If $z_{1}, z_{2}$ are two distinct complex numbers such that $\left|\frac{z_{1}-2z_{2}}{\frac{1}{2}-z_{1}\bar{z}_{2}}\right|=2$, then

(D) either $z_{1}$ lies on a circle of radius 1 or $z_{2}$ lies on a circle of radius $\frac{1}{2}$

(A) Both $\mathrm{z}_{1}$ and $\mathrm{z}_{2}$ lie on the same circle

(B) Either $z_{1}$ lies on a circle of radius $\frac{1}{2}$ or $z_{2}$ lies on a circle of radius 1

(C) $z_{1}$ lies on a circle of radius $\frac{1}{2}$ and $z_{2}$ lies on a circle of radius 1

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Complex Numbers

z_{1}, z_{2}

are two distinct complex numbers such that

\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2

, then

(A)Both $z_{1}$ and $z_{2}$ lie on the same circle
(B)Either $z_{1}$ lies on a circle of radius $\frac{1}{2}$ or $z_{2}$ lies on a circle of radius 1
(C) $z_{1}$ lies on a circle of radius $\frac{1}{2}$ and $z_{2}$ lies on a circle of radius 1
(D)either $z_{1}$ lies on a circle of radius 1 or $z_{2}$ lies on a circle of radius $\frac{1}{2}$ ✓ Correct

Answer: (D)

Squaring

\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2

gives

∣ z_{1} - 2 z_{2} ∣^{2} = 4 \frac{1}{2} - z_{1} \overset{z}{ˉ}_{2}^{2}

. Expanding both sides, the cross terms

z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2}

cancel, leaving

∣ z_{1} ∣^{2} + 4∣ z_{2} ∣^{2} = 1 + 4∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2}

, i.e.

(∣ z_{1} ∣^{2} - 1) (1 - 4∣ z_{2} ∣^{2}) = 0

. So either

∣ z_{1} ∣ = 1

∣ z_{2} ∣ = \frac{1}{2}

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option