A) \[{{z}_{1}}={{z}_{2}}\]
B) \[{{\left| {{z}_{2}} \right|}^{2}}={{z}_{1}}{{z}_{2}}\]
C) \[{{z}_{1}}{{z}_{2}}=1\]
D) None of these
Correct Answer: B
Solution :
Let\[{{z}_{1}}={{\gamma }_{1}}(\cos {{\theta }_{1}}+i\sin {{\theta }_{i}})\]. Given that,\[\left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|=1\] \[\Rightarrow \]\[|{{z}_{1}}|\,\,=\,\,|{{z}_{2}}|\] \[\Rightarrow \] \[|{{z}_{1}}|\,\,=\,\,|{{z}_{2}}|\,\,={{\gamma }_{1}}\] Now, \[\arg \,\,({{z}_{1}}{{z}_{2}})=0\Rightarrow \arg ({{z}_{1}})+\arg ({{z}_{2}})=0\] \[\Rightarrow \] \[\arg ({{z}_{2}})=-{{\theta }_{1}}\] Therefore, \[{{z}_{2}}={{\gamma }_{1}}(\cos (-{{\theta }_{1}})+i\sin (-{{\theta }_{1}})\] \[={{\gamma }_{1}}(\cos {{\theta }_{1}}-i\sin {{\theta }_{1}})={{\bar{z}}_{1}}\] \[\Rightarrow \] \[{{\bar{z}}_{2}}=({{\bar{z}}_{1}})={{z}_{1}}\] \[\Rightarrow \] \[{{z}_{2}}{{\bar{z}}_{2}}={{z}_{1}}{{z}_{2}}\] \[\Rightarrow \,\,\,\,\,|z{{|}^{2}}={{z}_{1}}{{z}_{2}}\]
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