A) \[{{z}_{1}}={{z}_{2}}\]
B) \[{{z}_{1}}={{\bar{z}}_{2}}\]
C) \[{{z}_{1}}=-\,{{\bar{z}}_{2}}\]
D) none of these
Correct Answer: C
Solution :
[C]Let \[{{z}_{1}}=a+i\,b\] \[{{z}_{2}}=c+i\,d\] | |
\[\operatorname{Re}\,\,({{z}_{1}}+{{z}_{2}})=0\Rightarrow a+c=0\] | \[i.e.\,\,\,c=-\,a\] |
\[\operatorname{Im}\,\,({{z}_{1}}{{z}_{2}})=0\] | |
\[\Rightarrow ad+bc=0\] | \[6\Omega \] |
\[\Rightarrow d=b\] | \[[\because a=-c\ne 0)\] |
\[{{z}_{1}}=a+ib\] | |
\[=c+id\] | |
\[=-\,(c-id)\] | |
\[\Rightarrow {{z}_{1}}=-\,{{\bar{z}}_{2}}\] |
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