A) 1
B) \[-1\]
C) 2
D) \[-2\]
Correct Answer: C
Solution :
Given, \[\left| \frac{({{z}_{1}}-2{{z}_{2}})}{(2-{{z}_{1}}{{\overline{z}}_{2}})} \right|=1,|{{z}_{2}}|\ne 1\] \[\Rightarrow \] \[|{{z}_{1}}-2{{z}_{2}}{{|}^{2}}=|2-{{z}_{1}}\overline{z}{{|}^{2}}\] \[\Rightarrow \]\[({{z}_{1}}-2{{z}_{2}})(\overline{{{z}_{1}}-2{{z}_{2}}})=(2-{{z}_{1}}{{\overline{z}}_{2}})(\overline{2-{{z}_{1}}{{\overline{z}}_{2}}})\] \[\Rightarrow \]\[({{z}_{1}}-2{{z}_{2}})({{\overline{z}}_{1}}-2{{\overline{z}}_{2}})=(2-{{z}_{1}}{{\overline{z}}_{2}})(2-{{z}_{1}}{{\overline{z}}_{2}})\] \[\Rightarrow \]\[{{z}_{1}}{{\overline{z}}_{1}}-2{{z}_{1}}{{\overline{z}}_{2}}-2{{\overline{z}}_{1}}{{z}_{2}}+4{{z}_{2}}{{\overline{z}}_{2}}\] \[=4-2{{\overline{z}}_{1}}{{z}_{2}}-2{{z}_{1}}{{\overline{z}}_{2}}+{{z}_{1}}{{\overline{z}}_{1}}{{z}_{2}}{{\overline{z}}_{2}}\] \[\Rightarrow \] \[|{{z}_{1}}{{|}^{2}}+4|{{z}_{2}}{{|}^{2}}=4+|{{z}_{1}}{{|}^{2}}|{{z}_{2}}{{|}^{2}}\] \[\Rightarrow \] \[|{{z}_{1}}{{|}^{2}}+4|{{z}_{2}}{{|}^{2}}-4-|{{z}_{1}}{{|}^{2}}|{{z}_{2}}{{|}^{2}}=0\] \[\Rightarrow \] \[|{{z}_{1}}{{|}^{2}}(1-|{{z}_{2}}{{|}^{2}})-4(1-|{{z}_{2}}{{|}^{2}})=0\] \[\Rightarrow \] \[(1-|{{z}_{2}}{{|}^{2}})(|{{z}_{2}}{{|}^{2}}-4)=0\] \[\Rightarrow \] \[1-|{{z}_{2}}{{|}^{2}}=0|{{z}_{1}}{{|}^{2}}-4=0\] \[\Rightarrow \] \[|{{z}_{2}}|\ne 1,|{{z}_{1}}|=2\]
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