Let $a=-\frac{1}{2}(3+4i)=-\frac{3}{2}-2i.$ Then the given expression becomes $\left|z+\frac{1}{2}(3+4i)\right|=|z-a|.$

The condition $|z|\le 1$ means that the point representing $z$ lies inside or on the circle centered at the origin with radius $1$.

We need the minimum distance from a point $a=-\frac{3}{2}-2i$ to the closed disk $|z|\le 1$.

First find the distance of $a$ from the origin: $|a|=\left|\frac{1}{2}(3+4i)\right|=\frac{1}{2}|3+4i|=\frac{1}{2}\cdot 5=\frac{5}{2}.$

Since the disk has radius $1$, and the point $a$ is outside the disk, the minimum distance from $a$ to the disk is $|a|-1=\frac{5}{2}-1=\frac{3}{2}.$

Equivalently, $\min_{|z|\le 1}|z-a|=|a|-1.$

Therefore, $\min \left|z+\frac{1}{2}(3+4i)\right|=\frac{3}{2}.$

Checking options:

• A: $2$ ❌
• B: $\frac{5}{2}$ ❌
• C: $\frac{3}{2}$ ✅
• D: $3$ ❌