Key Concepts and Formulas

• Roots of Unity: Numbers that produce 1 when raised to some integer power (i.e., solutions to $x^n = 1$ for some integer $n$).
• Modular Arithmetic: $a \equiv b \pmod{m}$ means $a$ and $b$ have the same remainder when divided by $m$. If $a \equiv b \pmod{m}$, then $x^a \equiv x^b \pmod{m}$.
• Algebraic Manipulation: Multiplying by a strategic factor can simplify polynomial equations.

Step-by-Step Solution

Step 1: Find a Key Property of $\alpha$

We are given that $\alpha$ is a root of $1 + x^2 + x^4 = 0$. This means $1 + \alpha^2 + \alpha^4 = 0.$ To find a useful property of $\alpha$, we multiply both sides of the equation by $(\alpha^2 - 1)$: $(\alpha^2 - 1)(1 + \alpha^2 + \alpha^4) = (\alpha^2 - 1) \cdot 0$ The left side simplifies to a difference of squares/cubes: $\alpha^6 - 1 = 0$ Thus, we have $\alpha^6 = 1.$ This tells us that $\alpha$ is a 6th root of unity. We need to confirm that $\alpha^2 \neq 1$, otherwise $\alpha^2 - 1 = 0$, which would mean we multiplied by zero. If $\alpha^2 = 1$, then substituting into the original equation yields $1 + 1 + 1 = 3 = 0$, which is false. Thus, $\alpha^2 \neq 1$.

Why this step is important: Finding the relationship $\alpha^6 = 1$ allows us to simplify higher powers of $\alpha$ using modular arithmetic.

Step 2: Simplify the Exponents Using $\alpha^6 = 1$

We want to simplify the expression $\alpha^{1011} + \alpha^{2022} - \alpha^{3033}$. We can simplify the exponents by finding their remainders when divided by 6.

• Simplifying $\alpha^{1011}$: We need to find $1011 \pmod 6$. Since $1011 = 6 \cdot 168 + 3$, we have $1011 \equiv 3 \pmod 6$. Therefore, $\alpha^{1011} = \alpha^3.$

• Simplifying $\alpha^{2022}$: We need to find $2022 \pmod 6$. Since $2022 = 6 \cdot 337 + 0$, we have $2022 \equiv 0 \pmod 6$. Therefore, $\alpha^{2022} = \alpha^0 = 1.$

• Simplifying $\alpha^{3033}$: We need to find $3033 \pmod 6$. Since $3033 = 6 \cdot 505 + 3$, we have $3033 \equiv 3 \pmod 6$. Therefore, $\alpha^{3033} = \alpha^3.$

Why this step is important: Reducing the exponents modulo 6 allows us to work with smaller, more manageable powers of $\alpha$.

Step 3: Evaluate the Expression

Now we substitute these simplified powers back into the original expression: $\alpha^{1011} + \alpha^{2022} - \alpha^{3033} = \alpha^3 + 1 - \alpha^3.$ The $\alpha^3$ terms cancel, so we are left with: $\alpha^3 + 1 - \alpha^3 = 1.$

Why this step is important: This step shows how the simplifications lead to a straightforward calculation of the final result.

Common Mistakes & Tips

• Check Divisibility by 6: A number is divisible by 6 if it's divisible by both 2 and 3. This can speed up finding remainders.
• Verify the Multiplier: Always check that multiplying by a factor doesn't introduce extraneous solutions or invalidate the original equation.
• Roots of Unity: Recognize that equations of the form $1 + x^2 + x^4 = 0$ are closely related to roots of unity.

Summary

Given that $\alpha$ is a root of $1 + x^2 + x^4 = 0$, we found that $\alpha^6 = 1$. Using this property, we reduced the exponents in the expression $\alpha^{1011} + \alpha^{2022} - \alpha^{3033}$ modulo 6, resulting in $\alpha^3 + 1 - \alpha^3$. This simplifies to 1.

The final answer is \boxed{1}, which corresponds to option (A).