# 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.

• The 'distinct' part fails for $n=2$, since $(x-1/2)^4+(x+1/2)^4+1=\frac{(4x^2+3)^2}{8}$ has multiple roots. – user547557 Apr 2 '18 at 0:32
• @yixing Thank you. I overlooked the special character of $\pm\frac{i\sqrt{3}}{2}$ root. Corrected. – user Apr 2 '18 at 2:56
• Idea. Once you find the set $Z_L$ of zeros of $P_n$ on the left arc $\{ \frac{1}{2} - e^{i\theta} : -\frac{\pi}{3} < \theta < \frac{\pi}{3} \}$, then $$Z_R = \{ -z : z \in Z_0 \}$$ is the set of zeros of $P_n$ on the right arc and $$Z_I = \left\{ \frac{1}{\frac{1}{2}+z} - \frac{1}{2} : z \in Z_0\right\}$$ is the set of zeros of $P_n$ on the imaginary axis, possibly except for $\pm \frac{i\sqrt{3}}{2}$. (In particular, $Z_R, Z_L, Z_I$ are disjoint.) Now it seems not terribly hard to show that $|Z_R| \geq 2\lfloor n/3 \rfloor$, and this easily leads to the desired claim. – Sangchul Lee Apr 2 '18 at 5:43
• I would very appreciate if the five people who closed my question as "off-topic" let me know the reason for such strange decision. I would like also to thank other five people who have reopened it. – user Apr 2 '18 at 20:31

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 $\mathsf{Z}_{*}$.

1. $\mathsf{S}_{*}$ are disjoint for different symbols $* \in \{ T, L, R, I \}$. In particular, it follows that

$$|\mathsf{Z}_L| + |\mathsf{Z}_R| + |\mathsf{Z}_I| + |\mathsf{Z}_T| \leq 2n.$$

2. 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 $\frac{1}{z+\frac{1}{2}}-\tfrac{1}{2} = \frac{i}{2}\cot\left(\frac{t}{2}\right)$.

3. 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 $|\mathsf{Z}_L| = |\mathsf{Z}_R| = |\mathsf{Z}_I|$.

4. 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.

• Thank you very much for the splendid proof. I believe however there is a misprint in "4". $P_n^{(1)}$ should be equal to $4ni\sin\frac{(2n-1)\pi}{3}$. Could you suggest an intuitively understandable reason why the "arc" roots of the polynomial are so extremly close to $\pm\frac{1}{2}\mp\exp\left(\frac{2k+1}{2n}\pi i\right)$? – user Apr 2 '18 at 20:18
• @user, You are right, I made some mistake when simplifying the derivatives. As for the additional question, those zeros come from $t_k$'s and since the RHS of $\text{(*)}$ decays exponentially to $0$, it loosely follows that $t_k$'s become close to the zero of $\cos(nt) = 0$ in the interval $\left(\frac{(k-1)\pi}{n}, \frac{k\pi}{n} \right)$. I guess this explains the phenomena. – Sangchul Lee Apr 2 '18 at 21:29
• The explanation sounds quite reasonable. Thank you once more. – user Apr 3 '18 at 12:10