The Geometry of the Complex Plane and the Zeros of Polynomials

Let $z_1, z_2, \ldots , z_n$ be arbitrary points lying in the finite part of the complex plane and let $m_1, m_2, \ldots , m_n$ be non-negative masses whose sum is $1$ which are placed at the points $z_1, z_2, \ldots , z_n$ respectively. In any linear transformation of the complex plane that carries the point at infinity into itself, the center of gravity $\zeta$ of such a mass-distribution is mapped by the same transformation, i.e. if we again place the point mass $m_{\nu }$ at the image-point $z_{\nu }'$ of $z_{\nu }$ ($1 \leq \nu \leq n$), then the center of gravity $\zeta '$ of this new mass distribution is identical with the image of the point $\zeta$.

In the sequel (102--156) we mean by a point'' of the complex plane an arbitrary finite point or the point at infinity, by a circle'' the circumference of a circle or a straight line---the latter passes through the point at infinity---and by a ``circular domain'' a domain whose boundary is a circle. A circular domain is either the closed interior or the closed exterior of a circle or a closed half-plane, according as it does not contain the point at infinity or contains it in its interior or on its boundary.

Let the points $z_1, z_2, \ldots , z_n, z$ and the masses $m_1, m_2, \ldots , m_n$ be given. Let $z$ be different from all the points $z_1, z_2, \ldots , z_n$ while of the latter some may coincide. Let $m_1 \geq 0, m_2 \geq 0, \ldots , m_n \geq 0, m_1+m_2+\cdots +m_n=1$.

Find a point $\zeta$ situated in such a manner that, in a linear transformation of the plane that maps the $n+2$ points

into the points

respectively, $\zeta '$ becomes the center of gravity of the mass-distribution determined by the masses $m_{\nu }$ placed at the points $z_{\nu }'$ ($\nu =1,2,3,\ldots ,n$).

There are infinitely many linear transformations that map $z$ into $\infty$; the point $\zeta$ however is independent of the choice of the underlying transformation.

The point $\zeta =\zeta_z$, defined above, which is determined uniquely by the system of points $z_1, z_2, \ldots , z_n$, by the masses $m_1, m_2, \ldots , m_n$, $m_1+m_2+\cdots +m_n=1$, placed there, and by the ``point of reference'' $z$ which must be distinct from all the points $z_{\nu }$, is called the center of gravity of the mass distribution $m_1, m_2, \ldots , m_n$ with respect to $z$. If $z$ is the point at infinity, $z=\infty$, then $\zeta_z=\zeta_{\infty }$ is the ordinary center of gravity.

Consider all possible mass distributions with total mass $1$ over the fixed points $z_1, z_2, \ldots , z_n$ and let $z$ be a point distinct from $z_1, z_2, \ldots , z_n$. The centers of gravity $\zeta_z$ of all mass distributions of this kind with respect to $z$ cover a ``circular-arc polygon'' $\mathfrak {C}_z$.

Let $\mathfrak {C}_z$ be the circular-arc polygon defined in 103 and $w_1, w_2$ two points of $\mathfrak {C}_z$. Consider the circle through $w_1, w_2$ and $z$. The arc of this circle with end-points $w_1, w_2$, that does not contain $z$, is contained in $\mathfrak {C}_z$, in other words $\mathfrak {C}_z$ is circularly-convex with respect to $z$. (We call $\mathfrak {C}_z$ the smallest circularly-convex domain with respect to $z$ that contains the points $z_1, z_2, \ldots , z_n$.)

If the points $z_1, z_2, \ldots , z_n$ all belong to a circular domain $C$ and if the point $z$ lies outside $C$, then the circular-arc polygon $\mathfrak {C}_z$ [103] lies completely inside $C$.

Let $z_1, z_2, \ldots , z_n$ be arbitrary points and $z$ a point distinct from these. Let the same mass $1/n$ be placed at each of the points $z_{\nu }$. In the sequel we consider only the center of gravity $\zeta =\zeta_z$ of this special mass distribution and call it for brevity the center of gravity of the points $z_1, z_2, \ldots , z_n$ with respect to $z$.

