Q64 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics

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

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

Solution

Let $z_1 = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2} = \cos\left(\dfrac{\pi}{6}\right) + i\sin\left(\dfrac{\pi}{6}\right) = e^{i\pi/6}$.

Let $z_2 = \dfrac{\sqrt{3}}{2} - \dfrac{i}{2} = \cos\left(-\dfrac{\pi}{6}\right) + i\sin\left(-\dfrac{\pi}{6}\right) = e^{-i\pi/6}$.

The given expression is $z = z_1^5 + z_2^5$.

Using De Moivre's Theorem, $z_1^5 = \cos\left(\dfrac{5\pi}{6}\right) + i\sin\left(\dfrac{5\pi}{6}\right)$ and $z_2^5 = \cos\left(-\dfrac{5\pi}{6}\right) + i\sin\left(-\dfrac{5\pi}{6}\right)$.

Since $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$, we have $z_2^5 = \cos\left(\dfrac{5\pi}{6}\right) - i\sin\left(\dfrac{5\pi}{6}\right)$.

Adding these, $z = 2\cos\left(\dfrac{5\pi}{6}\right) + i\left(\sin\left(\dfrac{5\pi}{6}\right) - \sin\left(\dfrac{5\pi}{6}\right)\right) = 2\cos\left(\dfrac{5\pi}{6}\right)$.

Since $\cos\left(\dfrac{5\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}$, we get $z = 2\left(-\dfrac{\sqrt{3}}{2}\right) = -\sqrt{3}$.

Thus, $\operatorname{Re}(z) = -\sqrt{3}$ and $\operatorname{Im}(z) = 0$.

Answer: $\operatorname{Im}(z)=0$

Hence, the correct answer is option: 'B'.

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