Finite number of points inside a disk

Question

Let $n\ge 2$ and suppose that $z_1, z_2, \ldots, z_n$ are distinct points in the interior of some disk $D$ in the plane. Why is it true that there exists a smaller disk $D'\subseteq D$ such that $D'$ contains exactly $n-1$ of these points above?

In other words, we want $D'$ to contain $n-1$ of the points $z_1, z_2, \ldots, z_n$, and miss one of these points. This fact above was used in the proof of Cauchy's Residue Theorem in my complex analysis class. However, the proof wasn't provided, and the justification for it was kind of handwaving. I would love to see a rigorous proof for this interesting fact.

Any help is much appreciated!

P.S. What area of mathematics would this problem fall? (Topology? Geometry?)

Answer 1

Pick all pairs of two pints $(z_i, z_j)$. Draw the perpendicular bisector on this segment.

This way you get at most $\binom{n}{2}$ lines. Pick a point on none of these lines (measure theory shows such a point exists, but there mist be a simpler argument).

This point will have pairwise distinct distances to the $n$ points, and you can easily show that some circle with the centre here will work.

Answer 2

For ease of notation, let's assume $D$ is centered around the origin, and has radius $r$. Also, w.l.o.g. $z_1$ has maximal modulus (although not necessarily uniquely so).

Consider the disc $D^\prime = D_{\frac{|z_1|+r}{2}}(\frac{z_1}{|z_1|}\frac{|z_1|-r}{2})$.

Note that for all $z\not\in D^\prime$ we have $$|z|\geq \left|z-\frac{z_1}{|z_1|}\frac{|z_1|-r}{2}\right| - \left|\frac{z_1}{|z_1|}\frac{|z_1|-r}{2}\right| \geq \frac{|z_1|+r}{2} - \left|\frac{|z_1|-r}{2}\right| = |z_1|,$$ and equality holds only when $z,z_1$ have the same argument (but then, only for $z=z_1$). It follows that $z_1\not\in D^\prime$, and is alone in having this property among $z_1,\ldots,z_n$.

Answer 3

After a suitable Möbius transform we can assume that $\partial D$ is the real axis and that the $z_k=x_k+i y_k$ $\>(1\leq k\leq n)$ lie in the upper half plane, whereby $$y_k=1\quad(1\leq k\leq r),\qquad y_k>1\quad(r+1\leq k\leq n)$$ for some $r\geq1$. When $r=1$ the "circle" $y=1+\epsilon$ for some small $\epsilon$ will do. When $r\geq2$ we may further assume $-a=x_1<\ldots<x_r=a$ for some $a>0$. A sufficiently large circle with center $iM$, $\>M\gg1$, through the points $z_1=-a+i$ and $z_r=a+i$ will then lie in the upper half plane and contain all the other $z_k$ in its interior. Now move this circle a small amount $\epsilon>0$ to the right, and it will contain all $z_k$ with $k>1$ in its interior, whereas the point $z_1$ will lie on the outside.