Question

What is the value of $\sqrt{-1-\sqrt{-1-\sqrt{-1...}}}\infty$?

• A. $1$
• B. $\omega or{\omega }^2$
• C. $\omega$
• D. ${\omega }^2$

Answer 1

The correct option is B

$\omega or{\omega }^2$

According to the question, this is a repetitive root problem

In this question, we assume the repetitive part as some variable other than the first part

Assume $\chi$ $=$$\sqrt{-1-\sqrt{-1-\sqrt{-1...}}}\infty$

or we can say that, $\chi =\sqrt{-1-\chi }$

On squaring both sides, ${\left(\chi \right)}^2=\left(-1-\chi \right)$

We get a quadratic equation, ${\chi }^2+\chi +1=0$

On solving this equation we get,

$\chi =\frac{-1\pm \sqrt{1-4}}{2}$

$=\frac{-1\pm \sqrt{-3}}{2}$

$\frac{-1+3\mathfrak{i}}{2}or\frac{-1-3\mathfrak{i}}{2}$ $\left(\sqrt{-1}=\mathfrak{i}\right)$2459

$=\omega or{\omega }^2$ $\left[\mathbf{\because }{\mathit{\omega }}^{\mathbf{2}}\mathbf{}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{3}\mathbf{i}}{\mathbf{2}}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{\omega }\mathbf{=}\mathbf{}\frac{\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{3}\mathbf{i}}{\mathbf{2}}\right]$

Hence, option $\left(B\right)$ is the correct answer