Question:medium

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...

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Put \(x=\sqrt{-1-x}\), square it, and get \(x^2+x+1=0\).
Updated On: Oct 1, 2026
  • \(1\)
  • \(-1\)
  • \(-ω\)
  • \(ω^2\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Plan:
Test each option in the equation that the nested radical must satisfy.

Step 2: Equation:
Let the value be $x$. Then $x^2 = -1 - x$, i.e. $x^2 + x + 1 = 0$. Multiplying by $(x-1)$ gives $x^3 = 1$, so $x$ is a cube root of unity other than $1$.

Step 3: Pick from the list:
The value $1$ is excluded since it is the real cube root. $-1$ cubes to $-1$. $-\omega$ cubes to $-1$. Only $\omega^2$ cubes to $1$ and is non-real. So the value is $\omega^2$.

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