Question

If \(z=\frac{\sqrt{3}+i}{2}\), then the value of \(z^{69}\) is

• A. \(-i\)
• B. \(i\)
• C. \(-1\)
• D. \(1\)

Hint:

Write z in polar form with modulus 1 and argument 30 degrees, then apply De Moivre's theorem.

Answer 1

The Correct Option is A

Step 1: Powers of z
$z^{12}=\cos 2\pi+i\sin 2\pi=1$.

Step 2: Reduce the Exponent
$69=5\times12+9$, so $z^{69}=z^{9}$.

Step 3: Evaluate
$z^{9}=\cos\dfrac{3\pi}{2}+i\sin\dfrac{3\pi}{2}=-i$, option (A).

Final Answer:
Option (A), $-i$. \[ \boxed{\text{(A) }-i} \]