For the complex numbers $$z_1$$ and $$z_2$$ if $${\left| {1 - {{\bar z}_1}{z_2}} \right|^2} - {\left| {{z_1} - {z_2}} \right|^2} = k\left( {1 - {{\left| {{z_1}} \right|}^2}} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right)$$         then $$'k'$$ equals to

Solution :
$$\eqalign{ & {\left| {1 - {{\bar z}_1}{z_2}} \right|^2} - {\left| {{z_1} - {z_2}} \right|^2} \cr & = \left( {1 - {{\bar z}_1}{z_2}} \right)\left( {1 - {z_1}{{\bar z}_2}} \right) - \left( {{z_1} - {z_2}} \right)\left( {{{\bar z}_1} - {{\bar z}_2}} \right) \cr & = 1 - {z_1}{{\bar z}_2} - {{\bar z}_1}{z_2} + {z_1}{{\bar z}_1}{z_2}{{\bar z}_2} - \left( {{z_1}{{\bar z}_1} - {z_1}{{\bar z}_2} - {{\bar z}_1}{z_2} + {z_2}{{\bar z}_2}} \right) \cr & = 1 + {z_1}{{\bar z}_1}{z_2}{{\bar z}_2} - {z_1}{{\bar z}_1} - {z_2}{{\bar z}_2} \cr & = 1 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} - {\left| {{z_1}} \right|^2} - {\left| {{z_2}} \right|^2} \cr & = \left( {1 - {{\left| {{z_1}} \right|}^2}} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) \cr & \Rightarrow k = 1 \cr} $$