Let
\begin{align\*}
 p(z) & = \ \prod\_{j = 1}^{n} (z - z\_j) \, ; \ |z\_j| \ = \ 1 \\
 & = \ \prod\_{j = 1}^{l+1} (z - z\_j)^{M\_j}
\end{align\*}
be a non-constant complex polynomial with $l+1$ distinct zeros of multiplicity $M\_j$, which all lie on the unit circle. Then there are $l$ not necessarily distinct zeros $z'\_k$ of the derivative $p'(z)$ in the interior of the unit disc:
\begin{align\*}
 p'(z) & = \ n \, \prod\_{k = 1}^{n-1} (z - z'\_k) \, ; \qquad |z'\_k| \ < \ 1 \ : \ 1 \ \leq \ k \ \leq \ l \, ,\\
\end{align\*}
where $l \, = \, n-1$, if $p(z)$ has only simple zeros. The $z'\_k$ in the interior of the unit disc are known as critical points of the second kind, as p(z) itself is not zero there.




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The Poisson-kernel $k(a,b)$ of two complex numbers, where $|a| \ = \ 1$ and $|b| \ < \ 1$,
$$ 
k(a,b) \ = \ \frac{1 - |b|^2}{|a - b|^2} \, ,
$$
is well-known from the solution of the Dirichlet-Problem of the Laplace equation for the unit disc.




---


**Question, equation (1):**


Is the equation
\begin{equation}
 \sum^l\_{k = 1} \frac{1 - |z'\_k|^2}{|z\_j - z'\_k|^2} \ = \ \frac{n - M\_j}{M\_j}
\end{equation}
true for any zero $z\_j$ of any polynomial $p(z)$ with at least two distinct zeros, all of whose zeros lie on the unit circle?




---


**Some background:**




---


This question is related to the matrix


\begin{equation}
\mathbf{D\_{pu}} \ := \ 
\begin{pmatrix} 
\frac{M\_1}{n} & \ldots & \frac{M\_{l+1}}{n} \\
\frac{M\_1}{n} \, \frac{H(z'\_1)}{|z\_1 - z'\_1|^2} & \ldots & 
\frac{M\_{l+1}}{n} \, \frac{H(z'\_1)}{|z\_{l+1} - z'\_1|^2} \\
\vdots & & \vdots \\
\frac{M\_1}{n} \, \frac{H(z'\_l)}{|z\_1 - z'\_l|^2} & \ldots & 
\frac{M\_{l+1}}{n} \, \frac{H(z'\_l)}{|z\_{l+1} - z'\_l|^2} \\
\end{pmatrix} \, ,
\end{equation}
where $H(z'\_k)$ is the harmonic mean of the squares of the distances of the $z\_j$ to a critical point of the second kind $z'\_k$, defined as follows:
$$
H(z'\_k) \ := \ \frac{n}{\sum^{l+1}\_{j=1} \frac{M\_j}{|z\_j - z'\_k|^2}} \, .
$$
$D\_{pu}$ appears, with an immaterial difference in the definition, in the paper


"Borcea’s variance conjectures on the critical points of polynomials"
by D. Khavinson, R.Pereira, M. Putinar, E. Saff and S. Shimorin in Notions of Positivity and the Geometry of Polynomials, Series: Trends in Mathematics,
Birkhäuser, Basel, 2011, pp. 283 – 309, see <https://arxiv.org/abs/1010.5167>.


It is called "Augmented Gauss-Lucas matrix" there, as its rows 2 to l+1 give the coefficients in a convex combination of the zeros expressing a critical point of the second kind:


$D\_{pu}(z\_1, \ldots, z\_{l+1})^t \ = \ (S, z'\_1, \ldots, z'\_{l})^t$ ,


where $S$ is the centroid of the zeros. These convex combinations are used in an early and well-known proof of the Gauss-Lucas-Theorem, see


"Analytic Theory of Polynomials" by Q. I. Rahman and G. Schmeisser, Oxford University Press, 2002, section 2.6, notes to section 2.1,


hence the name of the matrix given in the paper above. The authors of the paper anticipated me in proving, that $D\_{pu}$ is orthostochastic if all zeros lie on a line and unistochastic if there are at most three different zeros. I arrived at these results independently with a slightly different approach.


