Question Number 27060 by math solver last updated on 01/Jan/18
Commented by math solver last updated on 02/Jan/18
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Answered by prakash jain last updated on 02/Jan/18
$$\frac{{k}−{z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} }{{z}_{\mathrm{1}} −{kz}_{\mathrm{2}} }×\frac{{k}−\bar {{z}}_{\mathrm{1}} {z}_{\mathrm{2}} }{\bar {{z}}_{\mathrm{1}} −{k}\bar {{z}}_{\mathrm{2}} }=\mathrm{1} \\ $$$${k}^{\mathrm{2}} +\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} −{k}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} +{z}_{\mathrm{2}} \bar {{z}}_{\mathrm{1}} \right) \\ $$$$=\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} +{k}^{\mathrm{2}} \mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} −{k}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} +{z}_{\mathrm{2}} \bar {{z}}_{\mathrm{1}} \right) \\ $$$${k}^{\mathrm{2}} +\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} =\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} +{k}^{\mathrm{2}} \mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \\ $$$$\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} =\mathrm{1} \\ $$