If Z1 and Z2 are two complex numbers such that Z12+Z22∈R and Z1Z12−3Z22=2 Z23Z12−Z22=11 then Z12+Z22=

Z1Z12−3Z22=2 and Z23Z12−Z22=11 ⇒z13-3z22z1+iZ23Z12−Z22=2+11i ⇒z13+3Z12iz2+3z1iz22+iz23=2+11i ⇒z1+iz23=2+11i ⇒z1+iz23=4+121 ⇒z12+z223=125 ⇒z12+z22=5

If Z1 and Z2 are two complex numbers such that Z12+Z22∈R and Z1Z12−3Z22=2 Z23Z12−Z22=11 then Z12+Z22=

Z1Z12−3Z22=2 and Z23Z12−Z22=11 ⇒z13-3z22z1+iZ23Z12−Z22=2+11i ⇒z13+3Z12iz2+3z1iz22+iz23=2+11i ⇒z1+iz23=2+11i ⇒z1+iz23=4+121 ⇒z12+z223=125 ⇒z12+z22=5

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If Z₁ and Z₂ are two complex numbers such that $Z_{1}^{2} + Z_{2}^{2} \in R and Z_{1} (Z_{1}^{2} - 3 Z_{2}^{2}) = 2$ $Z_{2} (3 Z_{1}^{2} - Z_{2}^{2}) = 11$ then $Z_{1}^{2} + Z_{2}^{2} =$

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Detailed Solution

$Z_{1} (Z_{1}^{2} - 3 Z_{2}^{2}) = 2 a n d Z_{2} (3 Z_{1}^{2} - Z_{2}^{2}) = 11 \Rightarrow {z_{1}}^{3} - 3 {z_{2}}^{2} z_{1} + i Z_{2} (3 Z_{1}^{2} - Z_{2}^{2}) = 2 + 11 i \Rightarrow {z_{1}}^{3} + 3 Z_{1}^{2} ({iz}_{2}) + 3 z_{1} {({iz}_{2})}^{2} + {({iz}_{2})}^{3} = 2 + 11 i \Rightarrow {(z_{1} + {iz}_{2})}^{3} = 2 + 11 i \Rightarrow |{(z_{1} + {iz}_{2})}^{3}| = \sqrt{4 + 121} \Rightarrow {({z_{1}}^{2} + {z_{2}}^{2})}^{3} = 125 \Rightarrow {z_{1}}^{2} + {z_{2}}^{2} = 5$