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Mathematics Question on Complex Numbers and Quadratic Equations

Find the value of kk, if for the complex numbers z1z_{1} and z2z_{2}, ∣1−zˉ1z2∣2−∣z1−z2∣2=k(1−∣z1∣2)(1−∣z2∣2)\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).

A

22

B

33

C

11

D

44

Answer

11

Explanation

Solution

L.H.S.L.H.S. =∣1−zˉ1z2∣2−∣z1−z2∣2=\left|1-\bar{z}_{1}z_{2}\right|^{2}-\left|z_{1}-z_{2}\right|^{2} =(1−zˉ1z2)(1−zˉ1z‾2)−(z1−z2)(z1−z2‾)=\left(1-\bar{z}_{1}z_{2}\right)\left(\overline{1-\bar{z}_{1} z}_{2}\right)-\left(z_{1}-z_{2}\right)\left(\overline{z_{1}-z_{2}}\right) =(1−zˉ1z2)(1−z1zˉ2)−(z1−z2)(zˉ1−zˉ2)=\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) =1+z1zˉ1z2zˉ2−z1zˉ1−z2zˉ2=1+z_{1}\bar{z}_{1} z_{2} \bar{z}_{2}-z_{1} \bar{z}_{1}-z_{2} \bar{z}_{2} =1+∣z1∣2⋅∣z2∣2−∣z1∣2−∣z2∣2=1+\left|z_{1}\right|^{2}\cdot\left|z_{2}\right|^{2}-\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2} =(1−∣z1∣2)(1−∣z2∣2)=\left(1-\left|z_{1}\right|^{2}\right)\left(1-\left|z_{2}\right|^{2}\right) R.H.S.=k(1−∣z1∣2)(1−∣z2∣2)R.H.S. =k\left(1-\left|z_{1}\right|^{2}\right)\left(1-\left|z_{2}\right|^{2}\right) Hence, equating L.H.S.L.H.S. and R.H.S.R.H.S., we get k=1k = 1.