Question

If $\omega $ is the cube root of the unity, then prove that $1+\omega +{{\omega }^{2}}=0$.

Answer 1

Hint: For solving this first we will find the value of cube roots of unity by solving the equation ${{z}^{3}}=1$ where $z$ is a complex number. After that, we will define $\omega $ , ${{\omega }^{2}}$ and find their values. Then, we will add them and prove the result $1+\omega +{{\omega }^{2}}=0$ easily.

Complete step-by-step solution -
Given:
It is given that, $\omega $ is the cube root of the unity and we have to prove that, $1+\omega +{{\omega }^{2}}=0$ .
Now, before we proceed we should know the result of “DE-MOIVERE’S Theorem”.
DE-MOIVERE’S THEOREM:
Statement: If $n\in Z$ (the set of integers), then ${{\left( \cos \theta +i\sin \theta \right)}^{k}}=\cos \left( k\theta \right)+i\sin \left( k\theta \right)$ .
Now, let $z$ be any complex number such that, ${{z}^{3}}=1$ .
Now, we can write $1=\cos \left( {{0}^{0}} \right)+i\sin \left( {{0}^{0}} \right)$ in the equation ${{z}^{3}}=1$ . Then,
$\begin{align}
  & {{z}^{3}}=1 \\
 & \Rightarrow {{z}^{3}}=\cos \left( {{0}^{0}} \right)+i\sin \left( {{0}^{0}} \right) \\
 & \Rightarrow z={{\left( \cos \left( {{0}^{0}} \right)+i\sin \left( {{0}^{0}} \right) \right)}^{\dfrac{1}{3}}} \\
\end{align}$
Now, there should be three values of $z$ so, we write $\cos \left( {{0}^{0}} \right)+i\sin \left( {{0}^{0}} \right)=\cos \left( 2r\pi \right)+i\sin \left( 2r\pi \right)$ , where $r=0,1,2$ in the above equation. Then,
$\begin{align}
  & z={{\left( \cos \left( {{0}^{0}} \right)+i\sin \left( {{0}^{0}} \right) \right)}^{\dfrac{1}{3}}} \\
 & \Rightarrow z={{\left( \cos \left( 2r\pi \right)+i\sin \left( 2r\pi \right) \right)}^{\dfrac{1}{3}}} \\
\end{align}$
Now, apply “DE-MOIVERE’S Theorem” in the above equation. Then,
$\begin{align}
  & z={{\left( \cos \left( 2r\pi \right)+i\sin \left( 2r\pi \right) \right)}^{\dfrac{1}{3}}} \\
 & \Rightarrow z=\cos \left( \dfrac{2r\pi }{3} \right)+i\sin \left( \dfrac{2r\pi }{3} \right) \\
\end{align}$
Now, as we know that ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ . Then,
$\begin{align}
  & z=\cos \left( \dfrac{2r\pi }{3} \right)+i\sin \left( \dfrac{2r\pi }{3} \right) \\
 & \Rightarrow z={{e}^{i\dfrac{2r\pi }{3}}} \\
\end{align}$
Now, for $r=0$ the value of $z=1$ , for $r=2$ the value of $z={{e}^{i\dfrac{2\pi }{3}}}$ and for $r=2$ the value of $z={{e}^{i\dfrac{4\pi }{3}}}$ . Moreover, we define $\omega ={{e}^{i\dfrac{2\pi }{3}}}$ and ${{\omega }^{2}}={{e}^{i\dfrac{4\pi }{3}}}$ as two non-real cube roots of unity.
Now, we will find the value of $\omega ={{e}^{i\dfrac{2\pi }{3}}}$ and ${{\omega }^{2}} = {{e}^{i\dfrac{4\pi }{3}}}$ with the help of the formula ${{e}^{i\theta }}=\cos \theta +i\sin \theta $. Then,
$\begin{align}
  & \omega ={{e}^{i\dfrac{2\pi }{3}}} \\
 & \Rightarrow \omega =\cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \\
 & \Rightarrow \omega =\cos \left( \pi -\dfrac{\pi }{3} \right)+i\sin \left( \pi -\dfrac{\pi }{3} \right) \\
 & \Rightarrow \omega =-\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3} \\
 & \Rightarrow \omega =-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \\
 & {{\omega }^{2}}={{e}^{i\dfrac{4\pi }{3}}} \\
 & \Rightarrow {{\omega }^{2}}=\cos \left( \dfrac{4\pi }{3} \right)+i\sin \left( \dfrac{4\pi }{3} \right) \\
 & \Rightarrow {{\omega }^{2}}=\cos \left( \pi +\dfrac{\pi }{3} \right)+i\sin \left( \pi +\dfrac{\pi }{3} \right) \\
 & \Rightarrow {{\omega }^{2}}=-\cos \dfrac{\pi }{3}-i\sin \dfrac{\pi }{3} \\
 & \Rightarrow {{\omega }^{2}}=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \\
\end{align}$
Now, from the above results, we got $\omega =-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ and ${{\omega }^{2}}=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$ . Then,
$\begin{align}
  & \omega +{{\omega }^{2}}=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \omega +{{\omega }^{2}}=-\dfrac{1}{2}-\dfrac{1}{2}+i\left( \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2} \right) \\
 & \Rightarrow \omega +{{\omega }^{2}}=-1+i\left( 0 \right) \\
 & \Rightarrow \omega +{{\omega }^{2}}+1=0 \\
\end{align}$
Now, from the above result, we conclude that $1+\omega +{{\omega }^{2}}=0$ .
Thus, if $\omega $ is the cube root of the unity, then $1+\omega +{{\omega }^{2}}=0$ .
Hence, proved.

Note: Here, the student should first understand and then proceed in the right direction to prove the result perfectly. After that, we should apply every fundamental result and theorem precisely without any error. Moreover, we should know that cube roots of the unity form a G.P. with a common ratio $\omega $. And for objective problems, we should remember this result.