JEE Mathematics 2026 Question with Solution

Given: $S = \{z : 3 \le |2z - 3(1+i)| \le 7\}$ and we need $\min_{z \in S} \left| z + \frac{1}{2}(5+3i) \right|$.

Find: The minimum distance from the point $z_1 = -\left(\frac{5}{2} + \frac{3}{2}i\right)$ to the annulus represented by $S$.

From

we divide by $2$ to get

So $S$ is an annulus with center $C = \left(\frac{3}{2}, \frac{3}{2}\right)$, inner radius $r_1 = \frac{3}{2}$ and outer radius $r_2 = \frac{7}{2}$.

Also,

where

Thus we need the minimum distance from the point $P = \left(-\frac{5}{2}, -\frac{3}{2}\right)$ to the annulus.

Now compute the distance from $P$ to the center $C$:

Since $CP = 5 > \frac{7}{2}$, the point $P$ lies outside the outer circle. Therefore the minimum distance from $P$ to the annulus is

So the working gives minimum value $\frac{3}{2}$. This matches option B. The solution states option A and final answer $2$, but that contradicts the extracted calculation.

Given: The set $S$ is the region between two concentric circles centered at $\left(\frac{3}{2}, \frac{3}{2}\right)$.

Find: The least value of the distance from $\left(-\frac{5}{2}, -\frac{3}{2}\right)$ to this region.

For a point outside a circle, the least distance to the circle is the distance from the center minus the radius. Since the point is outside the outer boundary of the annulus, the nearest point of the annulus lies on the outer circle along the line joining the center and the point.

Thus the least distance is

Hence the correct option is B.