Question

The $\sqrt{(-1-\sqrt{(1-\sqrt{-1-....\infty }}}$ equals:
A. $1$
B. $\omega$ or ${{\omega }^{2}}$
C. $\omega$
D. ${{\omega }^{2}}$

Answer 1

First give a variable to the required root such as $z$. Now $z$ is a complex number. Solve the equation for $z$ by taking squares on both sides and simplifying it. On simplifying it we get the value of $z$ which is the answer we require.

Complete step by step answer:
Now if we let $z$ be
$z=\sqrt{(-1-\sqrt{(1-\sqrt{-1-....\infty }}}$
Now since the equation continues we can notice that it starts being the same after a while which we can write as
$z=\sqrt{(-1-z)}$
To solve this further we can take square on both sides of the equation that is
${{z}^{2}}=-1-z$
Taking the whole equation on one side
${{z}^{2}}+z+1=0$
Now solving this we get that z has two values which are
$z=\dfrac{-1+\sqrt{3i}}{2}$ or $z=\dfrac{-1-\sqrt{3i}}{2}$
Therefore;
$\therefore z=\omega$ or $z={{\omega }^{2}}$

Hence the correct answer is option B.

Note: A complex number is a number that can be divided in the form of a+bi where a stands for the real part of any complex number and b stands for the imaginary part of the complex number. “i” is a symbol called an imaginary unit. It satisfies the equation that square if i is equal to $-1$. If a is equal to$0$ then the complex number is completely imaginary and if b is $0$ then the complex number is completely real. To explain what $\omega$ is we can basically explain it as the fact that it is the root of the equation ${{x}^{2}}+x+1=0$. We get the value of it by using Sridharacharya’s formula for solving a quadratic equation. Knowing the value of $\omega$ makes it very easier to solve multiple other questions for complex numbers.