A) \[{{z}_{1}}={{z}_{2}}\]
B) \[{{\bar{z}}_{1}}={{z}_{2}}\]
C) \[{{z}_{1}}+{{z}_{2}}=0\]
D) \[{{\bar{z}}_{1}}={{\bar{z}}_{2}}\]
Correct Answer: B
Solution :
Let \[{{z}_{1}}={{r}_{1}}(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})\] Then \[|{{z}_{1}}|\,=\,|{{z}_{2}}|\,\,\Rightarrow |{{z}_{2}}|={{r}_{1}}\] and \[arg({{z}_{1}})+arg({{z}_{2}})=0\]Þ \[arg({{z}_{2}})=-arg({{z}_{1}})=-{{\theta }_{1}}\] \[{{z}_{2}}={{r}_{1}}[\cos (-{{\theta }_{1}})-i\sin (-{{\theta }_{1}})]={{r}_{1}}(\cos {{\theta }_{1}}-i\sin {{\theta }_{1}})\] \[={{\bar{z}}_{1}}\]\[{{\bar{z}}_{1}}={{z}_{2}}\].
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