What is \(\sqrt {\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}} \) equal to, where ω is the cube root of unity?
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The question asks us to evaluate a complex expression involving \(\omega\), where \(\omega\) is a cube root of unity. We are given the expression \(\sqrt {\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}} \) and need to find its value among the given options.
A cube root of unity is a complex number \(z\) such that \(z^3 = 1\). The three cube roots of unity are 1, \(\omega\), and \(\omega^2\), where \(\omega\) is a non-real cube root of unity. These roots have specific properties that are crucial for solving this problem.
For the non-real cube root of unity, \(\omega\), the following properties hold true:
From the sum property, we can derive useful relations:
We need to simplify the expression \(\sqrt {\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}} \).
Let's first simplify the fraction inside the square root using the properties we just listed.
The fraction is \(\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}\).
Using the property \(1 + \omega^2 = -\omega\), we replace the numerator:
Numerator = \(-\omega\)
Using the property \(1 + \omega = -\omega^2\), we replace the denominator:
Denominator = \(-\omega^2\)
Now, substitute these into the fraction:
The fraction becomes \(\frac{{ - \omega }}{{ - \omega^2}}\)
Simplify the fraction:
\(\frac{{ - \omega }}{{ - \omega^2}} = \frac{\omega}{{\omega^2}}\)
We can simplify \(\frac{\omega}{{\omega^2}}\) using the property \(\omega^3 = 1\). From \(\omega^3 = 1\), we can divide both sides by \(\omega\) (since \(\omega \ne 0\)) to get \(\omega^2 = \frac{1}{\omega}\). Similarly, dividing by \(\omega^2\) gives \(\omega = \frac{1}{\omega^2}\).
So, \(\frac{\omega}{{\omega^2}} = \frac{1}{\omega}\). And since \(\frac{1}{\omega} = \omega^2\), the simplified fraction is \(\omega^2\).
Alternatively, \(\frac{\omega}{{\omega^2}} = \omega^{1-2} = \omega^{-1}\). Since \(\omega^3=1\), \(\omega^{-1} = \omega^{-1} \cdot \omega^3 = \omega^{-1+3} = \omega^2\).
So, the fraction \(\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}} \) simplifies to \(\omega^2\).
Now, we need to find the square root of this simplified fraction:
\(\sqrt {\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}} = \sqrt {{\omega ^2}} \)
The expression \(\sqrt{{\omega^2}}\) represents the complex numbers whose square is equal to \(\omega^2\). The numbers that satisfy this condition are \(\omega\) and \(-\omega\).
We are given four options for the value of the expression:
Comparing the possible values (\(\omega\) and \(-\omega\)) with the given options, we see that \(\omega\) is one of the options.
Therefore, the value of the expression is \(\omega\).
| Property | Application | Result |
|---|---|---|
| \(1 + \omega + \omega^2 = 0\) | Derive \(1 + \omega^2\) and \(1 + \omega\) | \(1 + \omega^2 = -\omega\) \(1 + \omega = -\omega^2\) |
| Substitution in fraction | \(\frac{{1 + {{\rm{\omega }}^2}}}{{1 + {\rm{\omega }}}}\) | \(\frac{{ - \omega }}{{ - \omega^2}}\) |
| Simplification of fraction | \(\frac{{ - \omega }}{{ - \omega^2}} = \frac{\omega}{{\omega^2}}\) | \(\frac{1}{\omega}\) |
| \(\omega^3 = 1\) | Substitute for \(\frac{1}{\omega}\) | \(\frac{1}{\omega} = \omega^2\) |
| Expression simplified | \(\sqrt{\omega^2}\) | \(\omega\) or \(-\omega\) |
The cube roots of unity are the solutions to the equation \(z^3 = 1\). In the complex plane, these roots are located on the unit circle at angles \(0\), \(2\pi/3\), and \(4\pi/3\) radians from the positive real axis.
Notice that \(\omega^2\) is also the complex conjugate of \(\omega\), denoted as \(\bar{\omega}\).
The properties \(1 + \omega + \omega^2 = 0\) and \(\omega^3 = 1\) are fundamental and frequently used in problems involving cube roots of unity in complex numbers.
Suppose ω 1and ω 2are two distinct cube roots of unity different from 1. Then what is (ω 1– ω 2) 2equal to?
If 1, ω, ω 2are the cube roots of unity, then the value of (1 + ω) (1 + ω 2) (1 + ω 4) (1 + ω 8) is
(x 3– 1) can be factorized as
Where ω is one of the cube roots of unity.
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