Let S=zC: 3|2z-3(1+i)|7.

Question

Let $S=\{z\in\mathbb{C}: 3\le|2z-3(1+i)|\le7\}$. Then $\min\limits_{z\in S}\left|z+\dfrac12(5+3i)\right|$ is equal to:

Accepted Answer

Dividing by 2, the condition becomes $\dfrac32\le\left|z-\left(\dfrac32+\dfrac32 i\right)\right|\le\dfrac72$, an annulus centered at $c=\dfrac32+\dfrac32i$ with inner radius $\dfrac32$ and outer radius $\dfrac72$.

We want the minimum distance from points of this annulus to the point $-\dfrac12(5+3i)=-\dfrac52-\dfrac32i$.

Distance from center to this point:
$$|c-(-\tfrac52-\tfrac32i)| = \left|\left(\tfrac32+\tfrac52\right)+\left(\tfrac32+\tfrac32\right)i\right|=|4+3i|=\sqrt{16+9}=5.$$

Since $5$ is larger than the outer radius $\tfrac72$, the closest point of the annulus to $-\tfrac12(5+3i)$ lies on the outer boundary, along the line joining the two points:

$$\text{Min distance} = 5-\frac72=\frac32.$$

Solutions · 1