A) \[{{z}_{1}}=-{{z}_{2}}\]
B) \[{{z}_{1}}={{\bar{z}}_{2}}\]
C) \[{{z}_{1}}=-{{\bar{z}}_{2}}\]
D) \[{{z}_{1}}={{z}_{2}}\]
Correct Answer: B
Solution :
Let \[{{z}_{1}}=a+ib,{{z}_{2}}=c+id\], then \[{{z}_{1}}+{{z}_{2}}\] is real Þ \[(a+c)+i(b+d)\]is real Þ \[b+d=0\] Þ \[d=-b\] .....(i) \[{{z}_{1}}{{z}_{2}}\] is real Þ \[(ad-bd)+i(ac+bc)\]is real Þ \[ad+bc=0\] Þ \[a(-b)+bc=0\]Þ \[a=c\] \\[{{z}_{1}}=a+ib=c-id={{\bar{z}}_{2}}\] \[(\because a=c\]and \[b=-d)\]
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