Question

Let $ \omega ={{e}^{i\dfrac{\pi }{3}}} $ and $ a,b,c,x,y,z $ be non-zero complex number such that
\[\begin{align}
  & a+b+c=x \\
 & a+b\omega +c{{\omega }^{2}}=y \\
 & a+b{{\omega }^{2}}+c\omega =z \\
\end{align}\]
Then the value of $ \dfrac{{{\left| x \right|}^{2}}+{{\left| y \right|}^{2}}+{{\left| z \right|}^{2}}}{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}}} $ is;\[\]

Answer 1

Hint: We recall the cube roots of unity $ 1,\omega ={{e}^{i\dfrac{\pi }{3}}},{{\omega }^{2}}={{e}^{i\dfrac{2\pi }{3}}} $ and the relationship between them. We use the property of any complex number $ {{\left| z \right|}^{2}}=z\overline{z} $ to have $ {{\left| x \right|}^{2}}=x\overline{x},{{\left| y \right|}^{2}}=y\overline{y},{{\left| z \right|}^{2}}=z\overline{z} $ and put the expressions for $ x,y,z $ in terms of $ a,b,c $ and then expand. \[\]

Complete step by step answer:
We know that we represent the cube roots of unity as $ 1,\omega =-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}={{e}^{i\dfrac{\pi }{3}}},{{\omega }^{2}}=\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}={{e}^{i\dfrac{2\pi }{3}}} $ . We know that sum of cube roots of unity is zero that is
\[1+\omega +{{\omega }^{2}}=0\]
We also know that the power $ n $ on cube roots of unity returns as follows;
\[{{\omega }^{n}}=\left\{ \begin{matrix}
   1 & \text{if }n\equiv 0\bmod \left( 3 \right) \\
   \omega & \text{if }n\equiv 1\bmod \left( 3 \right) \\
   {{\omega }^{2}} & \text{if }n\equiv 2\bmod \left( 3 \right) \\
\end{matrix} \right.\]
We know that the general form of a complex number is $ z=a+ib $ where $ a\in R $ is called the real part of $ z $ and $ b\in R $ is called the imaginary part of the complex number. We know that the modulus of a complex x number is defined as \[\left| z \right|=\left| a\pm ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}\]. We know the property that $ {{\left| z \right|}^{2}}=z\overline{z} $ . We also know the multiplicative and additive property of conjugate respectively as;
\[\begin{align}
  & \overline{xy}=\overline{x}\cdot \overline{y} \\
 & \overline{x+y+z}=\overline{x}+\overline{y}+\overline{z} \\
\end{align}\]
So we use the above property on the cube roots of unity and have;
\[\overline{\omega }={{\omega }^{2}},\overline{{{\omega }^{2}}}=\omega \]
 We are give in the question the cube root of unity $ \omega ={{e}^{i\dfrac{\pi }{3}}} $ that $ a,b,c,x,y,z $ are non-zero complex numbers such that,
\[\begin{align}
  & a+b+c=x \\
 & a+b\omega +c{{\omega }^{2}}=y \\
 & a+b{{\omega }^{2}}+c\omega =z \\
\end{align}\]
We use the property of $ {{\left| z \right|}^{2}}=z\overline{z} $ to have;
\[{{\left| x \right|}^{2}}=x\overline{x},{{\left| y \right|}^{2}}=y\overline{y},{{\left| z \right|}^{2}}=z\overline{z}\]
So we put the given expressions in $ {{\left| x \right|}^{2}} $ in terms of $ a,b,c $ for the numerators of the expression to be evaluated and have;
 \[{{\left| x \right|}^{2}}=x\overline{x}=\left( a+b+c \right)\left( \overline{a+b+c} \right)\]
We use additive property of conjugate to have;
\[\begin{align}
  & \Rightarrow {{\left| x \right|}^{2}}=\left( a+b+c \right)\left( \overline{a}+\overline{b}+\overline{c} \right) \\
 & \Rightarrow {{\left| x \right|}^{2}}=a\overline{a}+b\overline{b}+c\overline{c}+a\overline{b}+a\overline{c}+\overline{a}b+\overline{b}c+\overline{a}c+a\overline{c} \\
\end{align}\]
We use the property $ {{\left| z \right|}^{2}}=z\overline{z} $ to have;
\[\Rightarrow {{\left| x \right|}^{2}}={{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}}+a\overline{b}+\overline{a}b+\overline{b}c+\overline{b}c+a\overline{c}+\overline{a}c.........\left( 1 \right)\]
Similarly we evaluate $ {{\left| y \right|}^{2}} $ by putting $ y=a+b\omega +c{{\omega }^{2}} $ to have;
\[{{\left| y \right|}^{2}}=y\overline{y}=\left( a+b\omega +c{{\omega }^{2}} \right)\left( \overline{a+b\omega +c{{\omega }^{2}}} \right)\]
We use additive and multiplicative property of conjugate to have;
\[\begin{align}
