Find the value of $k$ if for the complex numbers $\;z_{1}\;$ and $\;z_{2}\;,|1-\overline{z_{1}}z_{2}|^{2} - |z_{1} - z_{2}|^{2}= k (1-|z_{1}|^{2}) (1-|z_{2}|^{2})$

Question

$(a)\;1\qquad(b)\;\pm6\qquad(c)\;\pm3\qquad(d)\;4$

Answer 1

Answer : 1

Explanation :

L.H.S = $|1-\overline{z_{1}}z_{2}|^{2} - |z_{1} - z_{2}|^{2}$

$= (1- \overline{z_{1}} z_{2}) \overline{ (1- \overline{z_{1}} z_{2}) }- (z_{1} -z_{2}) \overline{ (z_{1} -z_{2})}$

$= (1- \overline{z_{1}} z_{2}) (1-z_{1} \overline{z_{2}}) - (z_{1}-z_{2}) (\overline{z_{1}}-\overline{z_{2}})$

$= 1+z_{1} \overline{z_{1}} z_{2} \overline{z_{2}} - z_{1} \overline{z_{1}} - z_{2} \overline{z_{2}}$

$= 1+ |z_{1}|^{2}\;. |z_{2}|^{2} - |z_{1}|^{2} - |z_{2}|^{2} $

$= (1- |z_{1}|^{2} )(1- |z_{2}|^{2} )$

R.H.S $= k(1- |z_{1}|^{2} )(1- |z_{2}|^{2} )$

Therefore , k=1

Hence , equating L.H.S and R.H.S , we get k=1 .