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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?)

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    $\begingroup$ I don't see how this can be true if $r:=|z_1|=|z_2|=...=|z_n|$. Any disk of radius smaller than $r$ contains none of the points, any disk of radius greater or equal to $r$ contains all of them. Am I missing something? $\endgroup$ – Oliver Braun Sep 23 '13 at 20:01
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    $\begingroup$ @OliverBraun: The discs don't necessarily have the same centres $\endgroup$ – fretty Sep 23 '13 at 20:03
  • $\begingroup$ Ok, right, I was thinking about disks with center $0$. Thanks to fretty. Prism That's no problem, $r$ may be strictly smaller than the radius of the disk $D$. Then all the points are in the interior of $D$. But this is not very relevant since I was forgetting about disks with center other than $0$... $\endgroup$ – Oliver Braun Sep 23 '13 at 20:07
  • $\begingroup$ @OliverBraun: Yeah you are right. Sorry, my comment didn't make sense, so I removed it :) $\endgroup$ – Prism Sep 23 '13 at 20:10
  • $\begingroup$ If $\large\left\vert z_{1}\right\vert < \left\vert z_{2}\right\vert < \cdots \left\vert z_{n}\right\vert$, just choose a disk with radius $\large r$ such that $\large \left\vert z_{n - 1}\right\vert < r <\left\vert z_{n}\right\vert$ unless $\large \left\vert z_{n - 1}\right\vert = \left\vert z_{n}\right\vert$ as @OliverBraun already pointed out. $\endgroup$ – Felix Marin Sep 23 '13 at 20:27
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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.

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    $\begingroup$ @StefanH.: you draw the segment between $z_i$ and $z_j$. Find the midpoint and draw a line perpendicular to the segment through that point. It bisects the segment. $\endgroup$ – Ross Millikan Sep 23 '13 at 20:25
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    $\begingroup$ You haven't shown that the circle will be within the original disk $D$. It will be true for centers near enough the original center, but that needs some argument. A good approach, though. $\endgroup$ – Ross Millikan Sep 23 '13 at 20:27
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    $\begingroup$ @RossMillikan: enthusiastic (+1) for your second comment. I am having trouble proving that the circle we are constructing will lie inside the original disk $D$. :( If N.S. could expand on that, it would be awesome :) $\endgroup$ – Prism Sep 23 '13 at 20:30
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    $\begingroup$ @Prism: as all the points are inside $D$, there is one at minimum distance from the boundary. The same argument can be used to say we can choose the center of $D'$ within half that distance and we are home. $\endgroup$ – Ross Millikan Sep 23 '13 at 21:05
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    $\begingroup$ @Prism: It is half the distance from the center of $D$. That way the circle doesn't bulge out on one side. Since there are only finitely many lines, there are points not on any line that are arbitrarily close to the center of $D$ such that no pair of points are at the same distance. $\endgroup$ – Ross Millikan Sep 23 '13 at 22:35
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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 $z_1,\ldots,z_n$.

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  • $\begingroup$ Thanks for your answer. I don't quite understand this bit: "and equality holds only when $z, z_1$ have the same argument (but then, only for $z=z_1$)". Why does that necessarily follow? Suppose $z_2$ is another point with $|z_1|=|z_2|$, then I don't see why $z_2$ is forced to have the same argument as $z_1$. $\endgroup$ – Prism Sep 23 '13 at 21:32
  • $\begingroup$ @Prism, that's because for all $z,w\in\mathbb{C}$, $|z+w|=|z|+|w|$ only if $z,w$ are linearly dependent over $\mathbb{R}$. $\endgroup$ – Jonathan Y. Sep 23 '13 at 22:16
  • $\begingroup$ Ah right. That makes sense! (+1) $\endgroup$ – Prism Sep 24 '13 at 22:00
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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.

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  • $\begingroup$ I don't know much about Möbius transforms, but thanks for the answer! $\endgroup$ – Prism Sep 24 '13 at 21:59

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