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# Let $\omega =\cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right)$ and $\alpha =\omega +{{\omega }^{2}}+{{\omega }^{4}}$ and $\beta ={{\omega }^{3}}+{{\omega }^{5}}+{{\omega }^{6}}$.Find the value of $\alpha +\beta$.(a) 0(b) ─1(c) ─2(d) 1

Last updated date: 20th Jun 2024
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Hint: To solve this question, we will first try to find the value of $~{{\omega }^{2}}$, $~{{\omega }^{3}}$ and so on and try to get a general term for $~{{\omega }^{n}}$ , where n is any integer. If we are able to find such a term, we can add $\alpha$ and $\beta$ directly, because if we add these both, we get a series which is in geometric progression. Then we will use the formula for sum of n terms of geometric series given as ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$, where a is the first term of the geometric series and r is the common difference of the series. It is to be kept in mind that ${{i}^{2}}=-1$.

It is given to us that $\omega =\cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right)$.
To find $~{{\omega }^{2}}$, we shall square both sides of the equation.
\begin{align} & \Rightarrow {{\omega }^{2}}={{\left( \cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right) \right)}^{2}} \\ & \Rightarrow {{\omega }^{2}}={{\cos }^{2}}\left( \dfrac{2\pi }{7} \right)+{{\left( i \right)}^{2}}{{\sin }^{2}}\left( \dfrac{2\pi }{7} \right)+2i\cos \left( \dfrac{2\pi }{7} \right)\sin \left( \dfrac{2\pi }{7} \right) \\ & \Rightarrow {{\omega }^{2}}={{\cos }^{2}}\left( \dfrac{2\pi }{7} \right)-{{\sin }^{2}}\left( \dfrac{2\pi }{7} \right)+i\left( 2\sin \left( \dfrac{2\pi }{7} \right)\cos \left( \dfrac{2\pi }{7} \right) \right) \\ \end{align}
From trigonometry we know that ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta$ and we also know that $2\sin \theta \cos \theta =\sin 2\theta$.
$\Rightarrow {{\omega }^{2}}=\cos \left( 2\left( \dfrac{2\pi }{7} \right) \right)+i\sin \left( 2\left( \dfrac{2\pi }{7} \right) \right)......\left( 1 \right)$
Now to find the value of $~{{\omega }^{3}}$, we shall multiply the value of $~{{\omega }^{2}}$ with $~\omega$.
\begin{align} & \Rightarrow {{\omega }^{2}}.\omega =\left[ \cos \left( 2\left( \dfrac{2\pi }{7} \right) \right)+i\sin \left( 2\left( \dfrac{2\pi }{7} \right) \right) \right]\left[ \cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right) \right] \\ & \Rightarrow {{\omega }^{3}}=\cos \left( 2\left( \dfrac{2\pi }{7} \right) \right)\cos \left( \dfrac{2\pi }{7} \right)+i\cos \left( 2\left( \dfrac{2\pi }{7} \right) \right)\sin \left( \dfrac{2\pi }{7} \right)+i\sin \left( 2\left( \dfrac{2\pi }{7} \right) \right)\cos \left( \dfrac{2\pi }{7} \right)+{{\left( i \right)}^{2}}\sin \left( 2\left( \dfrac{2\pi }{7} \right) \right)\sin \left( \dfrac{2\pi }{7} \right) \\ & \Rightarrow {{\omega }^{3}}=\cos \left( 2\left( \dfrac{2\pi }{7} \right) \right)\cos \left( \dfrac{2\pi }{7} \right)-\sin \left( 2\left( \dfrac{2\pi }{7} \right) \right)\sin \left( \dfrac{2\pi }{7} \right)+i\left[ \cos \left( 2\left( \dfrac{2\pi }{7} \right) \right)\sin \left( \dfrac{2\pi }{7} \right)+i\sin \left( 2\left( \dfrac{2\pi }{7} \right) \right)\cos \left( \dfrac{2\pi }{7} \right) \right] \\ \end{align}
But from trigonometry, we know that $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$ and we also that $\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$. We will use these two properties in the above equation.
\begin{align} & \Rightarrow {{\omega }^{3}}=\cos \left( 2\left( \dfrac{2\pi }{7} \right)+\dfrac{2\pi }{7} \right)+i\sin \left( 2\left( \dfrac{2\pi }{7} \right)+\dfrac{2\pi }{7} \right) \\ & \Rightarrow {{\omega }^{3}}=\cos \left( 3\left( \dfrac{2\pi }{7} \right) \right)+i\sin \left( 3\left( \dfrac{2\pi }{7} \right) \right)......\left( 2 \right) \\ \end{align}
From (1) and (2), we can safely say that ${{\omega }^{n}}=\cos \left( n\left( \dfrac{2\pi }{7} \right) \right)+i\sin \left( n\left( \dfrac{2\pi }{7} \right) \right)......\left( 3 \right)$
We are given that $\alpha =\omega +{{\omega }^{2}}+{{\omega }^{4}}$ and $\beta ={{\omega }^{3}}+{{\omega }^{5}}+{{\omega }^{6}}$.
Therefore $\alpha +\beta =\omega +{{\omega }^{2}}+{{\omega }^{3}}+{{\omega }^{4}}+{{\omega }^{5}}+{{\omega }^{6}}$
We can see that this series is in geometric progression with 6 terms, where the first terms as $~\omega$ and common difference is $~\omega$.
We know that the sum of n terms of GP is given as ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$, where a is the first term of the geometric series and r is the common difference of the series.
Hence, $\omega +{{\omega }^{2}}+{{\omega }^{3}}+{{\omega }^{4}}+{{\omega }^{5}}+{{\omega }^{6}}=\dfrac{\omega \left( {{\omega }^{6}}-1 \right)}{\omega -1}$
$\Rightarrow \alpha +\beta =\dfrac{\left( {{\omega }^{7}}-\omega \right)}{\omega -1}$
Now, if we substitute n = 7, ${{\omega }^{7}}=\cos \left( 7\left( \dfrac{2\pi }{7} \right) \right)+i\sin \left( 7\left( \dfrac{2\pi }{7} \right) \right)$
$\Rightarrow {{\omega }^{7}}=1$
\begin{align} & \Rightarrow \alpha +\beta =\dfrac{\left( 1-\omega \right)}{\omega -1} \\ & \Rightarrow \alpha +\beta =\dfrac{-1\left( \omega -1 \right)}{\omega -1} \\ & \Rightarrow \alpha +\beta =-1 \\ \end{align}

So, the correct answer is “Option B”.

Note: In complex numbers, $\cos \theta +i\sin \theta$ is usually represented as ${{e}^{i\theta }}$. Hence, ${{\left( \cos \theta +i\sin \theta \right)}^{2}}={{e}^{2i\theta }}=\cos 2\theta +i\sin 2\theta$. Similarly, if we want to find ${{\left( \cos \theta +i\sin \theta \right)}^{n}}$, it can be directly represented by ${{\left( \cos \theta +i\sin \theta \right)}^{n}}={{e}^{ni\theta }}=\cos n\theta +i\sin n\theta$.