Let S_1=z_1 C: |z_1-3|=12 and S_2=z_2 C: |z_2-|z_2+1||=|z_2+|z_2-1||.

Question

Let $S_1=\left\{z_1\in\mathbb C: |z_1-3|=\dfrac12\right\}$ and $S_2=\{z_2\in\mathbb C: |z_2-|z_2+1||=|z_2+|z_2-1||\}$. Then, for $z_1\in S_1$ and $z_2\in S_2$, the least value of $|z_2-z_1|$ is:

Accepted Answer

Expanding the condition for $S_2$ and simplifying (squaring both sides and using $z\bar z=|z|^2$) leads to the requirement that either $z_2+\bar z_2=0$ (the imaginary axis) or $|z_2+1|+|z_2-1|=2$ (the real segment $[-1,1]$). So $S_2$ is the union of the imaginary axis and the real interval $[-1,1]$.

$S_1$ is the circle centred $(3,0)$ with radius $\dfrac12$.

The closest point of $S_2$ to this circle is $(1,0)$ (the right endpoint of the real segment), at distance $3-1=2$ from the centre $(3,0)$.

$$|z_2-z_1|_{\min} = 2-\frac12=\frac32.$$

Solutions · 1