A) \[{{z}_{1}}=-{{z}_{2}}\]
B) \[{{z}_{1}}={{z}_{2}}\]
C) \[{{z}_{1}}={{\overline{z}}_{2}}\]
D) None of these
Correct Answer: C
Solution :
Since,\[\operatorname{Im}({{z}_{1}}+{{z}_{2}})=0\]and \[\operatorname{Im}({{z}_{1}}{{z}_{2}})=0\] Since, \[({{z}_{1}}+{{z}_{2}})\] and \[{{z}_{1}}{{z}_{2}}\] both are real. Let \[{{z}_{1}}={{a}_{1}}+i{{b}_{1}},{{z}_{2}}={{a}_{2}}+i{{b}_{2}},\] then\[{{z}_{1}}+{{z}_{2}}\] is real. \[\Rightarrow \] \[{{b}_{2}}=-{{b}_{1}}\]and \[{{z}_{1}}{{z}_{2}}\]is real. \[\Rightarrow \] \[{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}}=0\] \[\Rightarrow \] \[-{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{1}}=0\] \[(\because {{b}_{2}}=-{{b}_{1}})\] \[\Rightarrow \] \[{{a}_{1}}={{a}_{2}}\] So, \[{{z}_{1}}={{a}_{1}}+i{{b}_{1}}={{a}_{2}}-i{{b}_{2}}={{\overline{z}}_{2}}\]
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