A) \[\operatorname{Im}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=0\]
B) \[\operatorname{Re}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=0\]
C) \[\operatorname{Re}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=\operatorname{Im}\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)\]
D) None of these
Correct Answer: A
Solution :
\[\because \] \[|{{z}_{1}}|=|{{z}_{2}}|+|{{z}_{1}}-{{z}_{2}}|\] \[\Rightarrow \] \[|{{z}_{1}}|-|{{z}_{2}}|=|{{z}_{1}}-{{z}_{2}}|\] \[\Rightarrow \] \[{{(|{{z}_{1}}|-|{{z}_{2}}|)}^{2}}=|{{z}_{1}}-{{z}_{2}}{{|}^{2}}\] \[\Rightarrow \,|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}-2|{{z}_{1}}||\,{{z}_{2}}|=|{{z}_{1}}{{|}^{2}}\,+|{{z}_{2}}{{|}^{2}}\] \[-2|{{z}_{1}}||{{z}_{2}}|\cos ({{\theta }_{1}}-{{\theta }_{2}})\] \[\Rightarrow \] \[\cos ({{\theta }_{1}}-{{\theta }_{2}})=1\] \[\Rightarrow \] \[{{\theta }_{1}}-{{\theta }_{2}}=0\] \[\Rightarrow \] \[\arg ({{z}_{1}})-\arg ({{z}_{2}})=0\] \[\Rightarrow \] \[\frac{{{z}_{1}}}{{{z}_{2}}}\]is purely real. \[\Rightarrow \] \[\ln \left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=0\]
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