Let $S=\{z: 3 \leq|2 z-3(1+i)| \leq 7\}$ be a set of complex numbers. Then $\operatorname{Min}_{z \in S}\left|\left(z+\frac{1}{2}(5+3 i)\right)\right|$ is equal to:

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Complex Numbers

Let

S = {z : 3 \leq ∣2 z - 3 (1 + i) ∣ \leq 7}

be a set of complex numbers. Then

Min_{z \in S} (z + \frac{1}{2} (5 + 3 i))

is equal to:

Answer: (B)

Rewriting:

3 \leq ∣2 z - 3 (1 + i) ∣ \leq 7

becomes

\frac{3}{2} \leq ∣ z - \frac{3}{2} (1 + i) ∣ \leq \frac{7}{2}

.

This represents an annular region centered at

C = \frac{3}{2} (1 + i)

with inner radius

\frac{3}{2}

and outer radius

\frac{7}{2}

.

We want to minimize

∣ z - P ∣

where

P = - \frac{5}{2} - \frac{3}{2} i

.

Distance

P C = ∣ \frac{3}{2} (1 + i) - (- \frac{5}{2} - \frac{3}{2} i) ∣ = ∣4 + \frac{9}{2} i ∣ = 16 + \frac{81}{4} = 5

.

The minimum distance from P to the annulus is

P C - \frac{7}{2} = 5 - \frac{7}{2} = \frac{3}{2}

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option