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.
Best Answer
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.