If `z=-2+2sqrt(3)i,` then `z^(2n)+2^(2n)*z^n+2^(4n)` is equal to

Options

A. 0
B. `-1`
C. 1
D. `3xx 2^(4n)` if n is multiply of 3

Answer

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

### Step 1: Convert \( z \) to Polar Form
First, we need to express \( z \) in polar form. We can write \( z \) as:
\[ z = -2 + 2\sqrt{3}i \]
To convert this to polar form, we calculate the modulus \( r \) and the argument \( \theta \).

The modulus \( r \) is given by:
\[ r = |z| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4 \]

Next, we find the argument \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{2\sqrt{3}}{-2}\right) = \tan^{-1}(-\sqrt{3}) = \frac{2\pi}{3} \quad (\text{since } z \text{ is in the second quadrant}) \]

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

### Step 2: Calculate \( z^{2n} \)
Now, we calculate \( z^{2n} \):
\[ z^{2n} = (4 e^{i \frac{2\pi}{3}})^{2n} = 4^{2n} e^{i \frac{4\pi n}{3}} = 16^n e^{i \frac{4\pi n}{3}} \]

### Step 3: Calculate \( 2^{2n} z^n \)
Next, we calculate \( 2^{2n} z^n \):
\[ z^n = (4 e^{i \frac{2\pi}{3}})^n = 4^n e^{i \frac{2\pi n}{3}} = 2^{2n} e^{i \frac{2\pi n}{3}} \]
Thus,
\[ 2^{2n} z^n = 2^{2n} \cdot 2^{2n} e^{i \frac{2\pi n}{3}} = 4^{n} e^{i \frac{2\pi n}{3}} \]

### Step 4: Calculate \( 2^{4n} \)
Now, we calculate \( 2^{4n} \):
\[ 2^{4n} = 4^{2n} \]

### Step 5: Combine All Parts
Now we can combine all parts:
\[ z^{2n} + 2^{2n} z^n + 2^{4n} = 16^n e^{i \frac{4\pi n}{3}} + 4^n e^{i \frac{2\pi n}{3}} + 4^{n} \]

### Step 6: Factor Out Common Terms
We can factor out \( 4^{n} \):
\[ = 4^{n} \left( 4^n e^{i \frac{4\pi n}{3}} + e^{i \frac{2\pi n}{3}} + 1 \right) \]

### Step 7: Evaluate the Expression
Now, we need to evaluate \( 4^n e^{i \frac{4\pi n}{3}} + e^{i \frac{2\pi n}{3}} + 1 \).

Using the properties of complex exponentials:
- If \( n \) is a multiple of 3, \( e^{i \frac{2\pi n}{3}} = 1 \) and \( e^{i \frac{4\pi n}{3}} = 1 \) as well.

Thus, we have:
\[ 4^n (4^n + 1 + 1) = 4^n (4^n + 2) \]

### Final Result
Thus, the final result is:
\[ z^{2n} + 2^{2n} z^n + 2^{4n} = 3 \cdot 2^{4n} \quad \text{(if \( n \) is a multiple of 3)} \]