Question

If $\omega \left( \ne 1 \right)$ is a cube root of unity and ${{\left( 1+\omega \right)}^{7}}=A+B\omega$, then A & B are respectively the numbers
(a). 0, 1
(b). 1, 1
(c). 1, 0
(d). -1, 1

Answer 1

Hint: As $\omega$ is cube root of unity so use the property of it to find the value ${{\left( 1+\omega \right)}^{7}}$ by using fact that $1+\omega +{{\omega }^{2}}=0$ or $\left( 1+\omega \right)=-{{\omega }^{2}}$. So, ${{\left( 1+\omega \right)}^{7}}$ is ${{\left( -{{\omega }^{2}} \right)}^{7}}$ or $-{{\omega }^{14}}$. After that multiply it by ${{\omega }^{15}}$ or 1 as the value will not be changed because ${{\omega }^{3}}$ is 1. Hence compare to get find values of A and B.

Complete step-by-step answer:

In the question we are given that wise cube root of unity and also if ${{\left( 1+\omega \right)}^{7}}$ value is represented by $A+B\omega$ then we have to find the value of A & B respectively.
In order to find the cube roots of unity we need to factorize the following cubic equation:
\Rightarrow $$$${{x}^{3}}-1=0
Now we know that, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.
So, we can re – write the above equation as,
$\Rightarrow \left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0$
The given above equation gives one cube root or value of x as 1 while we have to find other two from the equation,
$\Rightarrow {{x}^{2}}+x+1$
Now we can solve it using formula,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
If the given quadratic equation is $a{{x}^{2}}+bx+c=0$.
Here the given is ${{x}^{2}}+x+1$. So, the value of a, b, c is 1.
So, $x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\times 1\times 2}}{2\times 1}$
Or, $x=\dfrac{-1\pm \sqrt{-3}}{2}$
Now as we know that value of $\sqrt{-1}$ is i.
So, the value of x is $\dfrac{-1\pm \sqrt{3i}}{2}$.
If one of the found values of x other than 1this considered as $\omega$ then other will be ${{\omega }^{2}}$.
As $\omega$ is the root of the quadratic equation ${{x}^{2}}+x+1$. So, if we substitute x as $\omega$ we get,
$\Rightarrow {{\omega }^{2}}+\omega +1=0$
Now in the given question it was given that ${{\left( 1+\omega \right)}^{7}}$. So, as we know that ${{\omega }^{2}}+\omega +1=0$.
$\Rightarrow {{\omega }^{2}}+\omega +1=0$
$\Rightarrow \omega +1=-{{\omega }^{2}}$
So, the value of $\omega +1$ can be written as $-{{\omega }^{2}}$.
Hence ${{\left( 1+\omega \right)}^{7}}$ can be written as ${{\left( -{{\omega }^{2}} \right)}^{7}}$ or $-{{\omega }^{2\times 7}}$ or $-{{\omega }^{14}}$.
Now as we know that ${{\omega }^{3}}=1$ so if we multiply $-{{\omega }^{14}}$ by ${{\omega }^{3}}$ as 5 is a multiple of 3, therefore the answer will not change.
So, ${{\omega }^{-14}}\times {{\omega }^{15}}$ is equal to $\omega$.
So, the value of ${{\left( 1+\omega \right)}^{7}}$ is $\omega$.
Now we were given that, ${{\left( 1+\omega \right)}^{7}}=A+B\omega$ and the value of ${{\left( 1+\omega \right)}^{7}}$ is $\omega$. So,
$\omega =A+B\omega$.
Now on comparing we can say that the value of A is 0 and B is 1.
Hence the correct option is (a).

Note: Here $\omega$ is a complex number which is considered as the cube root of unity which is generally used to find the values of higher powers of $\omega$. The value of $\omega$ is $\dfrac{-1\pm \sqrt{3i}}{2}$ as it is said that the value of $\dfrac{-1\pm \sqrt{3i}}{2}$ is 1.