This question just came to my mind and I have no idea as to how to approach it. Let $z_1,z_2,\dots,z_n$ be $n$ be any complex numbers in the unit disc $|z| \leq 1.$ Consider a complex function on the unit disc wiith real values $$ f(z)=\sum_{i=1}^n \frac{|z-z_i|}{n}. $$ My questions:

  • Does there exist a $z \in |w| \leq 1 $ so that $f(z)=|z|$?
  • If not can we make $f(z)$ arbitrarily close to $|z|$ for some $ z \in |w| \leq 1 $?
  • What about the maximum and the minimum value of $f(z)$?

Lots of thanks for any responces\hints\suggestions

  • 1
    $\begingroup$ If $f(z)$ can be made arbitrarily close to $|z|$, then shouldn’t compactness of the closed unit disc and continuity of $f$ guarantee that equality is attained somewhere? $\endgroup$
    – Vik78
    Mar 28, 2022 at 9:47
  • $\begingroup$ @yes indeed.I thought that might be easier approach to reach to the possibility of equality $\endgroup$ Mar 28, 2022 at 9:58
  • $\begingroup$ The minimum of the expression is a very famous problem called the Fermat-Weber location problem. See here for some history ssabach.net.technion.ac.il/files/2015/12/BS2015.pdf $\endgroup$
    – Vlad Matei
    Mar 28, 2022 at 11:22
  • $\begingroup$ The notation $z\in|z|\leq 1$ should probably be more like $z\in\{w:|w|\leq 1\}$. $\endgroup$ Mar 28, 2022 at 15:21
  • $\begingroup$ Yes ,it should be .Thanks for pointing it out $\endgroup$ Mar 28, 2022 at 17:23

1 Answer 1


The answer is no. E.g., let $n=3$ and $z_j=e^{i(j-1)2\pi/3}$ for $j=1,2,3$. Then $f(z)>|z|+15/100>|z|$ if $|z|\le1$.

This counterexample generalizes to any $n\ge3$. Indeed, take any $n\ge3$ and let $z_j=e^{i(j-1)2\pi/n}$ for $j=1,\dots,n$. Then $$f(z)=\frac1n\,\sum_{j=1}^n f_j(z),$$ where $f_j(z):=|z-z_j|$. Note that for each $j$ the function $f_j$ is convex and, moreover, $f_j$ is strictly convex on any straight line not through the point $z_j$. Since $n\ge3$, there is no straight line passing through all the points $z_1,\dots,z_n$. So, the function $f=\frac1n\,\sum_{j=1}^n f_j$ is strictly convex.

Also, for any real $a$ and $b$, the derivative of $f$ at $0$ along the vector $(a,b)=a+ib$ is $$\begin{aligned}&\frac d{ds}\,f(s(a+ib))\Big|_{s=0} \\ &=-\frac1n\,(a,b)\cdot\sum_{j=1}^n \Big(\cos\frac{(j-1)2\pi}n,\sin\frac{(j-1)2\pi}n\Big) \\ &=-\frac1n\,(a,b)\cdot(0,0)=0, \end{aligned}$$ where $\cdot$ denotes the dot product.

Thus, the strictly convex function $f$ has a unique minimum at $0$, and the minimum value of $f$ is $1$. That is, $f(0)=1<f(z)$ for all $z$ such that $0<|z|\le1$.

So, for all $z$ such that $0<|z|\le1$, we have $f(z)>1\ge|z|$ and hence $f(z)\ne|z|$. Also, $f(0)=1\ne0$. We conclude that there is no $z$ such that $|z|\le1$ and $f(z)=|z|$, which proves the claim.

On the other hand, the answer is yes for $n=1$ and $n=2$.

  • $\begingroup$ @losif Pinels ,you have taken the points on the unit circle and not general points in the unit disc.Could you kindly clarify ? $\endgroup$ Mar 29, 2022 at 13:36
  • $\begingroup$ @sajjadveeri : You asked if for any $n$ points in the unit disk there is some $z$ in the unit disk with $f(z)=|z|$. My answer proves that this is not true; that is, for some $n$ points in the unit disk there is no $z$ in the unit disk with $f(z)=|z|$. If you had asked if for some $n$ points in the unit disk there is some $z$ in the unit disk with $f(z)=|z|$, then the answer would have trivially been yes -- by letting e.g. $z_j=0$ for all $j$ and then choosing $z=0$. I hope you can understand this logic. $\endgroup$ Mar 29, 2022 at 13:59
  • $\begingroup$ Previous comment continued: Try not to use vague terms such as "general points"; use quantifiers "for all/any" and "there exist(s)" instead. $\endgroup$ Mar 29, 2022 at 13:59
  • $\begingroup$ Thank you for your edifying comment.Valuable Advice noted! $\endgroup$ Mar 29, 2022 at 14:19

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