<s>[INST] Complex number inequality $|z-1| \ge \frac{2}{n-1}$ 
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$

I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq  -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n-1}+z^{n-2}+...+z^2+z+1|}$$
Also $|z|=1$, so if I use triangle inequality, I find:
$$|z-1|\ge \frac{2}{|z^{n-1}|+...+|z|+1}=\frac{2}{n}$$
but this is lower than $\frac{2}{n-1}$. Does this help in any way?
 [/INST] If $z^n = -1$, we have
$$
z = \exp\left(\frac{(2k + 1)i \pi}{n}\right),
\qquad k \in \{0, \ldots,n - 1\}
$$
Denote for $k \in \{0, \ldots,n - 1\}$
$$
a_k := \frac{(2k + 1) \pi}{n}
$$
Then
\begin{align}
| z - 1 |^2
& = | \cos(a_k) + i \sin(a_k) - 1 |^2
= (\cos(a_k) - 1)^2 + \sin^2(a_k) \\
& = \cos^2(a_k) + \sin^2(a_k) - 2 \cos(a_k) + 1 \\
& = 2 - 2 \cos(a_k),
\end{align}
therefore
$$
| z - 1 |
= \sqrt{2} \sqrt{1 - \cos(a_k)}.
$$
As $\sqrt{2} \left| \sin\left(\frac{x}{2}\right)\right| = \sqrt{1 - \cos(x)}$, we have
$$
| z - 1| = 2 \left| \sin\left(\frac{(2k + 1)\pi}{2n}\right) \right|.
$$
Therefore it remains to show that
$$
\left| \sin\left(\frac{\left(k + \frac{1}{2}\right)\pi}{n}\right) \right|
\ge \frac{1}{n - 1}, \qquad
k \in \{0, \ldots,n - 1\}.
$$
Also, we can exclude one $k$, which is the $k$ for which $a_k = 0$.
Notice that $\left| \sin\left(\frac{\left(k + \frac{1}{2}\right)\pi}{n}\right) \right|$ is smallest (exercise!) for $k = 0$ and $k = n - 1$ (they yield the same result since $\frac{\pi}{2n} = \frac{(n - 1/2) \pi}{n}$), so it suffices to show it for $k = 0$, which is to say
$$
\left| \sin\left(\frac{\pi}{2n}\right) \right|
\ge \frac{1}{n - 1}.
$$
But this follows from the concavity (similar to here) of $x \mapsto \sin(x)$ on $\left[0, \frac{\pi}{6}\right]$ and $n \ge 3$.
</s>