Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Given: ${{z}_{1}}+{{z}_{2}}+{{z}_{3}}=A;{{z}_{1}}+{{z}_{2}}w+{{z}_{3}}{{w}^{2}}=B;{{z}_{1}}+{{z}_{2}}{{w}^{2}}+{{z}_{3}}w=C$ , where ‘w’ is the cube root of unity. Prove: ${{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}}=3\left\{ {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}} \right\}$ .

Last updated date: 20th Jun 2024
Total views: 413.7k
Views today: 11.13k
Verified
413.7k+ views
Hint: At first take the absolute value of the square of A, B, C using the formula ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca.$ Now add them up and use the fact ${{w}^{3}}=1,\left| w \right|=1,1+w+{{w}^{2}}=0$ to get the desired result.

In the question given that $A={{z}_{1}}+{{z}_{2}}+{{z}_{3}},B={{z}_{1}}+{{z}_{2}}w+{{z}_{3}}{{w}^{2}}$ and $C={{z}_{1}}+{{z}_{2}}{{w}^{2}}+{{z}_{3}}w$ also ‘w’ is the cube root of 1, so we can write ${{w}^{3}}=1$ .
Now we will use formula,
${{\left( a+b+c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \right)..........(i)$
At first, we will prove it by,
${{\left( a+b+c \right)}^{2}}=\left( a+b+c \right)\left( a+b+c \right)$
Now multiplying it, we will get,
$a\left( a+b+c \right)+b\left( a+b+c \right)+c\left( a+b+c \right)$
$\Rightarrow {{a}^{2}}+ab+ac+{{b}^{2}}+ab+bc+ac+bc+{{c}^{2}}$
Now, adding and rearranging, we will get,
$\Rightarrow {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$
So, now using it we will find the value of ${{\left| A \right|}^{2}}$ ,
${{\left| A \right|}^{2}}={{\left( {{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)}^{2}}$
Now using the formula from equation (i), we get
${{\left| A \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}+2{{z}_{1}}{{z}_{2}}+2{{z}_{2}}{{z}_{3}}+2{{z}_{3}}{{z}_{1}}.........(ii)$
Now using it we will find the value of ${{\left| B \right|}^{2}}$ ,
${{\left| B \right|}^{2}}={{\left( {{z}_{1}}+{{z}_{2}}w+{{z}_{3}}{{w}^{2}} \right)}^{2}}$
Now using the formula from equation (i), we get
${{\left| B \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}{{\left| w \right|}^{4}}+2{{z}_{1}}{{z}_{2}}w+2{{z}_{2}}{{z}_{3}}{{w}^{3}}+2{{z}_{1}}{{z}_{3}}{{w}^{2}}$
Now using the fact ${{w}^{3}}=1$ and then rearranging the terms, the above equation can be written as,
${{\left| B \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}\left| w \right|+2{{z}_{1}}{{z}_{2}}w+2{{z}_{2}}{{z}_{3}}+2{{z}_{1}}{{z}_{3}}{{w}^{2}}...........(iii)$
Now using it we will find the value of ${{\left| C \right|}^{2}}$ ,
${{\left| C \right|}^{2}}={{\left( {{z}_{1}}+{{z}_{2}}{{w}^{2}}+{{z}_{3}}w \right)}^{2}}$
Now using the formula from equation (i), we get
${{\left| C \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{4}}+{{\left| {{z}_{3}} \right|}^{2}}{{\left| w \right|}^{2}}+2{{z}_{1}}{{z}_{2}}{{w}^{2}}+2{{z}_{2}}{{z}_{3}}{{w}^{3}}+2{{z}_{1}}{{z}_{3}}w$
Now using the fact ${{w}^{3}}=1$ and then rearranging the terms, the above equation can be written as,
${{\left| C \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}\left| w \right|+{{\left| {{z}_{3}} \right|}^{2}}\left| w \right|+2{{z}_{1}}{{z}_{2}}{{w}^{2}}+2{{z}_{1}}{{z}_{3}}w+2{{z}_{2}}{{z}_{3}}.......(iv)$
Now we will be adding ${{\left| A \right|}^{2}},{{\left| B \right|}^{2}},{{\left| C \right|}^{2}}$ we get,
${{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}}$
Substituting values from equation (ii), (iii) and (iv), we get
\begin{align} & ={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}+2{{z}_{1}}{{z}_{2}}+2{{z}_{2}}{{z}_{3}}+2{{z}_{3}}{{z}_{1}}+{{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}{{\left| w \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}}\left| w \right|+2{{z}_{1}}{{z}_{2}}w+2{{z}_{2}}{{z}_{3}}+2{{z}_{1}}{{z}_{3}}{{w}^{2}} \\ & +{{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}\left| w \right|+{{\left| {{z}_{3}} \right|}^{2}}{{\left| w \right|}^{2}}+2{{z}_{1}}{{z}_{2}}{{w}^{2}}+2{{z}_{1}}{{z}_{3}}w+2{{z}_{2}}{{z}_{3}} \\ \end{align}
Now by rearranging and using the fact which is $\left| w \right|=1$ we get,
${{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}}=3\left\{ {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}} \right\}+2{{z}_{1}}{{z}_{2}}\left( 1+w+{{w}^{2}} \right)+2{{z}_{1}}{{z}_{3}}\left( 1+w+{{w}^{2}} \right)+2{{z}_{3}}{{z}_{1}}\left( 1+w+{{w}^{2}} \right)$So, now using fact $\left( 1+w+{{w}^{2}} \right)=0$ we get that,
${{\left| A \right|}^{2}}+{{\left| B \right|}^{2}}+{{\left| C \right|}^{2}}=3\left\{ {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+{{\left| {{z}_{3}} \right|}^{2}} \right\}$
Hence proved

Note: Students while breaking the square should be cautious as there are high chances of making mistakes of leaving any term. While adding they should also keep the fact such as ${{w}^{3}}=1,{{w}^{2}}+w+1=0$ . Students often make mistakes when considering the fact $\left( 1+w+{{w}^{2}} \right)=0$ .