If \(\alpha = \frac{-1+\sqrt{-3}}{2}\) then what is the value of \((1 + \alpha^{19} – \alpha^{35})^{100} – (1 – 3\alpha^{25} + \alpha^{38})^{50}\)?
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:
\((1 + \alpha^{19} – \alpha^{35}) = (1 + \alpha – \alpha^2)\)Now, use the property \(1 + \alpha + \alpha^2 = 0\). From this, we can write \(1 + \alpha = -\alpha^2\). Substituting this into the expression:
\((1 + \alpha – \alpha^2) = (-\alpha^2 – \alpha^2) = -2\alpha^2\)Now, we need to raise this to the power of 100:
\((-2\alpha^2)^{100} = (-2)^{100} (\alpha^2)^{100} = 2^{100} \alpha^{200}\)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:
\(2^{100} \alpha^2\)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:
\((1 – 3\alpha^{25} + \alpha^{38}) = (1 – 3\alpha + \alpha^2)\)Use the property \(1 + \alpha + \alpha^2 = 0\). We can rearrange this as \(1 + \alpha^2 = -\alpha\). Substituting this into the expression:
\((1 – 3\alpha + \alpha^2) = (1 + \alpha^2) - 3\alpha = (-\alpha) - 3\alpha = -4\alpha\)Now, we need to raise this to the power of 50:
\((-4\alpha)^{50} = (-4)^{50} \alpha^{50} = 4^{50} \alpha^{50}\)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:
\(2^{100} \alpha^2\)Calculating the Final Value
Now, subtract the simplified second term from the simplified first term:
\((1 + \alpha^{19} – \alpha^{35})^{100} – (1 – 3\alpha^{25} + \alpha^{38})^{50} = (2^{100} \alpha^2) - (2^{100} \alpha^2)\) \(= 0\)The value of the expression is 0.
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