Question:

Step 1: Express $\omega$ in terms of $z$ and identify its properties. Given the relationship $2\omega + 1 = z$ and $z = \sqrt{-3}$, we can substitute the value of $z$: $2\omega + 1 = \sqrt{-3}$ $2\omega + 1 = i\sqrt{3}$ Now, isolate $\omega$: $2\omega = -1 + i\sqrt{3}$ $\omega = \frac{-1 + i\sqrt{3}}{2}$ This value of $\omega$ is one of the complex cube roots of unity, specifically $e^{2\pi i/3}$. Therefore, it satisfies the following fundamental properties: $\omega^3 = 1$ $1 + \omega + \omega^2 = 0$ Step 2: Simplify the entries of the given determinant using the properties of $\omega$. We use the properties $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$ to simplify the entries of the determinant. The entry $\omega^7$ simplifies to: $\omega^7 = \omega^{3 \cdot 2 + 1} = (\omega^3)^2 \cdot \omega = (1)^2 \cdot \omega = \omega$ The entry $-\omega^2 - 1$ simplifies using $1 + \omega + \omega^2 = 0 \implies - \omega^2 - 1 = \omega$: $-\omega^2 - 1 = \omega$ Substitute these simplified expressions into the given determinant: $$ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -\omega^2-1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix} $$ Step 3: Calculate the value of the determinant. Let the simplified determinant be $D$. We expand the determinant along the first row: $$ D = 1 \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & \omega \end{vmatrix} - 1 \begin{vmatrix} 1 & \omega^2 \\ 1 & \omega \end{vmatrix} + 1 \begin{vmatrix} 1 & \omega \\ 1 & \omega^2 \end{vmatrix} $$ Now, calculate each $2 \times 2$ determinant: $$ D = (\omega \cdot \omega - \omega^2 \cdot \omega^2) - (1 \cdot \omega - \omega^2 \cdot 1) + (1 \cdot \omega^2 - \omega \cdot 1) $$ $$ D = (\omega^2 - \omega^4) - (\omega - \omega^2) + (\omega^2 - \omega) $$ Using the property $\omega^3 = 1$, we know that $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$. Substitute this into the expression for $D$: $$ D = (\omega^2 - \omega) - (\omega - \omega^2) + (\omega^2 - \omega) $$ Combine the like terms: $$ D = \omega^2 - \omega - \omega + \omega^2 + \omega^2 - \omega $$ $$ D = 3\omega^2 - 3\omega $$ Factor out 3: $$ D = 3(\omega^2 - \omega) $$ Step 4: Determine the value of $k$ and relate it to $z$. We are given that the determinant is equal to $3k$. From Step 3, we found the determinant $D = 3(\omega^2 - \omega)$. Therefore, we have: $3k = 3(\omega^2 - \omega)$ Divide by 3: $k = \omega^2 - \omega$ From the property of cube roots of unity, $1 + \omega + \omega^2 = 0$. We can express $\omega^2$ as $-(1 + \omega)$. Substitute this into the expression for $k$: $k = -(1 + \omega) - \omega$ $k = -1 - \omega - \omega$ $k = -1 - 2\omega$ Factor out $-1$: $k = -(1 + 2\omega)$ From Step 1, we were given the relationship $2\omega + 1 = z$. Substituting this into the expression for $k$: $k = -z$ The final answer is $k = -z$. The correct option is (2).