On the roots of the polynomial $(z-\frac12)^{2n}+(z+\frac12)^{2n}+1$ Consider a polynomial:
$$P_n(z)=\left(z-\frac{1}{2}\right)^{2n}+\left(z+\frac{1}{2}\right)^{2n}+1$$
where $n$ is a positive integer.
Obviously the polynomial has no real roots. The following properties are valid by numerical evidence:
At least one of the statements:
$$
\left|z_i-\frac{1}{2}\right|=1,\;\left|z_i+\frac{1}{2}\right|=1, \text{ or }
\Re(z_i)=0
$$
is valid for any root $z_i$ of the polynomial $P_n(z)$.
Further,  all roots of the polynomial are distinct except for $\pm\frac{i\sqrt{3}}{2}$, which have multiplicity $n\;\text{mod}\;3$, and thus are two-fold degenerate for $n=2\;\text{mod}\;3$.
I would appreciate any hint on a proof of these properties.
 A: Here is a more detailed explanation of my idea. We first introduce four sets
\begin{align*}
\mathsf{S}_T &= \left\{ -\frac{i\sqrt{3}}{2}, \frac{i\sqrt{3}}{2} \right\} \\
\mathsf{S}_L &= \{ z \in \mathbb{C} : \text{$|z - \tfrac{1}{2}| = 1$ and $\operatorname{Re}(z) < 0$} \} \\
\mathsf{S}_R &= \{ z \in \mathbb{C} : \text{$|z + \tfrac{1}{2}| = 1$ and $\operatorname{Re}(z) > 0$} \} \\
\mathsf{S}_I &= \{ z \in \mathbb{C} : \text{$\operatorname{Re}(z) = 0$ and $|z| > 1$} \}
\end{align*}
Likewise, define $ \mathsf{Z}_{*} $ as the set of zeros of $P_n$ lying on $\mathsf{S}_{*}$, counted with multiplicity, for each symbol $* \in \{ T, L, R, I \} $. (Here, $T=$ triple junction, $L=$ left arc, $R=$ right arc, $I=$ imaginary axis.)
We first make some observations on the sets $\mathsf{S}_{*}$ and $\mathsf{Z}_{*}$.


*

*$\mathsf{S}_{*}$ are disjoint for different symbols $* \in \{ T, L, R, I \} $. In particular, it follows that
$$|\mathsf{Z}_L| + |\mathsf{Z}_R| + |\mathsf{Z}_I| + |\mathsf{Z}_T| \leq 2n. $$

*The followings are equivalent:
$$ z \in \mathsf{S}_L
\qquad \Leftrightarrow \qquad -z \in \mathsf{S}_R
\qquad \Leftrightarrow \qquad  \frac{1}{z+\frac{1}{2}}-\tfrac{1}{2} \in \mathsf{S}_I. $$
Equivalence of first two is clear. For equivalence of the first one and the third one, notice that $z = \frac{1}{2} - e^{it}$ if and only if $\frac{1}{z+\frac{1}{2}}-\tfrac{1}{2} = \frac{i}{2}\cot\left(\frac{t}{2}\right)$.

*We have
$$ \left( z+\tfrac{1}{2} \right)^{2n}P_n \left( \frac{1}{z+\frac{1}{2}}-\tfrac{1}{2} \right) = P_n(z) = P_n(-z). $$
In particular, together with the previous part, we find that $|\mathsf{Z}_L| = |\mathsf{Z}_R| = |\mathsf{Z}_I|$.

*As OP has already observed, we have $|\mathsf{Z}_T| = 2(n \text{ mod } 3)$. This follows by investigating
\begin{align*}
P_n\left(\pm \frac{i\sqrt{3}}{2} \right) &= 1 + 2\cos\left(\frac{2n\pi}{3} \right) \\
P_n^{(1)}\left(\pm \frac{i\sqrt{3}}{2} \right) &= 4ni\sin\left(\frac{(2n-1)\pi}{3} \right) \\
P_n^{(2)}\left(\pm \frac{i\sqrt{3}}{2} \right) &= 4n(2n-1)\cos\left(\frac{(2n-2)\pi}{3} \right)
\end{align*}
In particular, we find that $|\mathsf{Z}_L| \leq 2 \lfloor n/3 \rfloor$. To conclude, it suffices to show the following claim:

Claim. We have $|\mathsf{Z}_L| \geq 2 \lfloor n/3 \rfloor$.

Indeed, let us plug $z = \frac{1}{2} - e^{it}$ with $|t| < \frac{\pi}{3}$ to the equation $P_n(z) = 0$. Upon simplification, we obtain
$$ \cos(nt) = \frac{(-1)^{n-1}}{2} (2\sin(t/2))^{2n}. \tag{*} $$
Notice that the RHS of $\text{(*)}$ is $<\frac{1}{2}$ whenever $|t| < \frac{\pi}{3}$. Hence for each $1 \leq k \leq \lfloor n/3 \rfloor$, there exists at least one $t = t_k \in \left( \frac{(k-1)\pi}{n}, \frac{k\pi}{n} \right)$ such that the equation $\text{(*)}$ is satisfied by the intermediate value theorem. Since $t = -t_k$ also solves $\text{(*)}$, it follows that
$$ \mathsf{Z}_L \supseteq \{ \tfrac{1}{2} - e^{it} : \text{$t = \pm t_k$ for some $1 \leq k \leq \lfloor n/3\rfloor$} \} $$
and therefore $|\mathsf{Z}_L| \geq 2\lfloor n/3 \rfloor$ as desired.
