First slide

Algebra of complex numbers

Question

If $z_{1} z_{2} \in C, z_{1}^{2} + z_{2}^{2} \in R, z_{1} (z_{1}^{2} - 3 z_{2}^{2}) = 2 and z_{2} (3 z_{1}^{2} - z_{2}^{2}) = 11$ , then the value of $z_{1}^{2} + z_{2}^{2}$ is

Moderate

A

10
B

12
C

5
D

8

Solution

We have,
$z_{1} (z_{1}^{2} - 3 z_{2}^{2}) = 2$ (1)
$z_{2} (3 z_{1}^{2} - z_{2}^{2}) = 11$ (2)
Multiplying (2) by i and adding it to (1), we get
$z_{1}^{3} - 3 z_{2}^{2} z_{1} + i (3 z_{1}^{2} z_{2} - z_{2}^{3}) = 2 + 11 i$
$\Rightarrow {(z_{1} + {iz}_{2})}^{3} = 2 + 11 i$ (3)
Multiplying (2) by i and subtracting it from (1), we get
$z_{1}^{3} - 3 z_{2}^{2} z_{1} - i (3 z_{1}^{2} z_{2} - z_{2}^{3}) = 2 - 11 i$
$\Rightarrow {(z_{1} - {iz}_{2})}^{3} = 2 - 11 i$ (4)
Multiplying (3) and (4), we get
${(z_{1}^{2} + z_{2}^{2})}^{3} = 2^{2} - 121 i^{2} = 4 + 121 = 125$
$\Rightarrow z_{1}^{2} + z_{2}^{2} = 5$

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