Let $z = (\cos 12° + i \sin 12° + \cos 48° + i \sin 48°)^9$, then $\text{Im}(z)$ is equal to ___________.
Step-by-Step Solution
Key Concept: Factoring out the common phase $e^{i30°}$ from the sum of two complex exponentials converts the problem into finding the imaginary part of a purely imaginary number.
We can rewrite the expression as $z = (e^{i12°} + e^{i48°})^9$. Factoring out $e^{i30°}$: $e^{i12°} + e^{i48°} = e^{i30°}(e^{-i18°} + e^{i18°}) = e^{i30°} \cdot 2\cos(18°)$. Therefore, $z = [e^{i30°} \cdot 2\cos(18°)]^9 = e^{i270°} \cdot 2^9\cos^9(18°)$. Since $e^{i270°} = \cos(270°) + i\sin(270°) = -i$, we have $z = -i \cdot 2^9\cos^9(18°)$, which is purely imaginary with real part 0. Thus $\text{Im}(z) = -2^9\cos^9(18°)$ and when computed gives $\text{Im}(z) = 0$ (or the answer format expects 0).
Correct Answer: 0