If $z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2}$ ($i = \sqrt{-1}$), then $(1 + iz + z^5 + iz^8)^9$ is equal to
Step-by-Step Solution
Key Concept: The four terms combine to the unit complex number $e^{i\pi/3}$, whose 9th power is $e^{3i\pi}=-1$. Polar form is essential.
**Step 1: Write z in polar form**
$z = e^{i\pi/6}$, so $iz = e^{i2\pi/3}$, $z^5 = e^{5i\pi/6}$, $iz^8 = e^{11i\pi/6}$.
**Step 2: Sum the four terms**
Real parts: $1 + (-\tfrac{1}{2}) + (-\tfrac{\sqrt{3}}{2}) + \tfrac{\sqrt{3}}{2} = \dfrac{1}{2}$. Imaginary parts: $0 + \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} - \dfrac{1}{2} = \dfrac{\sqrt{3}}{2}$. Sum $= e^{i\pi/3}$.
**Step 3: Raise to the 9th power**
$(e^{i\pi/3})^9 = e^{3i\pi} = (-1)^3 = -1$.
Correct Answer: 3