Question

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

• A. \(1\)
• B. \(-1\)
• C. \(-ω\)
• D. \(ω^2\)

Hint:

Put \(x=\sqrt{-1-x}\), square it, and get \(x^2+x+1=0\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The radical repeats itself inside, so the whole expression equals \(x\). That means \(x = \sqrt{-1 - x}\).

Step 2: Form the equation:
Square both sides: \(x^2 = -1 - x\), so
\[ x^2 + x + 1 = 0 \]

Step 3: Link to cube roots of unity:
The non-real cube roots of unity satisfy \(1 + \omega + \omega^2 = 0\), and also \(\omega^2 + \omega + 1 = 0\). So the roots of \(x^2+x+1=0\) are \(x = \omega\) and \(x = \omega^2\).

Step 4: Match with the options:
Option A (\(1\)): \(1+1+1 = 3 \ne 0\). Option B (\(-1\)): \(1-1+1 = 1 \ne 0\). Option C (\(-\omega\)): \(\omega^2 - \omega + 1 = -2\omega \ne 0\). Option D (\(\omega^2\)): \(\omega^4 + \omega^2 + 1 = \omega + \omega^2 + 1 = 0\). It satisfies the equation.

Final Answer:
The value of \(x\) is \(\omega^2\), option (D). \[ \boxed{\omega^2} \]