Question

If \( n \) is an integer which leaves remainder one when divided by three, then \( (1+\sqrt{3}i)^{n} + (1-\sqrt{3}i)^{n} \) equals

• A. $-2^{n+1}$
• B. $2^{n+1}$
• C. $-(-2)^{n}$
• D. $-2$

Hint:

$1+\omega+\omega^2=0$ and $\omega^3=1$.

Answer 1

The Correct Option is C

Step 1: Polar form
Rewrite $1 \pm \sqrt{3}i$ in terms of cube roots of unity: $1+\sqrt{3}i = -2\omega^2$ and $1-\sqrt{3}i = -2\omega$, where $\omega = \frac{-1+\sqrt{3}i}{2}$.
Step 2: Substitution
The expression becomes $(-2\omega^2)^n + (-2\omega)^n = (-2)^n (\omega^{2n} + \omega^n)$.
Step 3: Remainder condition
Since $n = 3r + 1$, then $\omega^n = \omega^{3r+1} = \omega$ and $\omega^{2n} = \omega^{6r+2} = \omega^2$.
Step 4: Final value
Sum $= (-2)^n (\omega^2 + \omega) = (-2)^n (-1) = -(-2)^n$.
Final Answer: (C)