  & \Rightarrow {{\left| y \right|}^{2}}=\left( a+b\omega +c{{\omega }^{2}} \right)\left( \overline{a}+\overline{b}\overline{\omega }+\overline{c}\overline{{{\omega }^{2}}} \right) \\
 & \Rightarrow {{\left| y \right|}^{2}}=\left( a+b\omega +c{{\omega }^{2}} \right)\left( \overline{a}+\overline{b}{{\omega }^{2}}+\overline{c}\omega \right) \\
 & \Rightarrow {{\left| y \right|}^{2}}=a\overline{a}+a\overline{b}{{\omega }^{2}}+a\overline{c}\omega +\overline{a}b\omega +b\overline{b}{{\omega }^{3}}+b\overline{c}{{\omega }^{2}}+\overline{a}c{{\omega }^{2}}+\overline{b}c{{\omega }^{2}}+c\overline{c}{{\omega }^{3}} \\
\end{align}\]
We put $ {{\omega }^{3}}=1 $ in the above step to have;
\[\Rightarrow {{\left| y \right|}^{2}}={{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}}+a\overline{b}{{\omega }^{2}}+\overline{a}b\omega +b\overline{c}{{\omega }^{2}}+\overline{b}c{{\omega }^{2}}+a\overline{c}\omega +\overline{a}c{{\omega }^{2}}........\left( 2 \right)\]
Similarly we evaluate $ {{\left| z \right|}^{2}} $ by putting $ z=a+b{{\omega }^{2}}+c\omega $ to have;
\[{{\left| z \right|}^{2}}=z\overline{z}=\left( a+b{{\omega }^{2}}+c\omega \right)\left( \overline{a+b{{\omega }^{2}}+c\omega } \right)\]
 We use additive and multiplicative property of conjugate to have;
 \[\begin{align}
  & \Rightarrow {{\left| z \right|}^{2}}=\left( a+b{{\omega }^{2}}+c\omega \right)\left( \overline{a}+\overline{b}\overline{{{\omega }^{2}}}+\overline{c}\overline{\omega } \right) \\
 & \Rightarrow {{\left| z \right|}^{2}}=\left( a+b{{\omega }^{2}}+c\omega \right)\left( \overline{a}+\overline{b}\omega +\overline{c}{{\omega }^{2}} \right) \\
 & \Rightarrow {{\left| z \right|}^{2}}=a\overline{a}+a\overline{b}\omega +a\overline{c}{{\omega }^{2}}+\overline{a}b{{\omega }^{2}}+b\overline{b}{{\omega }^{3}}+b\overline{c}{{\omega }^{4}}+\overline{a}c\omega +\overline{b}c{{\omega }^{2}}+c\overline{c}{{\omega }^{3}} \\
\end{align}\]
We put $ {{\omega }^{3}}=1,{{\omega }^{4}}={{\omega }^{3}}\cdot \omega =1\cdot \omega =\omega $ in the above step to have;
\[\Rightarrow {{\left| z \right|}^{2}}={{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}}+a\overline{b}\omega +\overline{a}b{{\omega }^{2}}+b\overline{c}\omega +\overline{b}c{{\omega }^{2}}+a\overline{c}{{\omega }^{2}}+\overline{a}c\omega .......\left( 3 \right)\]
We add respective sides of equation (1),(2), (3) to have;
\[\begin{align}
  & {{\left| x \right|}^{2}}+{{\left| y \right|}^{2}}+{{\left| z \right|}^{2}}=3\left( {{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}} \right)+a\overline{b}\left( 1+\omega +{{\omega }^{2}} \right)+\overline{a}b\left( 1+\omega +{{\omega }^{2}} \right)+b\overline{c}\left( 1+\omega +{{\omega }^{2}} \right) \\
 & +\overline{b}c\left( 1+\omega +{{\omega }^{2}} \right)+a\overline{c}\left( 1+\omega +{{\omega }^{2}} \right)+\overline{a}c\left( 1+\omega +{{\omega }^{2}} \right) \\
\end{align}\]
We put $ 1+\omega +{{\omega }^{2}}=0 $ in the above step to have;
\[{{\left| x \right|}^{2}}+{{\left| y \right|}^{2}}+{{\left| z \right|}^{2}}=3\left( {{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}} \right)\]
We divide both sides by $ \left( {{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}} \right) $ to have;
\[\dfrac{{{\left| x \right|}^{2}}+{{\left| y \right|}^{2}}+{{\left| z \right|}^{2}}}{{{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}}}=3\]

Note:
 We note that the division by $ \left( {{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}} \right) $ because modulus is always non-negative and since we are given $ a,b,c $ are non-zero complex number $ \left( {{\left| a \right|}^{2}}+{{\left| b \right|}^{2}}+{{\left| c \right|}^{2}} \right) $ cannot be zero. We note that we can also represent any complex number in the form of $ {{e}^{i\theta }}=\cos \theta +i\sin \theta $ . That is why the cue root unity is $ {{e}^{i\dfrac{\pi }{3}}}=\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}=\omega $ . Similarly we can express the $ {{n}^{\text{th}}} $ root of unity as $ {{e}^{i\dfrac{\pi }{n}}} $ .