Let $z$ be a number in the unit disc. I would like to know a lower bound of the form


$$\frac{1}{N}\sum\_{k=1}^N|z-\omega^k|\geq |z|+c$$


when $\omega$ is a primitive $N$'th root and $N>3$, where $c$ is independent of $z$ and $N$. The lower bound $|z|$ follows from the triangle inequality. 



Average distance to roots of unity
Such bound exists for certain $c>0$, but I do not know which $c$ is optimal. 


Denote $$f\_n(z):=\frac{1}{n}\sum\_{k=1}^n|z-\omega^k|,$$
then $f$ is 1-Lip as the average of 1-Lip functions. 


Denote $y=z/|z|$ (or $y=1$ if $z=0$), then $|y|=1$, $|z-y|=1-|z|$ and $f\_n(z)-|z|\geqslant f\_n(y)-|z-y|-|z|=f\_n(y)-|y|$, thus it suffices to consider $z$ with $|z|=1$. For such values of $z$ our sum is the integral sum for $\frac1{2\pi}\int\_0^{2\pi}|1-e^{it}|dt=\frac4{\pi}>1$. Therefore for any fixed $c\_0\in (0,4/\pi-1)$ the inequality $f\_n(z)\geqslant 1+c\_0$ holds whenever $|z|=1$ and $n$ is large enough. For finitely many small values of $n$ we note that $f\_n(z)>1$ for all $z$ with $|z|=1$ by triangle inequality, and by compactness there exists certain $c\_1>0$ such that $f\_n(z)>1+c\_1$ for all $z$ with $|z|=1$ and all these $n$. Then $c=\min(c\_0,c\_1)$ works.


