Question

The value of \((\frac{-1+i\sqrt{3}}{2})^{18}+(\frac{-1-i\sqrt{3}}{2})^{18}\) is

• A. \(2\)
• B. \(35\)
• C. \(\sqrt{3}\)
• D. \(\frac{\sqrt{3}}{2}\)

Hint:

The two numbers are the complex cube roots of unity, so use w cubed equals 1.

Answer 1

The Correct Option is A

Step 1: Use polar form:
$\frac{-1+i\sqrt{3}}{2} = \cos 120^{\circ} + i\sin 120^{\circ}$ and the other number is $\cos 120^{\circ} - i\sin 120^{\circ}$.

Step 2: De Moivre:
The 18th power gives $\cos(2160^{\circ}) \pm i\sin(2160^{\circ})$. Since $2160^{\circ} = 6 \times 360^{\circ}$, each is $1$.
So the sum is $1 + 1 = 2$.

Final Answer:
The sum is $2$, option (A). \[ \boxed{2} \]