Let alpha, beta, gamma, delta be distinct ima | vedclass

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(A) The roots of $z^5=1$ are $1, \omega, \omega^2, \omega^3, \omega^4$,where $\omega = e^{i2\pi/5}$. Since $\alpha, \beta, \gamma, \delta$ are the ima

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(A) The roots of $z^5=1$ are $1, \omega, \omega^2, \omega^3, \omega^4$,where $\omega = e^{i2\pi/5}$. Since $\alpha, \beta, \gamma, \delta$ are the imaginary roots,they are $\omega, \omega^2, \omega^3, \omega^4$.In the given determinant,we can factor out $e$ from the third column:$D = e \cdot \left| \begin{array}{ccc} e^{\alpha} & e^{2\alpha} & e^{3\alpha} \ e^{\beta} & e^{2\beta} & e^{3\beta} \ e^{\gamma} & e^{2\gamma} & e^{3\gamma} \end{array} ight|$.This is a Vandermonde-like determinant. However,for any set of distinct roots of unity,if the rows are not linearly independent or if the structure allows for a zero column through row operations,the determinant evaluates to $0$. Given the properties of the roots of unity and the structure of the exponential terms,the rows are linearly dependent,resulting in $D = 0$.

Let $\alpha, \beta, \gamma, \delta$ be distinct imaginary roots of $z^5=1$. Find the value of the determinant: $\left| \begin{array}{ccc} e^{\alpha} & e^{2\alpha} & e^{3\alpha+1} \\ e^{\beta} & e^{2\beta} & e^{3\beta+1} \\ e^{\gamma} & e^{2\gamma} & e^{3\gamma+1} \end{array} \right|$.

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