Question

If \( z = \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^5 + \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)^5 \), then

• A. \( \operatorname{Re}(z) > 0, \operatorname{Im}(z) < 0 \)
• B. \( \operatorname{Im}(z) = 0 \)
• C. \( \operatorname{Re}(z) = 0 \)
• D. \( \operatorname{Re}(z) > 0, \operatorname{Im}(z) > 0 \)

Hint:

When conjugate complex numbers are added after equal powers, their imaginary parts cancel and the result is real.

Answer 1

The Correct Option is B

Step 1: Recognize trigonometric form.
We know:
\[ \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6} \]
and
\[ \frac{1}{2} = \sin \frac{\pi}{6} \]

Step 2: Write both complex numbers in polar form.
\[ \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos \frac{\pi}{6} + i\sin \frac{\pi}{6} \]
\[ \frac{\sqrt{3}}{2} - \frac{i}{2} = \cos \frac{\pi}{6} - i\sin \frac{\pi}{6} \]

Step 3: Apply De Moivre's theorem.
\[ \left(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}\right)^5 = \cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6} \]
\[ \left(\cos \frac{\pi}{6} - i\sin \frac{\pi}{6}\right)^5 = \cos \frac{5\pi}{6} - i\sin \frac{5\pi}{6} \]

Step 4: Add both expressions.
\[ z = \left(\cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6}\right) + \left(\cos \frac{5\pi}{6} - i\sin \frac{5\pi}{6}\right) \]

Step 5: Cancel imaginary parts.
\[ +i\sin \frac{5\pi}{6} - i\sin \frac{5\pi}{6} = 0 \]
So:
\[ z = 2\cos \frac{5\pi}{6} \]

Step 6: Evaluate real value.
\[ \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \]
\[ z = 2\left(-\frac{\sqrt{3}}{2}\right) \]
\[ z = -\sqrt{3} \]

Step 7: Final conclusion.
Since \( z = -\sqrt{3} \), it is purely real. Therefore:
\[ \operatorname{Im}(z) = 0 \]
\[ \boxed{\operatorname{Im}(z)=0} \]