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.
  • $\begingroup$ At least a fairly simple way to quickly count the number of roots of $(x+1)^p-x^p-1$ on the “critical line”: compute $$2^{-\frac1p} \cos\left(\frac{2 \pi n}{p-1}\right)^{1-\frac1p}$$ for $n=0, \ldots, \lfloor(p-1)/4\rfloor$. Every value in $(0, \tfrac12]$ adds four roots and $0$ adds two roots. $\endgroup$ – WimC Jul 17 '18 at 19:29
  • $\begingroup$ Sounds interesting @WimC. I'm afraid I don't see it. Can you please elaborate? $\endgroup$ – Jyrki Lahtonen Jul 17 '18 at 19:51
  • 1
    $\begingroup$ See my answer below. It is based on the observation that solutions correspond to points $z$ on the ciritical line such that $z^p$ is also on the critical line. Then a bit of computation leads to the stated result. $\endgroup$ – WimC Jul 17 '18 at 19:55
  • 1
    $\begingroup$ @JyrkiLahtonen May be a method for calculating the number of real zeros can be used Let $\,x=-1/2 + i z\,$, then $\,P(x)=A(z) + i B(z)\,$ with $\,A,B \in \mathbb{R}[\text{x}]\,$ has roots with real part $\,-1/2\,$ iff $\,\gcd(A,B)\,$ has real roots. The computations would not be pretty, though. $\endgroup$ – dxiv Jul 17 '18 at 20:55

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.


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:

5th root of the critical line under

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.

  • 1
    $\begingroup$ Please add the details :-) $\endgroup$ – Jyrki Lahtonen Jul 18 '18 at 6:30
  • 1
    $\begingroup$ Added my derivation of the root counting method. Things work out so nicely that one cannot help but wonder if there is not a much more natural approach to all this. But until then... $\endgroup$ – WimC Jul 18 '18 at 10:34
  • $\begingroup$ it is a lovely idea, indeed, that $z$ from the critical line is a solution if and only if $z^p$ is also on the critical line. I don't doubt your details at all, but I need to set aside a little time to check evrything. Meanwhile, THANK YOU! $\endgroup$ – Jyrki Lahtonen Jul 18 '18 at 10:39

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.