Let $\zeta_z$ be the center of gravity of $z_1, z_2, \ldots , z_n$ with respect to $z$. Every circle through $z$ and $\zeta_z$ separates the points $z_1, z_2, \ldots , z_n$, i.e. either both circular domains bounded by the circle contain some of the points $z_1, z_2, \ldots , z_n$ in their interior or all the points $z_1, z_2, \ldots , z_n$ lie on the circle.

If all the points $z_1, z_2, \ldots , z_n$ belong to a circular domain $C$, then the points $z$ and $\zeta_z$ cannot both lie outside $C$. If one of them, e.g. $z$, lies outside $C$, then the other, $\zeta_z$, lies in the interior of $C$, except in the case where all the points $z_1, z_2, \ldots , z_n$ coincide at a boundary point of $C$ and $\zeta_z$ then also lies there.

Let the fixed points $w_1, w_2, \ldots , w_k, z$ be pair-wise distinct. Let the complex numbers $z_1, z_2, \ldots , z_n$ take only the $k$ different values $w_1, w_2, \ldots , w_k$. The centers of gravity with respect to $z$ of all systems $z_1, z_2, \ldots , z_n$ of this kind lie everywhere dense in the smallest domain, circularly convex with respect to $z$, that contains the points $w_1, w_2, \ldots , w_k$.

The points $z_1, z_2, \ldots , z_n$ are fixed and the variable point $z$ converges to $z_1$. To what limiting position does the center of gravity $\zeta_z$ of the points $z_1, z_2, \ldots , z_n$ with respect to $z$ tend?

The center of gravity $\zeta_z$ of the finite fixed points $z_1, z_2, \ldots , z_n$ with respect to the variable point $z$ may for sufficiently large $z$ be expanded in decreasing powers of $z$. Calculate the first two terms of the expansion.

In the sequel (111--156) we shall mean by a rational integral function (polynomial) of degree $n$ a function $f(z)$ not vanishing identically of the form

Here $a_n$ is not necessarily different from $0$. In the special case when $a_n \neq 0$, we say that $n$ is the precise degree of $f(z)$. If $a_n=a_{n-1}=\cdots =a_{n-k+1}=0$, $a_{n-k} \neq 0$, we shall say that $f(z)$ has the $k$-fold zero $z=\infty$. With this interpretation every rational integral function of degree $n$ has a total of $n$ zeros, of which some may possibly lie at infinity. Thus there is determined by $f(z)$ a system of points $z_1, z_2, \ldots , z_n$, in the complex plane, the zeros of $f(z)$. Conversely, to every system of $n$ points there belongs a polynomial, whose coefficients are uniquely determined up to a constant of proportion. We shall often disregard this unimportant constant factor and speak of the polynomial, whose zeros are $z_1, z_2, \ldots , z_n$.

By the center of gravity $\zeta$ of a polynomial we mean the center of gravity of its zeros. Express the center of gravity of $f(z)$ with respect to a variable point $z$: (1) in terms of $f(z)$ and $f'(z)$, (2) in terms of the coefficients of $f(z)$. Consider the special cases $z=\infty$ and $f(z)=z^n$.

Let $\zeta$ be the center of gravity of $f(z)$ with respect to $z$. A necessary and sufficient condition that $f(z)$ has only real zeros is that the imaginary parts of $z$ and $\zeta$ have opposite signs if they do not both vanish however $z$ may vary. ($z=\infty$ counts as a real zero.)

With respect to which points $z$ of the plane does the center of gravity of $f(z)$ lie at infinity? What do the statements in 106 and 107 say about these points?

Let $f(z)$ be a polynomial of precise degree $n$, $C$ a circular domain that contains the zeros of $f(z)$ and $c \neq 0$. The zeros of the derivative of the transcendental function $e^{-z/c}f(z)$ lie either in $C$ or in $C+nc$, i.e. in the circular domain that is obtained from $C$ by parallel displacement by the vector $nc$.

