Methods for calculating the number of zeros of a polynomial with a specified real part?

Question

Given a polynomial with real coefficients is there a method (e.g. from algebra or complex analysis) to calculate the number of complex zeros with a specified real part?

Background. This question is motivated by my tests related to this problem.

Let $p>3$ be a prime number. Let $G_p(x)=(x+1)^p-x^p-1$, and let $$F_p(x)=\frac{(x+1)^p-x^p-1}{px(x+1)(x^2+x+1)^{n_p}}$$ where the exponent $n_p$ is equal to $1$ (resp. $2$) when $p\equiv-1\pmod 6$ (resp. $p\equiv1\pmod 6$).

The answer by Lord Shark the Unknown (loc. linked) implies that $F_p(x)$ is a monic polynomial with integer coefficients. The degree of $F_p$ is equal to $6\lfloor(p-3)/6\rfloor$. I can show that the complex zeros of $F_p(x)$ come in groups of six. Each of the form $\alpha,-\alpha-1,1/\alpha,-1/(\alpha+1),-\alpha/(\alpha+1),-(\alpha+1)/\alpha.$ That is, orbits of a familiar group (isomorphic to $S_3$) of fractional linear transformations.

My conjecture. Exactly one third of the zeros of $F_p(x)$ have real part equal to $-1/2$.

I tested this with Mathematica for a few of the smallest primes and it seems to hold. Also, each sextet of zeros of the above form seems to be stable under complex conjugation, and seems to contain a complex conjugate pair of numbers with real part $=-1/2$. Anyway, I am curious about the number of zeros $z=s+it$ of the polynomial $F_p(x)$ on the line $s=-1/2$.

Summary and thoughts.

• Any general method or formula is welcome, but I will be extra grateful if you want to test a method on the polynomial $G_p(x)$ or $F_p(x)$ :-)
• My first idea was to try the following: Given a polynomial $P(x)=\prod_i(x-z_i)$ is there a way of getting $R(x):=\prod_i(x-z_i-\overline{z_i})$? If this can be done, then we get the answer by calculating the multiplicity of $-1$ as a zero of $R(x)$.
• May be a method for calculating the number of real zeros can be used with suitable substitution that maps the real axes to the line $s=-1/2$ (need to check on this)?
• Of course, if you can prove that $F_p(x)$ is irreducible it is better that you post the answer to the linked question. The previous bounty expired, but that can be fixed.

Answer 1

I realized that my comment actually answers your question completely for $(x+1)^p-x^p-1$. After raising to the power $p/(p-1)$ the counting criterion simplifies quite a bit: Compute $$ \cos\left(\frac{2 \pi n}{p-1}\right)$$ for $n=0, \ldots, \lfloor(p-1)/4\rfloor$. Every value in $(0, \tfrac12]$ adds four roots and $0$ adds two roots. In other words: every $ (p-1)/6 \leq n < (p-1)/4$ counts for four roots and $n=(p-1)/4$ (if $p\equiv 1 \pmod 4$) counts for two roots.

Here is how I derived this root counting method. To avoid my sign mistakes I make the substitution $x \leftarrow -x$ and investigate the roots of $(1-x)^p + x^p - 1$ for odd $p$ on the critical line $\operatorname{Re}(z) = \tfrac12$. Note that on the critical line $1-z=\overline{z}$ so $z$ is a root if and only if $\operatorname{Re}(z^p) = \tfrac12$. The strategy is now to investigate the image of the critical line under all branches of $z^{1/p}$ and see how often this image intersects the critical line. Let $f_0(z)=z^{1/p}$ indicate the principle branch. The other branches are then $$f_m(z)=\exp\left(\frac{2 \pi \mathrm{i}\,m}p\right)f_0(z)$$ for integral $m$. Here $m$ will be restricted to $[0, (p-1)/4]$, i.e. those $m$ for which the primitive $p$-th root lies in the upper right quadrant. Now parameterise the critical line by $$z = \tfrac12(1 + \mathrm{i}\tan(\alpha))$$ for $\alpha \in (-\pi/2, \pi/2)$. A straight forward computation shows that $$N_m(z) = (\operatorname{Re}f_m(z))^p = \frac{\cos^p((\alpha + 2 \pi m)/p)}{2 \cos(\alpha)}.$$ Another straight forward calculation shows that $N_m$ attains its extremal value at $$\alpha = \frac{2 \pi m}{p-1}$$ with extremal value $$N_m(z) = \tfrac12\cos^{p-1}\left(\frac{2 \pi m}{p-1}\right).$$

