Evaluating a Complex Number Expression with Powers of Alpha

The question asks for the value of the expression $(1 + \alpha^{19} – \alpha^{35})^{100} – (1 – 3\alpha^{25} + \alpha^{38})^{50}$, where $\alpha = \frac{-1+\sqrt{-3}}{2}$.

Understanding Alpha and its Properties

The given value of $\alpha$ is one of the complex cube roots of unity. The complex cube roots of unity are $1$, $\omega$, and $\omega^2$. In this case, $\alpha$ represents $\omega$. The key properties of these roots are:

• $ \alpha^3 = 1 $
• $ 1 + \alpha + \alpha^2 = 0 $

These properties are crucial for simplifying powers of $\alpha$. We can use the property $ \alpha^3 = 1 $ to reduce any power of $\alpha$ to one of $ \alpha^0 $, $ \alpha^1 $, or $ \alpha^2 $ by looking at the remainder of the exponent when divided by 3.

Simplifying the First Term: $ (1 + \alpha^{19} – \alpha^{35})^{100} $

Let's simplify the powers of $\alpha$ inside the first parenthesis:

• $ \alpha^{19} $: Divide 19 by 3. $ 19 = 3 \times 6 + 1 $. So, $ \alpha^{19} = \alpha^{3 \times 6 + 1} = (\alpha^3)^6 \cdot \alpha^1 = 1^6 \cdot \alpha = \alpha $.
• $ \alpha^{35} $: Divide 35 by 3. $ 35 = 3 \times 11 + 2 $. So, $ \alpha^{35} = \alpha^{3 \times 11 + 2} = (\alpha^3)^{11} \cdot \alpha^2 = 1^{11} \cdot \alpha^2 = \alpha^2 $.

Substitute these simplified powers back into the expression:

Now, use the property $ 1 + \alpha + \alpha^2 = 0 $. From this, we can write $ 1 + \alpha = -\alpha^2 $. Substituting this into the expression:

Now, we need to raise this to the power of 100:

Simplify $ \alpha^{200} $. Divide 200 by 3. $ 200 = 3 \times 66 + 2 $. So, $ \alpha^{200} = \alpha^{3 \times 66 + 2} = (\alpha^3)^{66} \cdot \alpha^2 = 1^{66} \cdot \alpha^2 = \alpha^2 $.

Therefore, the first term simplifies to:

Simplifying the Second Term: $ (1 – 3\alpha^{25} + \alpha^{38})^{50} $

Let's simplify the powers of $\alpha$ inside the second parenthesis:

• $ \alpha^{25} $: Divide 25 by 3. $ 25 = 3 \times 8 + 1 $. So, $ \alpha^{25} = \alpha^{3 \times 8 + 1} = (\alpha^3)^8 \cdot \alpha^1 = 1^8 \cdot \alpha = \alpha $.
• $ \alpha^{38} $: Divide 38 by 3. $ 38 = 3 \times 12 + 2 $. So, $ \alpha^{38} = \alpha^{3 \times 12 + 2} = (\alpha^3)^{12} \cdot \alpha^2 = 1^{12} \cdot \alpha^2 = \alpha^2 $.

Substitute these simplified powers back into the expression:

Use the property $ 1 + \alpha + \alpha^2 = 0 $. We can rearrange this as $ 1 + \alpha^2 = -\alpha $. Substituting this into the expression:

Now, we need to raise this to the power of 50:

We know that $ 4^{50} = (2^2)^{50} = 2^{100} $. Now simplify $ \alpha^{50} $. Divide 50 by 3. $ 50 = 3 \times 16 + 2 $. So, $ \alpha^{50} = \alpha^{3 \times 16 + 2} = (\alpha^3)^{16} \cdot \alpha^2 = 1^{16} \cdot \alpha^2 = \alpha^2 $.

Therefore, the second term simplifies to:

Calculating the Final Value

Now, subtract the simplified second term from the simplified first term:

The value of the expression is 0.