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(D) Given $x = \sqrt{-1 - \sqrt{-1 - \sqrt{-1 - ... \infty}}}$.Squaring both sides, we get $x^2 = -1 - \sqrt{-1 - \sqrt{-1 - ... \infty}}$.Since the e

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$\omega^2$

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(D) Given $x = \sqrt{-1 - \sqrt{-1 - \sqrt{-1 - ... \infty}}}$.Squaring both sides, we get $x^2 = -1 - \sqrt{-1 - \sqrt{-1 - ... \infty}}$.Since the expression repeats, we can write $x^2 = -1 - x$.Rearranging the terms, we get $x^2 + x + 1 = 0$.We know that the roots of the equation $x^2 + x + 1 = 0$ are the non-real cube roots of unity, which are $\omega$ and $\omega^2$.However, for the expression $x = \sqrt{-1 - x}$, $x$ must satisfy the condition that the value under the square root is non-negative in the real sense, but here we are dealing with complex numbers.Given $x^2 + x + 1 = 0$, we know $\omega^2 + \omega + 1 = 0$. Thus, $x$ can be $\omega$ or $\omega^2$.By testing the convergence or the specific structure, the value is $\omega^2$.

If $x = \sqrt{-1 - \sqrt{-1 - \sqrt{-1 - ... \infty}}}$, where $\omega$ is a non-real complex cube root of unity, then the value of $x$ is...

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