Question

$\sqrt{-1-\sqrt{-1-\sqrt{-1-\cdots \text{to}\infty }}}$ is equal to :

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

Answer 1

The correct option is A ${\omega }^2$

Let $x=\sqrt{-1-\sqrt{-1-\sqrt{-1-\cdots \text{to}\infty }}}$
Then, $x=\sqrt{-1-x}$ or $x^2=-1-x$
$\Rightarrow x^2+x+1=0$
$\therefore x=\frac{-1\pm \sqrt{1-4\cdot 1\cdot 1}}{2\cdot 1}=\frac{-1\pm \sqrt{-3}}{2}$
$=\frac{-1\pm \sqrt{3}i}{2}=\omega$ or ${\omega }^2$