A) 10
B) 12
C) 14
D) 16
Correct Answer: B
Solution :
\[\because \] \[z_{1}^{2}+z_{2}^{2}=4\] \[\Rightarrow \,\,\overline{z}_{1}^{2}+\overline{z}_{2}^{2}=4\] \[\therefore \,\,{{({{z}_{1}}+{{z}_{1}})}^{2}}+{{({{z}_{2}}+{{\overline{z}}_{2}})}^{2}}\] \[=z_{1}^{2}+\overline{z}_{1}^{2}+2{{z}_{1}}{{\overline{z}}_{1}}+z_{2}^{2}+\overline{z}_{2}^{2}+2{{z}_{2}}{{\overline{z}}_{2}}\] \[=\left( z_{1}^{2}+z_{2}^{2} \right)+\left( z_{1}^{2}+z_{2}^{2} \right)+4=12\].You need to login to perform this action.
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