Question

Let $\alpha ,\beta$ denote the cube roots of unity other than 1. Let $s=\sum\limits_{n=0}^{2}{{{\left( -1 \right)}^{n}}{{\left( \dfrac{\alpha }{\beta } \right)}^{n}}}$ .Then the value of s is
(a) either $-2\omega$ or $-2{{\omega }^{2}}$
(b) either $-2\omega$ or $2{{\omega }^{2}}$
(c) either $2\omega$ or $-2{{\omega }^{2}}$
(d) either $2\omega$ or $2{{\omega }^{2}}$

Answer 1

Hint: Cube roots of infinity are given as $1,\omega ,{{\omega }^{2}}$ .Take two cases by supposing value of $\alpha$ and $\beta$ as $\omega$ and ${{\omega }^{2}}$ and another case by supposing $\alpha$ and $\beta$ as ${{\omega }^{2}}$ and $\omega$ respectively. Use the relation among the cube roots of unity, given as
${{\omega }^{3}}=1$ and $1+\omega +{{\omega }^{2}}=0$
Simplify the given expression to get the value of it.

Complete step-by-step answer:

As we know the cube roots of unity are given as $1,\omega ,{{\omega }^{2}};$ where $\omega$ and ${{\omega }^{2}}$ are given as
$\omega =\dfrac{-1+\sqrt{3}i}{2},{{\omega }^{2}}=\dfrac{-1-\sqrt{3}i}{2}$
Now it is given that $\alpha ,\beta$ is denoting cube roots of unity other than ‘1’. So, $\alpha$ and $\beta$ can take values $\omega$ and ${{\omega }^{2}}$ or ${{\omega }^{2}}$ and $\omega$ .The relation among the cube roots of unity is given as

& {{\omega }^{3}}=1.......\left( i \right) \\\ & 1+\omega +{{\omega }^{2}}=0.........\left( ii \right) \\\ \end{aligned}$$ So, there are two possibility of $\left( \alpha ,\beta \right)$ i.e. $\omega $ and ${{\omega }^{2}}$ or ${{\omega }^{2}}$ and $\omega $ . Hence, we need to observe two cases here. Case 1: $\alpha =\omega $ and $\beta ={{\omega }^{2}}$ So, we get value of s as $s=\sum\limits_{n=0}^{2}{{{\left( -1 \right)}^{n}}{{\left( \dfrac{\alpha }{\beta } \right)}^{n}}}$ Now, we can write s as $$\begin{aligned} & s={{\left( -1 \right)}^{0}}{{\left( \dfrac{\alpha }{\beta } \right)}^{0}}+{{\left( -1 \right)}^{1}}{{\left( \dfrac{\alpha }{\beta } \right)}^{1}}+{{\left( -1 \right)}^{2}}{{\left( \dfrac{\alpha }{\beta } \right)}^{2}} \\\ & s=1-\dfrac{\alpha }{\beta }+{{\left( \dfrac{\alpha }{\beta } \right)}^{0}} \\\ \end{aligned}$$ Now, we can put values of $\alpha $ and $\beta $ to the above equation, we get $\begin{aligned} & s=1-\dfrac{\omega }{{{\omega }^{2}}}+{{\left( \dfrac{\omega }{{{\omega }^{2}}} \right)}^{2}} \\\ & s=1-\dfrac{1}{\omega }+{{\left( \dfrac{1}{\omega } \right)}^{2}} \\\ & s=1-\dfrac{1}{\omega }+\dfrac{1}{{{\omega }^{2}}}.............\left( iii \right) \\\ \end{aligned}$ Now, we can replace 1 by ${{\omega }^{3}}$ to the above equation using the relation of the equation(i). So, we get $\begin{aligned} & s=1-\dfrac{{{\omega }^{3}}}{\omega }+\dfrac{{{\omega }^{3}}}{{{\omega }^{2}}} \\\ & s=1-{{\omega }^{2}}+\omega \\\ \end{aligned}$ Now, we can add and subtract ${{\omega }^{2}}$ to the above equation. Hence, we get $\begin{aligned} & s=1-{{\omega }^{2}}+\omega +{{\omega }^{2}}-{{\omega }^{2}} \\\ & s=1+\omega +{{\omega }^{2}}-2{{\omega }^{2}} \\\ \end{aligned}$ Now, we can replace $1+\omega +{{\omega }^{2}}$ by 0 from the relation given in equation(ii). So, we get $\begin{aligned} & s=0-2{{\omega }^{2}} \\\ & s=-2{{\omega }^{2}}........\left( iv \right) \\\ \end{aligned}$ Case 2: $\alpha ={{\omega }^{2}}$ and $\beta =\omega $ Now, we can get value of s as $s=1-\left( \dfrac{\alpha }{\beta } \right)+{{\left( \dfrac{\alpha }{\beta } \right)}^{2}}$ Putting the values of $\alpha $ and $\beta $ in the above equation, we get $$\begin{aligned} & s=1-\left( \dfrac{{{\omega }^{2}}}{\omega } \right)+{{\left( \dfrac{{{\omega }^{2}}}{\omega } \right)}^{2}} \\\ & s=1-\omega +{{\left( \omega \right)}^{2}}=1-\omega +{{\omega }^{2}} \\\ \end{aligned}$$ Now, we can add and subtract $'\omega '$ to the above equation. We get $\begin{aligned} & s=1-\omega +{{\omega }^{2}}-\omega +\omega \\\ & s=1+\omega +{{\omega }^{2}}-2\omega \\\ \end{aligned}$ Now, we can use the relation of equation (ii) and hence can replace $1+\omega +{{\omega }^{2}}$ by 0 in the above equation. So, we get $\begin{aligned} & s=0-2\omega =-2\omega \\\ & s=-2\omega \\\ \end{aligned}$ Hence, two positive values of the given expression ‘s’ are $-2\omega $ and $-2{{\omega }^{2}}$ . So, option(a) is the correct answer. Note: One may put the direct values of $\alpha $ and $\beta $ to the given expression i.e. value of $\alpha $ and $\beta $as $\dfrac{-1-\sqrt{3}i}{2}$ and $\dfrac{-1+\sqrt{3}i}{2}$ or vice-versa as well. But it will make the expression complex and may not be easy for some of the students. Answer will remain same by this approach as well, but try to always use $\omega $ and ${{\omega }^{2}}$ not their values, because we know the relations between them i.e. ${{\omega }^{3}}=1$ and $1+\omega +{{\omega }^{2}}=0$. So, take care of it, with these kinds of questions. Don’t miss any of the cases, otherwise you will not get multiple possible values of the given expression.