JEE Main 2024 — Mathematics Algebra

Given: ${|z-{z}_{0}|}^{2}=4$∣z−z0​∣2=4

$$\Rightarrow (z-{z}_{0})(z-{z}_{0})=4$$⇒(z−z0​)(z−z0​)=4

$$\Rightarrow zz-{z}_{0}z-{z}_{0}z+{|{z}_{0}|}^{2}=4$$⇒zz−z0​z−z0​z+∣z0​∣2=4

$$\Rightarrow {|z|}^{2}-{z}_{0}z-{z}_{0}z=2{\text{as}{|{z}_{0}|}^{2}=2}$$⇒∣z∣2−z0​z−z0​z=2as∣z0​∣2=2

It is given that, $\alpha$α lies on ${|z-{z}_{0}|}^{2}=4$∣z−z0​∣2=4.

$$\Rightarrow {|\alpha |}^{2}-\alpha {z}_{0}-{z}_{0}\alpha -2=0...(i)$$⇒∣α∣2−αz0​−z0​α−2=0...(i)

We know that, $\alpha \cdot \alpha ={|\alpha |}^{2}={|\alpha |}^{2}$α⋅α=∣α∣2=∣α∣2

$${|z-{z}_{0}|}^{2}=16$$∣z−z0​∣2=16

$$\Rightarrow (\frac{1}{\bar{\alpha }}-{z}_{0})(\frac{1}{\alpha }-{\bar{z}}_{0})=16$$⇒(αˉ1​−z0​)(α1​−zˉ0​)=16

$$\Rightarrow (1-\bar{\alpha }{z}_{0})(1-\alpha {\bar{z}}_{0})=16|\alpha {|}^{2}$$⇒(1−αˉz0​)(1−αzˉ0​)=16∣α∣2

$$\Rightarrow 1-\bar{\alpha }{z}_{0}-\alpha {\bar{z}}_{0}+|\alpha {|}^{2}{|{z}_{0}|}^{2}=16|\alpha {|}^{2}$$⇒1−αˉz0​−αzˉ0​+∣α∣2∣z0​∣2=16∣α∣2

$$\Rightarrow 1-\bar{\alpha }{z}_{0}-\alpha {\bar{z}}_{0}=14|\alpha {|}^{2}...(\mathrm{ii})$$⇒1−αˉz0​−αzˉ0​=14∣α∣2...(ii)

Using, $(i)\text{and}(\mathrm{ii})$(i)and(ii)

$$\Rightarrow 5|\alpha {|}^{2}=1$$⇒5∣α∣2=1

$$\Rightarrow 100|\alpha {|}^{2}=20$$⇒100∣α∣2=20