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