Now the central observation is this: The image of the critical line under $f_m$ looks a bit like a hyperbola. See this picture for $p=5$ which shows all branches:

So if at this extremal angle $0 < N_m(z) \leq 2^{-p}$ then $0 < \operatorname{Re}f_m(z) \leq \tfrac12$ and the image of $f_m$ will intersect the critical line in two places (counting multiplicity) since the branch is located at the right of the extremal value. By conjugate symmetry this $m$ accounts for four zeroes on the critical line.

for $m=(p-1)/4$ the situation is a bit different: the image now has the imaginary axis as one of its asymptotes. (As visible in the picture for $p=5$. The extremal value $N_m(z)$ is $0$ in this case.) This branch clearly intersects the critical line only at a single point, accounting for two zeroes on the critical line in total.

Answer 2

How about Cauchy argument variation

$$\int_{C} \frac{f'(z)}{f(z)}dz=2\pi i N$$

around a rectangular contour $a \pm \epsilon + bi$ where $a$ is specified and $b$ leaps $\pm \infty$?

---

If it doesn't help maybe for Rouche's theorem you can choose some dominant function.

Answer 3

Note The part below that states that $g_c$ can only lose roots on the critical line for increasing $c$ is not rigorous yet, even though it must be true given my other answer...

Another approach. Let $p \geq 3$ be odd and consider $g_c(z) = (1-z)^p + z^p - c$ for real $c \geq 0$. Note that the zeroes of $g_c$ are conjugate symmetric and symmetric in the critical line $\operatorname{Re}(z) = \tfrac12$. Now start at $c=0$ and track what happens with the roots on the critical line when $c$ is increased.

Note that $z$ is a root of $g_c$ on the critical line if and only if $2 \operatorname{Re}(z^p) = c$. For $c=0$ the situation is easy: $g_0$ has all its $p-1$ roots on the critical line and all roots are simple. By symmetry, the number of roots on the critical line can only change if $g_c$ has a double root on the critical line. Either a new double root appears or two roots disappear from such $c$ upward.

So when does $g_c$ have a double root on the critical line? Exactly if $g_c$ and $g_c'$ have a common root there. Now $$p^{-1}(1-z)\, g_c'(z )+ g_c(z) = z^{p-1} - c$$ so a double root is a positive multiple of a $(p-1)$-th root of unity. This shows that $g_c$ has a double root on the critical line precisely at $$\frac12\left(1 \pm \mathrm{i}\,\tan\left(\frac{2 \pi m}{p-1}\right)\right)$$ for some integer $m$ when $$c=\frac1{2^{p-1}\cos^{p-1}\left(\frac{2 \pi m}{p-1}\right)}.$$ For $c \in(0,1)$ this happens for $m \in [0, (p-1)/6)$ and $g_c$ loses two ($m=0$) or four ($m>0$) roots on the critical line every time $c$ passes such a point. Conclusion $g_1(z)$ has $$p+1 - 4\lceil\frac{p-1}6\rceil$$ roots on the critical line.

Answer 4

For the zeros of $G_p(z)=(z+1)^p-z^p-1$ on $\Re z=-\frac12$, let $z=-\frac12+iy$. Then we have $$T_p(u)=2^{p-1}u^p\tag1$$ where $u=\frac1{\sqrt{y^2+4}}$ and $T_p(u)$ is the $p$-th Chebyshev polynomial. To count the real solutions, we intersect the graphs of $y=T_p(u)$ and $y=2^{p-1}u^p.$ Since, $-\frac12<u<\frac12$ and the zeros of $T_p(u)$ are in the form $\cos\frac{(2k-1)\pi}{2p}$, the number of k's satisfying $$\frac\pi 3<\left|\frac{(2k-1)\pi}{2p}\right|\leq\frac\pi2$$ is approximately (exactly?) the number of real zeros of $(1)$.