\frac{\sqrt{3}}{2}+\frac{i}{2}=-i\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=-i \omega$

where $\omega$ is imaginary cube root of unity.

Now, $\left(1+i z+z^{5}+i z^{8}\right)^{9}$

$=\left(1+\omega-i \omega^{2}+i \omega^{2}\right)^{9}=(1+\omega)^{9}$

$=\left(-\omega^{2}\right)^{9}=-\omega^{18}=-1 \quad\left(\because 1+\omega+\omega^{2}=0\right)$