Upper bound of $\sum_{n=1}^N |1-z^n|$ where $|z| \leq 1$ How to derive an upper bound of
$$\sum_{n=1}^N |1-z^n|$$ where $z\in\mathbb{C}$ and $|z| \leq 1$?
A trivial upper bound would be $2N$ since each $|1-z^n| \leq 2$. But I am hoping for tighter bounds. I ran some numerical experiments and believe the bound should be $\frac{3}{2}N$ but don't know how to prove it.
 A: Here is an upper bound:
According to the link given by @Gerd in comment (Maximum of sum of finite modulus of analytic function.), the maximum of
$\sum_{n=1}^N |1 - z^n|$ is attained on $|z| = 1$.
Let $z = \mathrm{e}^{\mathrm{i}\theta}$
with $\theta \in [0, 2\pi]$. Using AM-QM, we have
\begin{align*}
 \sum_{n=1}^N |1 - z^n|
 &\le \sqrt{N\sum_{n=1}^N |1 - z^n|^2}\\
 &= \sqrt{N\sum_{n=1}^N (2 - 2\cos n\theta)}\\
 &= \sqrt{2N^2 + N \frac{\sin \frac{\theta}{2} - \sin\frac{(2N + 1)\theta}{2}}{\sin \frac{\theta}{2}}}. \tag{1}
\end{align*}
Fact 1: Let $x\in [0, \pi/2]$ and $N \ge 20$. Then
$$\frac{\sin x - \sin (2N + 1)x}{\sin x} \le \frac{N}{2}.$$
(The proof is not difficult.)
By Fact 1 and (1), we have, for all $N \ge 20$,
$$\sum_{n=1}^N |1 - z^n| \le \sqrt{5/2}\, N.$$

