Let $S_{1}=\left\{z_{1} \in C:\left|z_{1}-3\right|=\frac{1}{2}\right\}$ and $S_{2}=\left\{z_{2} \in C:\left|z_{2}-\left|z_{2}+1\right|\right|=\left|z_{2}+\left|z_{2}-1\right|\right|\right\}$. Then, for $\mathrm{z}_{1} \in \mathrm{~S}_{1}$ and $\mathrm{z}_{2} \in \mathrm{~S}_{2}$, the least value of $\left|\mathrm{z}_{2}-\mathrm{z}_{1}\right|$ is :

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Complex Numbers

Let

S_{1} = {z_{1} \in C : ∣ z_{1} - 3 ∣ = \frac{1}{2}}

and

S_{2} = {z_{2} \in C : ∣ z_{2} - ∣ z_{2} + 1 ∣ ∣ = ∣ z_{2} + ∣ z_{2} - 1 ∣ ∣}

. Then, for

z_{1} \in S_{1}

and

z_{2} \in S_{2}

, the least value of

∣ z_{2} - z_{1} ∣

is :

Answer: (C)

S_{1}

is a circle centered at

3

with radius

\frac{1}{2}

.

For

S_{2}

, squaring the given condition and simplifying yields

(z_{2} + \overline{z}_{2}) (∣ z_{2} - 1∣ + ∣ z_{2} + 1∣ - 2) = 0

. This means either

z_{2} + \overline{z}_{2} = 0

(imaginary axis) or

∣ z_{2} - 1∣ + ∣ z_{2} + 1∣ = 2

(real axis segment

[- 1, 1]

).

The minimum distance from the circle

S_{1}

to the imaginary axis is

3 - \frac{1}{2} = \frac{5}{2}

. The minimum distance to the segment

[- 1, 1]

on the real axis is

(3 - \frac{1}{2}) - 1 = \frac{3}{2}

.

Therefore, the least value is

\frac{3}{2}

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option