The connection to the Poisson-kernel for unimodular zeros arises by the equation:


$$ (2) \qquad H(z'\_k) \ = \ 1 - |z'\_k|^2 \, ,$$


which is valid, if all zeros are unimodular, see equation $(\ast)$ in my new question [Question about an exact expression for the root-mean-square of the distances of the critical points to a given zero of a polynomial](https://mathoverflow.net/q/402980/37855). I edited this one, because it linked to an old MathStackexchange question, which was deleted, because it drew no attention.


Equation (2) implies that the stochastic matrix $D\_{pu}$ is doubly stochastic for any polynomial $p(z)$ with unimodular zeros if and only if equation (1) is true. As $D\_{pu}$ is stochastic, equation (1) is true "in the mean". Double stochasticity would prove a (slight) refinement of Lemma 7.3.3 in Rahman/Schmeisser, showing that there is a critical point of the second kind in a certain circle associated with a zero, implying the correctness of Sendov's conjecture for the polynomial $p(z)$, [Sendov's conjecture](https://mathoverflow.net/q/360035/37855). According to Profs. Khavinson and Pereira, this connection was not recognized before.




---


All my (rather few) investigations of examples of low degree showed, that $D\_{pu}$ is doubly stochastic if and only if the zeros of $p(z)$ lie on a line or a circle, $p(z) = z^4 - z$ is an instructive counterexample, so a generalization of my problem above is the following




---


**Further Conjecture/Question (\*):**


$D\_{pu}$ is doubly stochastic if and only if all zeros of $p(z)$ lie on a line or a circle.


If this wrong: When is $D\_{pu}$ doubly stochastic?




---


The paper by Khavinson et al. also gives an example of a $D\_{pu}$ which is not doubly stochastic and where the zeros do not lie on a line or circle. The authors mention questions about the column-sums of $D\_{pu}$, which are of the same nature as my own, without the connection to the Poisson-kernel and conjecture (\*), though. According to Profs. Khavinson and Pereira, the problems have not been resolved up to now.




---


The following question is now discussed more throroughly in my new question cited above, I leave it here, because I refered to it there:


Is the inequality
\begin{equation}
 \sum^l\_{k = 1} \frac{H(z'\_k)}{|z\_j - z'\_k|^2} \ \leq \ \frac{n}{M\_j}
\end{equation}
true for any complex polynomial $p(z)$ with at least two distinct zeros, without restrictions on its zeros?



The Poisson-kernel in the plane and polynomials
Here is an easy algebraic proof of your identity. I present it in the case of a simple zero, a general case requires some obvious modifications. 


So, let $z\_1$ be a simple zero of $p$. Your equation (1) is a special case of more general equation: 
$$\sum\_{l=1}^{n-1}\frac{|z\_1|^2-|z\_l'|^2}{|z\_1-z\_l'|^2}= n-1 + 2\sum\_{k=2}^n\frac{|z\_1|^2-|z\_k|^2}{|z\_1-z\_k|^2}. \tag 3$$
To prove it, we write first 
$$\frac{p''(z\_1)}{p'(z\_1)}=\sum\_{l=1}^{n-1}\frac 1{z\_1-z\_l'}$$
which is true by partial fraction decomposition of $\frac{p''(z)}{p'(z)}$. 
On the other hand, 
$$\frac{p''(z\_1)}{p'(z\_1)}=2\lim\_{z\to z\_1}\left(\frac{p'(z)}{p(z)}-\frac 1{z-z\_1}\right) = 2\sum\_{k=2}^n\frac 1{z\_1-z\_k}$$
by partial fraction decomposition of $\frac{p'(z)}{p(z)}$. 


This gives the identity 
$$\sum\_{l=1}^{n-1}\frac 1{z\_1-z\_l'}=2\sum\_{k=2}^n\frac 1{z\_1-z\_k}.$$
It remains to multiply it by $z\_1$ and to take the real part using identity $2\text{Re}\left(\frac {z\_1}{z\_1-a}\right) =1+\frac{|z\_1|^2-|a|^2}{|z\_1-a|^2}$. 


