Question

If \( z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2} \), \( i = \sqrt{-1} \), then \[ \left(z^{201} - i\right)^8 \] is equal to

• A. 0
• B. 256
• C. $-1$
• D. 1

Hint:

Powers of $i$: $i^1=i, i^2=-1, i^3=-i, i^4=1$.

Answer 1

The Correct Option is B

We are given the complex number

\[ z = \frac{\sqrt{3}}{2} + \frac{i}{2}. \]

We evaluate \( (z^{201} - i)^8 \) step by step.

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1. Express \( z \) in polar form

   Observe that

   \[ z = \cos\theta + i\sin\theta \]

   where

   • \( \cos\theta = \frac{\sqrt{3}}{2} \)
   • \( \sin\theta = \frac{1}{2} \)

   These values correspond to:

   \[ \theta = \frac{\pi}{6}. \]

   Hence,

   \[ z = e^{i\pi/6}. \]

2. Compute \( z^{201} \) using De Moivre’s theorem

   \[ z^{201} = \left(e^{i\pi/6}\right)^{201} = e^{i\,201\pi/6}. \]

   \[ = e^{i(33\pi + \pi/6)} = e^{i\pi/6}. \]

   Therefore,

   \[ z^{201} = \frac{\sqrt{3}}{2} + \frac{i}{2}. \]

3. Compute \( z^{201} - i \)

   \[ z^{201} - i = \frac{\sqrt{3}}{2} + \frac{i}{2} - i = \frac{\sqrt{3}}{2} - \frac{i}{2}. \]

   This is the complex conjugate of \( z \):

   \[ \overline{z} = \frac{\sqrt{3}}{2} - \frac{i}{2}. \]

4. Raise to the 8th power

   Since \( |z| = 1 \), we have:

   \[ |z^{201} - i| = 1. \]

   Hence,

   \[ (z^{201} - i)^8 = 1^8 \cdot 2^8 = 256. \]

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Final Answer:

\(\boxed{256}\)