Complex Numbers
Rotation – De Moivre's Theorem
Complex Numbers_PYQ
Grade 11
Question:
Let $z = \left(\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}\right)^5 + \left(\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}\right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$, then
$R(z) > 0$ and $I(z) > 0$
$I(z) = 0$
$R(z) < 0$ and $I(z) > 0$
$R(z) = -3$
Step-by-Step Solution
Key Concept: The two bases are complex conjugates, so $z^n + \bar{z}^n = 2\,\text{Re}(z^n)$ is always real. De Moivre gives $\cos(5\pi/6) = -\sqrt{3}/2$, doubling to $-\sqrt{3}$.
**Step 1: Write bases in polar form**
$\dfrac{\sqrt{3}}{2}+\dfrac{i}{2} = e^{i\pi/6}$ and $\dfrac{\sqrt{3}}{2}-\dfrac{i}{2} = e^{-i\pi/6}$.
**Step 2: Apply De Moivre's theorem**
$z = e^{5i\pi/6}+e^{-5i\pi/6} = 2\cos\dfrac{5\pi}{6} = 2\cdot\left(-\dfrac{\sqrt{3}}{2}\right) = -\sqrt{3}$.
**Step 3: Read off real and imaginary parts**
$z = -\sqrt{3} \in \mathbb{R}$, so $R(z)=-\sqrt{3}<0$ and $I(z)=0$. Option (2) is correct.
Correct Answer: 2