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. \(-2^n\)

Hint:

When conjugate complex numbers are raised to power and added, imaginary parts cancel, leaving \(2^{n+1}\cos(n\theta)\).

Answer 1

The Correct Option is C

Step 1: Convert complex numbers to polar form.
\[ 1+\sqrt{3}i = 2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right) \]
\[ 1-\sqrt{3}i = 2\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right) =2\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right) \]
Step 2: Apply De Moivre’s theorem.
\[ (1+\sqrt{3}i)^n=2^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right) \]
\[ (1-\sqrt{3}i)^n=2^n\left(\cos\frac{n\pi}{3}-i\sin\frac{n\pi}{3}\right) \]
Step 3: Add the two expressions.
\[ (1+\sqrt{3}i)^n+(1-\sqrt{3}i)^n =2^n\left(2\cos\frac{n\pi}{3}\right) =2^{n+1}\cos\frac{n\pi}{3} \]
Step 4: Use condition \(n\equiv 1\pmod{3}\).
So \(n=3k+1\). Then:
\[ \cos\frac{n\pi}{3}=\cos\left(k\pi+\frac{\pi}{3}\right) =\cos(k\pi)\cos\frac{\pi}{3}-\sin(k\pi)\sin\frac{\pi}{3} \]
\[ =\cos(k\pi)\cdot\frac{1}{2} =\frac{(-1)^k}{2} \]
Thus:
\[ 2^{n+1}\cos\frac{n\pi}{3}=2^{n+1}\cdot\frac{(-1)^k}{2} =2^n(-1)^k \]
Since \(n=3k+1\Rightarrow (-1)^k = (-1)^n\).
So:
\[ =2^n(-1)^n = (-2)^n \]
Final Answer:
\[ \boxed{(-2)^n} \]