Complex Numbers
De Moivre's Theorem
Complex Numbers_PYQ
Grade 11
Question:
If $z=\left(\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}\right)^5+\left(\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}\right)^5$, then
$\text{Re}(z)=0$
$\text{Im}(z)=0$
$\text{Re}(z)>0,\;\text{Im}(z)>0$
$\text{Re}(z)>0,\;\text{Im}(z)<0$
Step-by-Step Solution
Key Concept: The two bases are complex conjugates, so their sum $z^n+\bar{z}^n=2\,\text{Re}(z^n)$ is always real. De Moivre gives $\cos(5\pi/6)=-\sqrt{3}/2$.
**Step 1: Write 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 the result**
$z=-\sqrt{3}\in\mathbb{R}$, so $\text{Re}(z)=-\sqrt{3}<0$ and $\text{Im}(z)=0$. Option (b) is correct.
Correct Answer: 2