Question

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

Answer 1

Hint: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.