Question

The value of $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{3}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{3}$ is equal to

• A. $3\sqrt{3}$
• B. $2\sqrt{3}$
• C. $2$
• D. $0$
• E. $-1$

Hint:

Logic Tip: Notice that $z$ and $\bar{z}$ are complex conjugates. The sum of any complex number and its conjugate ($z^n + \bar{z}^n$) will always result in a purely real number, as the imaginary components will cancel out.

Answer 1

The Correct Option is D

Concept:
De Moivre's Theorem states that for a complex number in polar form $z = r(\cos \theta + i\sin \theta)$, its $n$-th power is given by $z^n = r^n(\cos n\theta + i\sin n\theta)$.
Step 1: Convert the base complex numbers to polar form.
[cite_start]Let $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$[cite: 98]. The modulus $r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1$. The argument $\theta = \tan^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^{\circ} = \frac{\pi}{6}$. Thus, $z = \cos \frac{\pi}{6} + i\sin \frac{\pi}{6}$. The second term is the complex conjugate $\bar{z} = \frac{\sqrt{3}}{2} - \frac{i}{2} = \cos \left(-\frac{\pi}{6}\right) + i\sin \left(-\frac{\pi}{6}\right)$.
Step 2: Apply De Moivre's Theorem.
For the first term: $$z^3 = \left(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}\right)^3 = \cos\left(3 \cdot \frac{\pi}{6}\right) + i\sin\left(3 \cdot \frac{\pi}{6}\right)$$ $$z^3 = \cos \frac{\pi}{2} + i\sin \frac{\pi}{2} = 0 + i(1) = i$$ For the second term: $$\bar{z}^3 = \left(\cos \left(-\frac{\pi}{6}\right) + i\sin \left(-\frac{\pi}{6}\right)\right)^3 = \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)$$ $$\bar{z}^3 = 0 + i(-1) = -i$$
Step 3: Add the simplified terms together.
$$z^3 + \bar{z}^3 = i + (-i) = 0$$ [cite_start]This matches Option D[cite: 104, 107].