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.

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 $N > 21$,
$$\frac{\pi}{\sin \frac{\pi}{2N}}\, \frac{29}{40} + 1 < \frac{3}{2}N.$$
