Question

If ${{z}_{1}},{{z}_{2}}\in C$ , $z_{1}^{2}+z_{2}^{2}\in R$ , ${{z}_{1}}(z_{1}^{2}-3z_{2}^{2})=2$ and ${{z}_{2}}(3z_{1}^{2}-z_{2}^{2})=11$, then the $z_{1}^{2}+z_{2}^{2}=$
$A)5$
$B)6$
$C)7$
$D)8$

Answer 1

Hint : To solve this question we need to have knowledge about complex numbers and quadratic equations. In this question we will apply the algebra of complex numbers to find the value of the unknown. We will firstly multiply the second equation to iota ($i$) making the second equation a complex number. Then we will add and subtract the equation finding the value of $z_{1}^{2}+z_{2}^{2}$.

Complete step-by-step solution:
The question ask us to find the value of $z_{1}^{2}+z_{2}^{2}$ if ${{z}_{1}},{{z}_{2}}$ is given as the complex number and $z_{1}^{2}+z_{2}^{2}$ belongs to real number. The conditions which will help us in calculating the unknown is given as ${{z}_{1}}(z_{1}^{2}-3z_{2}^{2})=2$ and ${{z}_{2}}(3z_{1}^{2}-z_{2}^{2})=11$. The first step to solve the equation is to number the two equations:
${{z}_{1}}(z_{1}^{2}-3z_{2}^{2})=2$……………..(1)
${{z}_{2}}(3z_{1}^{2}-z_{2}^{2})=11$…………….(2)
We will multiply the second equation with “i” making the number complex. On doing this we get:
$\Rightarrow 3z_{1}^{2}{{z}_{2}}i-z_{2}^{3}i=11i$ …………. (3)
We will now add the equation (1) and (3). On doing this we get:
$\Rightarrow 3z_{1}^{2}{{z}_{2}}i-z_{2}^{3}i+z_{1}^{3}-3z_{2}^{2}{{z}_{1}}=2+11i$
We will apply the formula ${{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ to change the above expression.
$\Rightarrow {{\left( {{z}_{1}}+i{{z}_{2}} \right)}^{3}}=2+11i$……………..(4)
We will now subtract the equation (2) from equation (1). On doing this we get:
$\Rightarrow z_{1}^{3}-3z_{2}^{2}{{z}_{1}}-3z_{1}^{2}{{z}_{2}}i+z_{2}^{3}i=2-11i$
We will apply the formula ${{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}+3ab\left( a-b \right)$ to change the above expression.
$\Rightarrow {{\left( {{z}_{1}}-i{{z}_{2}} \right)}^{3}}=2-11i$……………..(5)
Now the most important step, which is to multiply the equation (4) and equation (5). On doing this we get:
$\Rightarrow {{\left( {{z}_{1}}-i{{z}_{2}} \right)}^{3}}{{\left( {{z}_{1}}+i{{z}_{2}} \right)}^{3}}=\left( 2-11i \right)\left( 2+11i \right)$
$\Rightarrow (\left( {{z}_{1}}-i{{z}_{2}} \right){{\left( {{z}_{1}}+i{{z}_{2}}) \right)}^{3}}=\left( 2-11i \right)\left( 2+11i \right)$
We will use the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ for the further calculation of the above equation:
$\Rightarrow {{\left( z_{1}^{2}+z_{2}^{2} \right)}^{3}}=\left( {{2}^{2}}-{{(11i)}^{2}} \right)$
We should know that ${{i}^{2}}=1$. So applying the same we get:
$\Rightarrow {{\left( z_{1}^{2}+z_{2}^{2} \right)}^{3}}=\left( 4+121 \right)$
$\Rightarrow {{\left( z_{1}^{2}+z_{2}^{2} \right)}^{3}}=125$
$\Rightarrow z_{1}^{2}+z_{2}^{2}=\sqrt[3]{125}$
The cube root of the $125$ is $5$.
$\Rightarrow z_{1}^{2}+z_{2}^{2}=5$
$\therefore $ The value of $z_{1}^{2}+z_{2}^{2}$ is $A)5$.

Note: The complex number is presented as $z=a+ib$ , here $a$ and $b$ are real and imaginary parts of the complex number respectively. Complex numbers can be real when there is no presence of an imaginary term or iota.