<p>If \(z + z^{-1} = 1\), then find the value of \(z^{100} + z^{-100}\).</p>
Step-by-Step Solution
Key Concept: When z + z⁻¹ = 1, recognize that z satisfies a quadratic equation whose roots are complex cube roots of unity (specifically ω and ω²). Use the periodicity property that z³ = 1 to reduce the exponent 100 modulo 3.
Step 1:
Given the equation $z + z^{-1} = 1$.
Multiply the entire equation by $z$:
$$z^2 + 1 = z$$
Rearranging the terms yields a quadratic equation:
$$z^2 - z + 1 = 0$$
Step 2:
Multiply the equation $z^2 - z + 1 = 0$ by $(z+1)$:
$$(z+1)(z^2 - z + 1) = (z+1) \cdot 0$$
This is the sum of cubes factorization, which simplifies to:
$$z^3 + 1 = 0$$
Therefore, we have:
$$z^3 = -1$$
Step 3:
From $z^3 = -1$, we can determine $z^6$:
$$z^6 = (z^3)^2 = (-1)^2 = 1$$
Step 4:
To find $z^{100}$, we use the property $z^6 = 1$. Divide $100$ by $6$:
$$100 = 16 \times 6 + 4$$
Thus, $z^{100}$ can be written as:
$$z^{100} = (z^6)^{16} \cdot z^4 = (1)^{16} \cdot z^4 = z^4$$
Step 5:
Similarly, to find $z^{-100}$:
$$z^{-100} = (z^6)^{-16} \cdot z^{-4} = (1)^{-16} \cdot z^{-4} = z^{-4}$$
Since $z^6 = 1$, we can multiply $z^{-4}$ by $z^6$ without changing its value:
$$z^{-4} = z^{-4} \cdot z^6 = z^{6-4} = z^2$$
Therefore, $z^{-100} = z^2$.
Step 6:
Substitute the simplified terms for $z^{100}$ and $z^{-100}$ into the expression $z^{100} + z^{-100}$:
$$z^{100} + z^{-100} = z^4 + z^2$$
From Step 2, we know $z^3 = -1$. We can rewrite $z^4$ as $z \cdot z^3$:
$$z^4 = z \cdot (-1) = -z$$
Substitute this back into the expression:
$$z^4 + z^2 = -z + z^2$$
From Step 1, we have the equation $z^2 - z + 1 = 0$. Rearranging this equation gives:
$$z^2 - z = -1$$
Therefore, the value of $z^{100} + z^{-100}$ is:
$$z^{100} + z^{-100} = -1$$
Correct Answer: -1