JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Complex Numbers

The condition $|z-3-2i|\le 2$ describes a closed disk centered at $(3,2)$ with radius 2.

We want to minimize $|2z-6+5i|=2|z-3+\frac{5i}{2}|=2|z-(3-\frac{5}{2}i)|$.

This is twice the distance from $z$ to the point $(3,-\frac{5}{2})$.

The minimum distance from the point $(3,-\frac{5}{2})$ to the disk occurs along the line joining $(3,-\frac{5}{2})$ to the center $(3,2)$.

The distance between these points is $|2-(-\frac{5}{2})|=\frac{9}{2}$.

The minimum distance from $(3,-\frac{5}{2})$ to the disk is $\frac{9}{2}-2=\frac{5}{2}$.

Therefore, the minimum value of $|2z-6+5i|$ is $2\cdot \frac{5}{2}=5$.