Interpret the given complex number

We have $z=\frac{\sqrt{3}}{2}+\frac{i}{2}.$

This is a complex number on the unit circle: $z=\cos\theta+i\sin\theta$ with $\cos\theta=\frac{\sqrt{3}}{2},\quad \sin\theta=\frac12.$ Hence, $\theta=\frac{\pi}{6}.$

So, $z=\operatorname{cis}\left(\frac{\pi}{6}\right).$

Find the required powers of $z$

Using De Moivre's theorem, $z^n=\operatorname{cis}\left(n\frac{\pi}{6}\right).$

Therefore, $z^5=\operatorname{cis}\left(\frac{5\pi}{6}\right)=\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}=-\frac{\sqrt3}{2}+\frac{i}{2}.$

Also, $z^8=\operatorname{cis}\left(\frac{8\pi}{6}\right)=\operatorname{cis}\left(\frac{4\pi}{3}\right)=-\frac12-\frac{\sqrt3}{2}i.$

Compute $iz$ and $iz^8$

First, $iz=i\left(\frac{\sqrt3}{2}+\frac{i}{2}\right)=\frac{\sqrt3}{2}i-\frac12=-\frac12+\frac{\sqrt3}{2}i.$

Next, $iz^8=i\left(-\frac12-\frac{\sqrt3}{2}i\right)=-\frac{i}{2}+\frac{\sqrt3}{2}=\frac{\sqrt3}{2}-\frac{i}{2}.$

Add the terms inside the bracket

We need $1+iz+z^5+iz^8.$

Substitute all values: $1+\left(-\frac12+\frac{\sqrt3}{2}i\right)+\left(-\frac{\sqrt3}{2}+\frac{i}{2}\right)+\left(\frac{\sqrt3}{2}-\frac{i}{2}\right).$

Combine real parts: $1-\frac12-\frac{\sqrt3}{2}+\frac{\sqrt3}{2}=\frac12.$

Combine imaginary parts: $\frac{\sqrt3}{2}i+\frac12 i-\frac12 i=\frac{\sqrt3}{2}i.$

Thus, $1+iz+z^5+iz^8=\frac12+\frac{\sqrt3}{2}i=z.$

Raise to the 9th power

The expression is $(1+iz+z^5+iz^8)^9=z^9.$

Since $z=\operatorname{cis}\left(\frac{\pi}{6}\right),$ we get $z^9=\operatorname{cis}\left(\frac{9\pi}{6}\right)=\operatorname{cis}\left(\frac{3\pi}{2}\right)=-i.$

Compare with the options

We obtained $(1+iz+z^5+iz^8)^9=-i.$

But the options are:

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

None of these equals $-i$.

Likely issue in the printed question

If the intended expression were $(1+iz+z^5+iz^8)^6,$ then $z^6=\operatorname{cis}(\pi)=-1,$ which matches option B.

So the stored answer $\text{B}$ appears inconsistent with the given exponent $9$.