If `alpha(1+isqrt3)/(2),` then what is the value of `1+alpha^(8)+alpha^(16)+alpha^(24)+alpha^(32)` ?

Options

A. 0
B. 1
C. `omega`
D. `-omega^(2)`

Explanation

To solve the problem, we start with the given expression for \( \alpha \):

\[ \alpha = \frac{1 + i\sqrt{3}}{2} \]

### Step 1: Rewrite \( \alpha \)
We can express \( \alpha \) in a different form. Notice that:

\[ \alpha = \frac{1}{2} + \frac{i\sqrt{3}}{2} \]

This can be recognized as a complex number in polar form. Specifically, it can be expressed in terms of the cube roots of unity.

### Step 2: Identify \( \alpha \) as a cube root of unity
The cube roots of unity are \( 1, \omega, \omega^2 \), where \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \) and \( \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \).

We can see that:

\[ \alpha = -\omega^2 \]

### Step 3: Calculate \( 1 + \alpha^8 + \alpha^{16} + \alpha^{24} + \alpha^{32} \)
Now, we need to evaluate:

\[ 1 + \alpha^8 + \alpha^{16} + \alpha^{24} + \alpha^{32} \]

Substituting \( \alpha = -\omega^2 \):

\[ 1 + (-\omega^2)^8 + (-\omega^2)^{16} + (-\omega^2)^{24} + (-\omega^2)^{32} \]

### Step 4: Simplify the powers of \( \alpha \)
Calculating the powers:

- \( (-\omega^2)^8 = \omega^{16} \)
- \( (-\omega^2)^{16} = \omega^{32} \)
- \( (-\omega^2)^{24} = \omega^{48} \)
- \( (-\omega^2)^{32} = \omega^{64} \)

### Step 5: Reduce the powers modulo 3
Since \( \omega^3 = 1 \), we can reduce the powers modulo 3:

- \( \omega^{16} \equiv \omega^{1} \) (since \( 16 \mod 3 = 1 \))
- \( \omega^{32} \equiv \omega^{2} \) (since \( 32 \mod 3 = 2 \))
- \( \omega^{48} \equiv \omega^{0} = 1 \) (since \( 48 \mod 3 = 0 \))
- \( \omega^{64} \equiv \omega^{1} \) (since \( 64 \mod 3 = 1 \))

### Step 6: Substitute back into the expression
Now substituting back, we have:

\[ 1 + \omega + \omega^2 + 1 + \omega \]

Combine like terms:

\[ 2 + 2\omega + \omega^2 \]

### Step 7: Use the property of cube roots of unity
We know that:

\[ 1 + \omega + \omega^2 = 0 \]

Thus, we can replace \( \omega + \omega^2 \) with \(-1\):

\[ 2 + 2\omega + \omega^2 = 2 + 2\omega - 1 = 1 + 2\omega \]

### Step 8: Final calculation
Since \( 1 + \omega + \omega^2 = 0 \), we have:

\[ 1 + 2\omega = 1 - 1 - \omega^2 = -\omega^2 \]

### Final Answer
The value of \( 1 + \alpha^8 + \alpha^{16} + \alpha^{24} + \alpha^{32} \) is:

\[ -\omega^2 \]