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