<s>[INST] 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.
 [/INST] 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
$$\sum_{k=1}^\infty \frac{4}{\pi(4k^2 - 1)}
 \frac{ - \sin \frac{(2N + 1)k \theta}{2}}{\sin \frac{k\theta}{2}}.$$
</s>