JEE Main 2015 — Mathematics Algebra

Given, $\frac{{z}_{1}-2{z}_{2}}{2-{z}_{1}{\bar{z}}_{2}}$2−z1​zˉ2​z1​−2z2​​ is unimodular.

$$\Rightarrow |\frac{{z}_{1}-2{z}_{2}}{2-{z}_{1}{\bar{z}}_{2}}|=1$$⇒∣2−z1​zˉ2​z1​−2z2​​∣=1

$$\Rightarrow |{z}_{1}-2{z}_{2}|=|2-{z}_{1}{\bar{z}}_{2}|$$⇒∣z1​−2z2​∣=∣2−z1​zˉ2​∣

Squaring both the sides, we get,

$${|{z}_{1}-{2z}_{2}|}^{2}={|2-{z}_{1}{z}_{2}|}^{2}$$∣z1​−2z2​∣2=∣2−z1​z2​∣2

$$\Rightarrow ({z}_{1}-2{z}_{2})({\bar{z}}_{1}-2{\bar{z}}_{2})=(2-{z}_{1}{\bar{z}}_{2})(2-{\bar{z}}_{1}{z}_{2})$$⇒(z1​−2z2​)(zˉ1​−2zˉ2​)=(2−z1​zˉ2​)(2−zˉ1​z2​)

$$(\because {|z|}^{2}=z\bar{z})$$(∵∣z∣2=zzˉ)

$$\Rightarrow {z}_{1}{\bar{z}}_{1}-2{z}_{1}{\bar{z}}_{2}-2{\bar{z}}_{1}{z}_{2}+4{\bar{z}}_{2}{\bar{z}}_{2}$$⇒z1​zˉ1​−2z1​zˉ2​−2zˉ1​z2​+4zˉ2​zˉ2​

$$=4-2{\bar{z}}_{1 }{z}_{2}-2{z}_{1 }{\bar{z}}_{2}+{z}_{1 }{\bar{z}}_{1 }{z}_{2 }{\bar{z}}_{2}$$=4−2zˉ1​z2​−2z1​zˉ2​+z1​zˉ1​z2​zˉ2​

$$\Rightarrow {|{z}_{1}|}^{2}+4{|{z}_{2}|}^{2}=4+{|{z}_{1}|}^{2 }{|{z}_{2}|}^{2}$$⇒∣z1​∣2+4∣z2​∣2=4+∣z1​∣2∣z2​∣2

$$\Rightarrow {|{z}_{1}|}^{2}-4+4{|{z}_{2}|}^{2}-{|{z}_{1}|}^{2 }{|{z}_{2}|}^{2}=0$$⇒∣z1​∣2−4+4∣z2​∣2−∣z1​∣2∣z2​∣2=0

$$\Rightarrow ({|{z}_{1}|}^{2}-4)(1-{|{z}_{2}|}^{2})=0$$⇒(∣z1​∣2−4)(1−∣z2​∣2)=0

$\Rightarrow |{z}_{1}|=2$⇒∣z1​∣=2 or $|{z}_{2}|=1$∣z2​∣=1

Given, ${z}_{2}$z2​ is not unimodular

$$\therefore |{z}_{1}|=2$$∴∣z1​∣=2

$\therefore$∴ Point ${z}_{1}$z1​ lies on a circle of radius $2$2.