Complex Numbers – Geometry – JEE Main 23 Jan 2026 Shift 1

Solution:

Simplify the inequality for $S$:
$$ 3 \le |2z – 3(1+i)| \le 7 $$
Divide by 2:
$$ \frac{3}{2} \le \left|z – \frac{3}{2}(1+i)\right| \le \frac{7}{2} $$

This represents the region between two concentric circles centered at $C\left(\frac{3}{2}, \frac{3}{2}\right)$.
Inner radius $r_1 = \frac{3}{2}$.
Outer radius $r_2 = \frac{7}{2}$.

We need to find the minimum distance from a point $P\left(-\frac{5}{2}, -\frac{3}{2}\right)$ to any point $z$ in this annular region.
First, find the distance between the center $C$ and the point $P$:
$$ CP = \sqrt{\left(\frac{3}{2} – \left(-\frac{5}{2}\right)\right)^2 + \left(\frac{3}{2} – \left(-\frac{3}{2}\right)\right)^2} $$

$$ CP = \sqrt{\left(\frac{8}{2}\right)^2 + \left(\frac{6}{2}\right)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = 5 $$

Since $CP = 5$ and the outer radius $r_2 = 3.5$, the point $P$ lies outside the annular region.
The minimum distance from an external point to an annulus is the distance to the outer boundary minus the outer radius? No.
The distance to the region is the distance to the center minus the outer radius if inside? No.
Let’s visualize. $P$ is far away. The closest point in the region is on the outer circle boundary on the line joining $P$ and $C$.
Wait, the inequality is $r_1 \le |z-C| \le r_2$.
The closest distance from $P$ to the set $S$ is the distance to the *nearest* point of the annulus.
Since $P$ is outside the outer circle ($5 > 3.5$), the closest point is on the outer circle.
Distance = $CP – r_2$.

$$ \text{Min Distance} = 5 – \frac{7}{2} = \frac{10 – 7}{2} = \frac{3}{2} $$

Ans. (2)