Problem. Find the loci of the complex numbers $w$ and $z$ which satisfy $|w+ 2+ 3i|\leqq |w- 3- 2i|$ and $|z- 3- 2i|\leqq 1$. For such $w$ and $z$, find the minimum value of $|z- w|$.
Solution. We will identify the complex $a+ bi$ with the point $(\!a,\,b\!)$. By definition, $|w+ 2+ 3i|$ is the distance from $w$ to $w_{1}= -2- 3i$ and $|w- 3- 2i|$ is the distance from $w$ to $w_{2}= 3+ 2i$. Therefore, the locus of $w$ such that $|w+ 2+ 3i|\leqq |w- 3- 2i|$ is the half-plane which contains $w_{1}$ and is determined by the perpendicular bisector $l$ of $w_{1}$ and $w_{2}$. Similarly, $|z- 3- 2i|$ is the distance from $z$ to $w_{2}= 3+ 2i$. Thus, the locus of $z$ such that $|z- 3- 2i|\leqq 1$ is the closed disk with center at $w_{2}= 3+ 2i$ and with radius $1$. It is easy to see that the line which goes through $w_{1}$ and $w_{2}$, say $l_{1}$, is perpendicular to $l$ and intersects $l$ at $\left ( \dfrac{1}{2},\,-\dfrac{1}{2} \right )$. Hence $|z- w|$ obtains its minimum value when $w= \left ( \dfrac{1}{2},\,-\dfrac{1}{2} \right )$ and $z$ is the lower intersection between $l_{1}$ and the circle $|z- 3- 2i|= 1$. And its minimum value is ($\lceil$ https://www.desmos.com/calculator/hglpvz3hmz $\rfloor$ is an image used) as follow
$$\sqrt{\left ( 3- \dfrac{1}{2} \right )^{2}+ \left ( 2+ \dfrac{1}{2} \right )^{2}}- 1= \dfrac{5}{2}\sqrt{2}- 1$$
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