JEE Main 2022 — Mathematics Algebra

Here $|{z}_{1}-3|=\frac{1}{2}$∣z1​−3∣=21​ represents a circle on argand plane with centre $(3,0)$(3,0) and radius $\frac{1}{2}$21​

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

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

$$\Rightarrow {z}_{2}(|{z}_{2}-1|+|{z}_{2}+1|)+{\bar{z}}_{2}(|{z}_{2}-1|+|{z}_{2}+1|)={|{z}_{2}+1|}^{2}-{|{z}_{2}-1|}^{2}$$⇒z2​(∣z2​−1∣+∣z2​+1∣)+zˉ2​(∣z2​−1∣+∣z2​+1∣)=∣z2​+1∣2−∣z2​−1∣2

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

$\Rightarrow$⇒ Either ${z}_{2}+{\bar{z}}_{2}=0$z2​+zˉ2​=0 or $|{z}_{2}+1|+|{z}_{2}-1|=2$∣z2​+1∣+∣z2​−1∣=2

i.e. ${z}_{2}$z2​ lies on imaginary axis or it lies on the line segment joining $(-1,0)$(−1,0) and $(1,0)$(1,0)

So, the minimum distance between {z}_{1}&{z}_{2} will be the distance between the points (1,0)&(\frac{5}{2},0)

Hence, ${|{z}_{1}-{z}_{2}|}_{\mathrm{min}}=\frac{3}{2}$∣z1​−z2​∣min​=23​