$If z_{1}, z_{2} are two complex numbers satisfying |\frac{z_{1} - 3 z_{2}}{3 - z_{1} {\bar{z}}_{2}}| = 1, |z_{1}| \neq 3, then |z_{2}| is equal to$

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Detailed Solution

$Given, |\frac{z_{1} - 3 z_{2}}{3 - z_{1} {\bar{z}}_{2}}| = 1, |z_{1}| \neq 3 \Rightarrow |z_{1} - 3 z_{2}| = |3 - z_{1} {\bar{z}}_{2}| | ∵ | \frac{z_{1}}{z_{2}} | = | \frac{z_{1}}{z_{2}} ∣] \begin{array}{l} \Rightarrow {|z_{1} - 3 z_{2}|}^{2} = {|3 - z_{1} {\bar{z}}_{2}|}^{2} \\ \Rightarrow (z_{1} - 3 z_{2}) ({\bar{z}}_{1} - 3 {\bar{z}}_{2}) = (3 - z_{1} {\bar{z}}_{2}) (3 - {\bar{z}}_{1} z_{2}) \\ [∵ {\bar{z}}_{2} = z_{2}] \\ \Rightarrow {|z_{1}|}^{2} - 3 z_{1} {\bar{z}}_{2} - 3 z_{2} {\bar{z}}_{1} + 9 {|z_{2}|}^{2} \\ \Rightarrow = 9 - 3 {\bar{z}}_{1} z_{2} - 3 z_{1} {\bar{z}}_{2} + {|z_{1}|}^{2} {|z_{2}|}^{2} \\ \Rightarrow {|z_{1}|}^{2} + 9 {|z_{2}|}^{2} - 9 - {|z_{1}|}^{2} {|z_{2}|}^{2} = 0 \\ \Rightarrow (9 - {|z_{1}|}^{2}) (1 - {|z_{2}|}^{2}) = 0 \\ \Rightarrow {|z_{1}|}^{2} = 9 or {|z_{2}|}^{2} = 1 \\ \Rightarrow |z_{1}| = 3 or |z_{2}| = 1 \\ But |z_{1}| \neq 3 |z_{2}| = 1 \end{array}$