Some thoughts:
We have
\begin{align}
 \sum_{n=1}^N |1 - z^n|
 &= \sum_{n=1}^N \sqrt{2 - 2\cos n\theta }\\
 &= \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|\\
 &= \frac{4}{\pi}N - \sum_{n=1}^N \sum_{k=1}^\infty \frac{8}{\pi(4k^2 - 1)}\cos kn\theta\\
 &= \frac{4}{\pi}N -  \sum_{k=1}^\infty \sum_{n=1}^N \frac{8}{\pi(4k^2 - 1)}\cos kn\theta\\
 &= \frac{4}{\pi}N +  \sum_{k=1}^\infty \frac{8}{\pi(4k^2 - 1)}
 \frac{\sin\frac{k\theta}{2} - \sin \frac{(2N + 1)k \theta}{2}}{2\sin \frac{k\theta}{2}}\\
 &= \frac{4}{\pi}N +  \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
 + \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
 \frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}\\
 &= \frac{4}{\pi}N + \frac{2}{\pi} + \sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
 \frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}
\end{align}
where we have used the identity
$$\left|\sin y\right| = \frac{2}{\pi}-\sum_{k = 1}^\infty \frac{4}{\pi(4k^2-1)}\cos(2k y).$$
(Note: The RHS is the Fourier expansion of the LHS.)
We need to find bounds for
$$\sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
 \frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}.$$
A: The behavior in the limit (as $N\to\infty$) looks interesting: the value $$L:=\lim_{N\to\infty}\frac1N\sup_{|z|\leqslant 1}\sum_{n=1}^N|1-z^n|\approx1.4492227\ldots\color{red}{<1.5}$$ equals $2\sin\lambda$, where $\lambda$ is the smallest positive root of $\cos\lambda+\lambda\sin\lambda=1$.
Here is a sketch towards a proof. As already noted, the $\sup_{|z|\leqslant 1}$ is attained at $|z|=1$: $$\sup_{|z|\leqslant 1}\sum_{n=1}^N|1-z^n|\underset{\big[z=e^{2it}\big]}{=}2\sup_{0\leqslant t\leqslant\pi/2}\sum_{n=1}^N|\sin nt|.$$ Suppose the maximum is attained at $t=t_N$, and let $\lambda_N=Nt_N$. Then it turns out that $\lambda_N<\pi$ for all $N$, and that the limit $\lambda=\lim_{N\to\infty}\lambda_N$ exists. Assuming this is proven, $t=t_N$ is the smallest positive root of $\sum_{n=1}^N n\cos nt=0$, that is of $(N+1)\cos Nt-N\cos(N+1)t=1$, so that $$\left(N+1-N\cos\frac{\lambda_N}N\right)\cos\lambda_N+N\sin\frac{\lambda_N}N\sin\lambda_N=1,$$ which gives $\cos\lambda+\lambda\sin\lambda=1$ after $N\to\infty$. And then $$L=\lim_{N\to\infty}\frac{\cos\frac{t_N}2-\cos\left(N+\frac12\right)t_N}{N\sin\frac{t_N}2}=\frac{1-\cos\lambda}{\lambda/2}=2\sin\lambda.$$
A: Some thoughts:
As already noted, the maximum of
$\sum_{n=1}^N |1 - z^n|$ is attained on $|z| = 1$.
Let $z = \mathrm{e}^{\mathrm{i}\theta}$
with $\theta \in [0, 2\pi]$.
We have
$$\sum_{n=1}^N |1 - z^n| = \sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|.$$
Conjecture 1: The maximum of $\sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|$ on $[0, 2\pi]$ is attained at some $x_0 \in [0, 2\pi/N]$.
(Note: Numerical experiment supports the claim. )
If Conjecture 1 is true, when
$\theta \in [0, 2\pi/N]$,
we have
$$
\sum_{n=1}^N 2\left|\sin \frac{n\theta}{2}\right|
= \sum_{n=1}^N 2\sin \frac{n\theta}{2} 
= \frac{2\sin^2 \frac{N\theta }{4} \cos \frac{\theta}{4}}{\sin \frac{\theta}{4}} + \sin \frac{N\theta}{2}
= \frac{2\sin^2 y \cos \frac{y}{N}}{\sin \frac{y}{N}} + \sin 2y
$$
where $y = N\theta/4 \in [0, \pi/2]$.
Then we have
$$\frac{2\sin^2 y \cos \frac{y}{N}}{\sin \frac{y}{N}} + \sin 2y
\le \frac{2\sin^2 y}{\frac{2y}{\pi}\sin \frac{\pi}{2N}} + 1 = \frac{\pi}{\sin \frac{\pi}{2N}}\frac{\sin^2 y}{y} + 1 \le \frac{\pi}{\sin \frac{\pi}{2N}}\, \frac{29}{40} + 1$$
where we have used
$\sin \frac{y}{N} \ge \frac{2y}{\pi}\sin \frac{\pi}{2N}
$ and $\frac{\sin^2 y}{y} \le \frac{29}{40}$ for all $y\in [0, \pi/2]$.
It is not difficult to prove that, for all $N > 21$,
$$\frac{\pi}{\sin \frac{\pi}{2N}}\, \frac{29}{40} + 1 < \frac{3}{2}N.$$
A: This is not an an exact answer: I just want to present a physical-geometrical approach
to the problem which can help to "visualize" and guide to a rigorous mathematical solution.
Please allow me to slightly change the notation and use a more practical small $n$ instead of $N$.
So let's consider
$$
\left\{ \matrix{
  S_n  = \sum\limits_{1 \le k \le n} {\left| {1 - z^k } \right|}
  = \sum\limits_{1 \le k \le n} {\left| {z^k  - 1} \right|}
  = \sum\limits_{0 \le k \le n} {\left| {z^k  - 1} \right|}  \hfill \cr 
  z = re^{i\alpha } \quad \left| {\;r \le 1} \right. \hfill \cr}  \right.
$$
Now, we can look at $\left| {w - 1} \right|$ as being the distance from $U=(1,0) = 1+i0$,
and thus the first moment of a unit mass placed at $w$ with respect to $U$.
Then
$$
\sum\limits_{0 \le k \le n} {\left| {e^{ik\alpha }  - 1} \right|} 
$$
is the sum of the moments of $n+1$ unit masses, placed on the unit circle at constant angular separation $\alpha$
with the first being at $U$.
If $r < 1$ the $n+1$ unit masses will be placed instead on the log-spiral
an starting from $U$.
$$
\rho  = \left| {r^k e^{ik\alpha } } \right|
 = \left| {r^{\left( {k\alpha } \right)/\alpha } e^{i\left( {k\alpha } \right)} } \right|
 = r^{\theta /\alpha }  = e^{{{\ln r} \over \alpha }\theta }  = e^{\ln r\;k} 
$$
again with constant angular separation, $\alpha$ or better $k$ if we eliminate $\alpha$ from the constant at the exponent,
and again starting from $U$ and proceeding towards the origin.

Therefore  $S_n$ is the moment of a mass $n+1$ placed at the barycenter of the configuration of the single masses.

It becomes then geometrically clear that

*

*For large $n$ and $r=1$ the largest distance of the baricenter from $U$ is attained when most of the points
are lying  in the 2nd and 3rd quadrant; this means to correspondingly decrease $\alpha$ so that the almost continuous
circular arc is about $270^{circ}$.

*In the limit, for $n \to \infty$,   $S_n /(n+1)$ will tend to the barycenter of an arc of unitary mass of which to maximize
the barycenter distance.

*For a spiral configuration, also the barycenter distance will be maximized when most of the points are in the 2nd and 3rd
quadrant

*It is also geometrically "visible" that given a maximized spiral configuration, we can radially expand it to the circle,
thereby increasing the distance from $U$ except for the points close to that.

