1. Key Concepts and Formulas

• Euler's Formula: $e^{i\theta} = \cos\theta + i\sin\theta$
• De Moivre's Theorem: $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$ or $(re^{i\theta})^n = r^n e^{in\theta}$
• Complex Conjugate: If $z = a + bi$, then $\bar{z} = a - bi$. If $|z| = 1$, then $\bar{z} = \frac{1}{z}$.

2. Step-by-Step Solution

Step 1: Convert the complex number $z$ to polar form.

We are given $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$. We need to find $r$ and $\theta$ such that $z = re^{i\theta} = r(\cos\theta + i\sin\theta)$.

• Calculate the modulus $r$: $r = |z| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1$. The modulus is the distance from the origin to the complex number in the complex plane.

• Calculate the argument $\theta$: Since both the real and imaginary parts of $z$ are positive, $z$ lies in the first quadrant. $\theta = \arg(z) = \arctan\left(\frac{1/2}{\sqrt{3}/2}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$. The argument is the angle between the positive real axis and the line connecting the origin to the complex number in the complex plane.

• Express $z$ in polar form: Therefore, $z = 1\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = e^{i\frac{\pi}{6}}$. This exponential form simplifies calculations involving powers of complex numbers.

Step 2: Express $i$ as a power of $z$.

We know that $i = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = e^{i\frac{\pi}{2}}$. We want to find an integer $n$ such that $z^n = i$. $z^n = \left(e^{i\frac{\pi}{6}}\right)^n = e^{i\frac{n\pi}{6}} = e^{i\frac{\pi}{2}}$. Equating the exponents, we get $\frac{n\pi}{6} = \frac{\pi}{2}$, which implies $n = 3$. Therefore, $i = z^3$. Recognizing this relationship will simplify the given expression.

Step 3: Substitute $i = z^3$ into the given expression and simplify.

The given expression is $(1 + iz + z^5 + iz^8)^9$. Substituting $i = z^3$, we have: $(1 + z^3z + z^5 + z^3z^8)^9 = (1 + z^4 + z^5 + z^{11})^9$.

Step 4: Evaluate the terms inside the parentheses in rectangular form.

Since $z = e^{i\frac{\pi}{6}}$, we have: $z^4 = e^{i\frac{4\pi}{6}} = e^{i\frac{2\pi}{3}} = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. $z^5 = e^{i\frac{5\pi}{6}} = \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + i\frac{1}{2}$. $z^{11} = e^{i\frac{11\pi}{6}} = \cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6} = \frac{\sqrt{3}}{2} - i\frac{1}{2}$.

Step 5: Substitute the rectangular forms back into the expression and simplify.

$1 + z^4 + z^5 + z^{11} = 1 + \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) + \left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2} - i\frac{1}{2}\right) = 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + i\frac{1}{2} + \frac{\sqrt{3}}{2} - i\frac{1}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.

Step 6: Convert the simplified expression back to polar form.

$\frac{1}{2} + i\frac{\sqrt{3}}{2} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = e^{i\frac{\pi}{3}}$.

Step 7: Raise the expression to the power of 9.

$\left(e^{i\frac{\pi}{3}}\right)^9 = e^{i\frac{9\pi}{3}} = e^{i3\pi} = \cos(3\pi) + i\sin(3\pi) = -1 + i(0) = -1$. Thus, $(1 + iz + z^5 + iz^8)^9 = -1$.

3. Common Mistakes & Tips

• Incorrect argument: Ensure you are in the correct quadrant when finding the argument of a complex number.
• Forgetting De Moivre's Theorem: Remember to raise both the modulus and apply the exponent to the argument when raising a complex number in polar form to a power.
• Sign Errors: Be careful with signs, especially when substituting and simplifying complex numbers.

4. Summary

By converting the complex number $z$ to polar form, expressing $i$ as a power of $z$, and simplifying the expression using De Moivre's Theorem, we found that $(1 + iz + z^5 + iz^8)^9 = -1$.

5. Final Answer

The final answer is \boxed{-1}, which corresponds to option (B).