If `z=-2 + 2 sqrt3i` then `z^(2n) + z^(n) + 2^(4n)` may be equal to

Options

A. `2^(2n)`
B. `0`
C. `3. 2^(4n)`
D. None of these

Answer

To solve the problem, we need to find the value of \( z^{2n} + z^n + 2^{4n} \) where \( z = -2 + 2\sqrt{3}i \).

### Step 1: Express \( z \) in polar form
Given \( z = -2 + 2\sqrt{3}i \), we can express it in polar form.

1. Calculate the modulus \( |z| \):
\[ |z| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \]

2. Calculate the argument \( \theta \):
\[ \tan(\theta) = \frac{2\sqrt{3}}{-2} = -\sqrt{3} \implies \theta = \frac{2\pi}{3} \text{ (since it's in the second quadrant)} \]

Thus, we can write:
\[ z = 4 \left( \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) \right) = 4 e^{i \frac{2\pi}{3}} \]

### Step 2: Calculate \( z^{2n} \) and \( z^n \)
Using the polar form:
\[ z^{2n} = (4 e^{i \frac{2\pi}{3}})^{2n} = 4^{2n} e^{i \frac{4\pi n}{3}} = 16^n e^{i \frac{4\pi n}{3}} \]
\[ z^n = (4 e^{i \frac{2\pi}{3}})^{n} = 4^n e^{i \frac{2\pi n}{3}} = 16^{n/2} e^{i \frac{2\pi n}{3}} \]

### Step 3: Substitute into the expression
Now substitute \( z^{2n} \), \( z^n \), and \( 2^{4n} \) into the expression:
\[ z^{2n} + z^n + 2^{4n} = 16^n e^{i \frac{4\pi n}{3}} + 4^n e^{i \frac{2\pi n}{3}} + 2^{4n} \]

### Step 4: Factor out \( 2^{4n} \)
Notice that \( 16^n = (2^4)^n = 2^{4n} \) and \( 4^n = (2^2)^n = 2^{2n} \):
\[ = 2^{4n} \left( e^{i \frac{4\pi n}{3}} + e^{i \frac{2\pi n}{3}} + 1 \right) \]

### Step 5: Evaluate the expression inside the parentheses
The expression \( e^{i \frac{4\pi n}{3}} + e^{i \frac{2\pi n}{3}} + 1 \) can be simplified using the roots of unity:
- The three terms represent the cube roots of unity, which sum to zero.

Thus, we have:
\[ e^{i \frac{4\pi n}{3}} + e^{i \frac{2\pi n}{3}} + 1 = 0 \]

### Final Step: Conclusion
Substituting back, we find:
\[ z^{2n} + z^n + 2^{4n} = 2^{4n} \cdot 0 = 0 \]

### Answer
The expression \( z^{2n} + z^n + 2^{4n} \) may be equal to **0**.