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 :
Let \[{{z}_{1}}=a+i\,\,b\] \[{{z}_{2}}=c+i\,\,d\] |
\[\operatorname{Re}\,\,({{z}_{1}}+{{z}_{2}})=0\] \[\Rightarrow \] \[a+c=0\] i.e. \[c=-\,a\] |
\[\text{lm(}{{z}_{1}}{{z}_{2}}\text{)=0}\] \[\Rightarrow \] \[ad+bc=0\] i.e. \[a\,(d-b)=0\]\[\Rightarrow \] \[d=b\] \[[\because a=-c\ne 0]\] |
\[{{z}_{1}}=a+ib=-\,c+id=-\,(c-id)\]\[\Rightarrow \] \[{{z}_{1}}=-\,{{\bar{z}}_{2}}\] |
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