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Let $z_1,z_2,...,z_n$ be distinct complex numbers such that $|z_i|\leq1$.

Is it true that there exists $z, |z|\leq1$ such that $\displaystyle\sum_{i=1}^n |z-z_i|\geq n$ ?

Thank you.

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    $\begingroup$ Do you mean $|z|\leq 1$? $\endgroup$ – Sarastro Apr 22 '15 at 12:19
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One of $z=1$ or $z=-1$ will always work. Since $|1-z_0| \ge |1-\Re(z_0)|$ and $|-1 -z_0| \ge |-1-\Re(z_0)|$ for any $z_0$ (draw a picture) wlog we can assume all $z_i$ are real. Then for $z = \pm 1$ all the $z-z_i$ are of the same sign, and the sum of their absolute values is the absolute value of the sum, which is $|\pm n-\sum z_i|$. For at least one choice of sign this will be $\ge n$

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