Let $z_1$ be a zero of the polynomial $f(z)$ of degree $n$, $x$ finite and $x \neq z_1$, and $f'(x)=0$. Then $f(z)$ has at least one zero in every circle that passes through both the points

Let $f(z)$ be a polynomial of degree $n$, whose zeros are all of absolute value $\geq 1$. Let $\alpha_1$ and $\alpha_2$ be arbitrary positive numbers, such that $n\alpha_1/\alpha_2 \neq 1$. Then all the zeros of $\alpha_1zf'(z)-\alpha_2f(z)$ are in absolute value

Let $f(z)$ be a polynomial of degree $n$, all of whose zeros lie in the annulus $r \leq \left|z\right| \leq R$. Let $\alpha_1$ and $\alpha_2$ be arbitrarily positive numbers such that $n\alpha_1/\alpha_2 \neq 1$. Then all the zeros of $\alpha_1zf'(z)-\alpha_2f(z)$ lie in the annulus

If $z_1$ is a finite zero of the polynomial $f(z)$ of degree $n$, then the center of gravity with respect to $z_1$ of the remaining $n-1$ zeros of $f(z)$ is

How is this formula to be modified if $z_1$ is at infinity?

The $n$th Hermite polynomial satisfies the differential equation

(This is the differential equation satisfied by the Hermite polynomial as defined via $e^{-z^2/2}H_n(z) = \tfrac {1}{n!}\tfrac {d^n}{dz^n}e^{-z^2/2}$ in problem 100(g) of Part Six's solution.) Show by means of 118 that the Hermite polynomials have only real zeros (problem 100(i) of Part Six's solution). [Consider the assumed complex zero with largest imaginary part.]

The polynomial of degree $n$ that satisfies the differential equation

(the $n$th Legendre polynomial, cf. problem 90 of Part Six), has only real zeros (problem 97 of Part Six).

Gauss's theorem (the circular-domain formulation given in the introduction to problem 31 of Part Three) is equivalent to the following: If all the zeros of a polynomial $f(z)$ lie in a circular disc $C$, then the zeros of its derivative $f'(z)$ also lie in $C$ [113]. Decide whether the following theorem is true or false: If all the zeros of a polynomial $f(z)$ lie in two circular discs $C_1, C_2$, then the zeros of its derivative $f'(z)$ also lie in $C_1$ or $C_2$.

Let $C_1$ and $C_2$ be two circular discs or, more generally, two arbitrary circular domains. Denoting by $n_1$ and $n_2$ fixed positive numbers, we call the set of all points

where $z_1$ and $z_2$ independently run through $C_1$ and $C_2$ respectively, a mean domain of $C_1$ and $C_2$, and use for it the following notation

$C$ is completely determined by $C_1, C_2$ and $n_1, n_2$. [Cf. H. Minkowski: Werke, Vol. 2, p. 176, Leipzig: B. G. Teubner 1911.]

Let $C_1$ and $C_2$ be two circular discs whose centers are $z_1^{(0)}, z_2^{(0)}$ and radii $r_1, r_2$ respectively. Show that then

is again a circular disc whose center $z^{(0)}$ and radius $r$ are given by the following equations:

$C$ is homothetic to $C_1$ and $C_2$ with respect to the same center of similitude. [H. S. M. Coxeter: Introduction to Geometry. New York: John Wiley Sons 1961, pp. 68--70.]

Let $C_1$ and $C_2$ be two half-planes with parallel boundary lines, one of which contains the other. The mean domain $C=n_1C_2+n_2C_1/(n_1+n_2)$ is then again a half-plane, whose boundary line runs parallel to those of $C_1$ and $C_2$ and divides the distance between them in the ratio $n_1:n_2$.

Let $f(z)=f_1(z)f_2(z)$, and $f_1(z)$ be a polynomial of degree $n_1$, $f_2(z)$ a polynomial of degree $n_2$. Let all the zeros of $f_1(z)$ lie in the circular domain $C_1$ and all the zeros of $f_2(z)$ in the circular domain $C_2$. Then the zeros of the derivative $f'(z)$ lie either in $C_1$ or in $C_2$ or in the mean domain

(By degree we mean the precise degree.)

