<s>[INST] Minimum of $E(z)=|z+1|+|z^{2n}+1|$ 
Let $n \in \mathbb{N}$. Find the minimum of the expression $E(z)=|z+1|+|z^{2n}+1|$ over $\mathbb{C}$.

I found this problem in a Romanian magazine with all sorts of math problems.
(Edit: ${\color{blue}{\textrm{This was suggested for a contest for 10th graders}}}$, According to @alexanderv's comment.)
What I've done so far(WRONG):
$E \geq 0$ so if we can find $z$ with $E(z)=0$ it's enough. If $n=0$, the minimum is achieved in $-1$, $E(z) =2$. If $n \geq 1$, let's take $z= \cos \theta + i\sin \theta$. Using $|1+\cos \alpha + i\sin \alpha|=2| \cos (\frac{\alpha}{2})|$, we get $E(z)=2(|cos(\frac{\theta}{2})|+|cos(n\theta)|)$. Then, taking the modulus and using $\cos(\pi-\alpha)=-\cos(\alpha)$ and the $\cos(x) + \cos(y)$ formula, we take $z$ so one of the product terms equals to $0$, achieving the minimum. But I'm not sure if it's right, can someone approve?
What's wrong above is $E(z)=0$ implies $z=-1$, that means $E(z)=2$, contradiction. I don't have any idea how to find the minimum.
 [/INST] shymilan@AoPS's elegant proof (I rewrote it):
Let $w_k = \mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi}, ~ k = 1, 2, \cdots, 2n$. Then $Q(z) := z^{2n} + 1
= (z - w_1)(z - w_2)\cdots (z - w_{2n})$.
We have the partial fraction decomposition
$$\frac{1}{z^{2n} + 1} = \frac{1}{Q(z)}
= \sum_{k=1}^{2n} \frac{1}{Q'(w_k)}\frac{1}{z - w_k}
= \sum_{k=1}^{2n} \frac{-w_k}{2n}\frac{1}{z - w_k}. \tag{1}$$
(easy to prove, e.g. see Partial_fraction_decomposition).
For $k = 1, 2, \cdots, 2n$, we have
\begin{align*}
 |z + 1| + |z^{2n} + 1| &= 
 |z - w_k + w_k + 1| + |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\
 &\ge |w_k + 1| - |z - w_k|
 +  |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\
 &= |w_k + 1| + |z - w_k|
 \left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right). \tag{2}
\end{align*}
Using (1), we have
\begin{align*}
 \sum_{k=1}^{2n} \left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right)
 &= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k}\right| - 2n \\
 &= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\
 &\ge \left|\sum_{k=1}^{2n} \frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\
 &= 0. \tag{3}
\end{align*}
From (2) and (3), there exists $k \in \{1, 2, \cdots, 2n\}$ such that $\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1 \ge 0$ and
$$|z + 1| + |z^{2n} + 1| \ge |w_k + 1|.$$
For $k = 1, 2, \cdots, 2n$, we have
$$|w_k + 1| = |\mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi} + 1| = 2\left|\sin \frac{(2n - 2k + 1)\pi}{4n}\right|
\ge 2 \sin \frac{\pi}{4n}.$$
Thus, we have
$$|z + 1| + |z^{2n} + 1| \ge 2 \sin \frac{\pi}{4n}.$$
Also, if $z = \mathrm{e}^{\mathrm{i}\frac{2n - 1}{2n}\pi}$,
we have $|z+1| + |z^{2n} + 1| = 2 \sin \frac{\pi}{4n}$.
Thus, the minimum of  $|z+1| + |z^{2n} + 1|$ is $2 \sin \frac{\pi}{4n}$.
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