Question

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

Answer 1

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 ${{i}^{2}}=-1$.

Complete step by step answer:
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.
$\begin{aligned} & \Rightarrow {{\omega }^{2}}={{\left( \cos \left( \dfrac{2\pi }{7} \right)+i\sin \left( \dfrac{2\pi }{7} \right) \right)}^{2}} \\\ & \Rightarrow {{\omega }^{2}}={{\cos }^{2}}\left( \dfrac{2\pi }{7} \right)+{{\left( i \right)}^{2}}{{\sin }^{2}}\left( \dfrac{2\pi }{7} \right)+2i\cos \left( \dfrac{2\pi }{7} \right)\sin \left( \dfrac{2\pi }{7} \right) \\\ & \Rightarrow {{\omega }^{2}}={{\cos }^{2}}\left( \dfrac{2\pi }{7} \right)-{{\sin }^{2}}\left( \dfrac{2\pi }{7} \right)+i\left( 2\sin \left( \dfrac{2\pi }{7} \right)\cos \left( \dfrac{2\pi }{7} \right) \right) \\\ \end{aligned}$
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 $~\omega$.

& \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{aligned}$$ But from trigonometry, we know that $\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$ and we also that $\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$. We will use these two properties in the above equation. $$\begin{aligned} & \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{aligned}$$ From (1) and (2), we can safely say that $${{\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$$ $\begin{aligned} & \Rightarrow \alpha +\beta =\dfrac{\left( 1-\omega \right)}{\omega -1} \\\ & \Rightarrow \alpha +\beta =\dfrac{-1\left( \omega -1 \right)}{\omega -1} \\\ & \Rightarrow \alpha +\beta =-1 \\\ \end{aligned}$ **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 $.