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 value of $z_1^2+z_2^2$ is

• A. $10$
• B. $12$
• C. $5$
• D. $8$

Answer 1

The correct option is C $5$

We have,
$z_1(z_1^2-3z_2^2)=2\dots \left[1\right]$
and $z_2(3z_1^2-z_2^2)=11\dots \left[2\right]$
Multiplying $\left[2\right]$ by $i$ and adding it to $\left[1\right]$, we get,
$z_1^3-3z_2^2{z}_1+i(3z_1^2{z}_2-z_2^3)=2+11i$
$\Rightarrow (z_1+i{z}_2{)}^3=2+11i\dots [3]$
Multiplying $\left[2\right]$ by $i$ and subtracting it from $\left[1\right]$, we get,
$z_1^3-3z_2^2{z}_1-i(3z_1^2{z}_2-z_2^3)=2-11i$
$\Rightarrow (z_1-i{z}_2{)}^3=2-11i\dots [4]$
Multiplying $\left[3\right]$ and $\left[4\right]$, we get
$(z_1^2+z_2^2{)}^3=2^2-121i^2=4+121=125$
$\Rightarrow z_1^2+z_2^2=5$