Let $f(z)$ be a rational (non-integral) function, whose numerator and denominator have no common factors. Let it have $n_1$ zeros (poles) which lie in the circular domain $C_1$ and $n_2$ poles (zeros) which lie in the circular domain $C_2$. (Only finite zeros and poles are considered.) Let $n_1 \geq n_2$. Then the zeros of $f'(z)$ lie either in $C_1$ or in $C_2$ or in the domain $C$ defined by the numbers

where $z_1$ and $z_2$ independently run through $C_1$ and $C_2$, for which we use the notation

If $n_1=n_2$ and the circular domains $C_1$ and $C_2$ have no common points, then all the zeros of $f'(z)$ lie either in $C_1$ or in $C_2$.

Let $f(z)$ be a polynomial, and let the equation

have all its roots in a convex domain (oval) $\mathfrak {O}_1$. Then all the numbers $a$ for which this is the case also lie in a convex domain $\mathfrak {O}_2$.

Let $f(z)$ be a polynomial and let its $a$-points (the zeros of $f(z)-a$) lie in a circular domain $C_1$ and its $b$-points in a circular domain $C_2$. If $n_1, n_2$ are positive numbers and we set $c=(n_1b+n_2a)/(n_1+n_2)$, those $c$-points of $f(z)$ that lie neither in $C_1$ nor in $C_2$, lie in the mean domain

[We may take $n_1, n_2$ to be integers.]

By the derivative of a polynomial $f(z)$ of degree $n$ with respect to the point $\zeta$, denoted by $A_{\zeta }f(z)$, we mean the ordinary derivative if $\zeta =\infty$, and

if $\zeta$ is finite. $A_{\zeta }f(z)$ is a polynomial of degree $n-1$. (When using homogeneous coordinates this expression is called the first polar, or, by older authors, emanant; cf. Laguerre, Oeuvres, Vol. 1, p. 48. Paris: Gauthier-Villars, 1898.)

Let the $n+1$ coefficients of $f(z)$ (in the notation of §2 preceding 111) be $a_0, a_1, a_2, \ldots , a_n$. What are the $n$ coefficients of $[A_{\zeta }f(z)]/n$?

Let $\zeta$ be arbitrary. Show that the operation $A_{\zeta }f(z)$ is distributive: If we denote by $f_1(z)$ and $f_2(z)$ two polynomials of degree $n$, and by $c_1$ and $c_2$ arbitrary constants, then we have

Let $\zeta$ be arbitrary and $f(z)$ a polynomial of degree $n$, $f(z)=g(z)h(z)$ a factorization of $f(z)$ into the two factors $g(z)$ and $h(z)$ of degree $k$ and $l$ respectively, $k+l=n$. Show that

Let $f(z)$ be a polynomial of degree $n$, $\zeta_1$ and $\zeta_2$ two arbitrary constants. The operations $A_{\zeta_1}f(z)$ and $A_{\zeta_2}f(z)$ are commutative, that is to say

Show that $A_{\zeta }f(z)$ vanishes identically if and only if all the zeros of $f(z)$ coincide with $\zeta$.

By the derived system of the $n$ points $z_1, z_2, \ldots , z_n$ with respect to an $(n+1)$th point $\zeta$ we mean the $n-1$ zeros $z_1', z_2', \ldots , z_{n-1}'$ of $A_{\zeta }f(z)$, where $f(z)$ is the polynomial of degree $n$ whose zeros are $z_1, z_2, \ldots , z_n$, all the zeros being correctly counted (§2, preceding 111). The derived system is completely determined, except in the single exceptional case where the $n+1$ points $z_1, z_2, \ldots , z_n, \zeta$ all coincide (cf. 132).

If $\zeta$ is finite, then to the derived system with respect to $\zeta$ in general there belong four different kinds of points:

(1) The finite points distinct from $z_1, z_2, \ldots , z_n, \zeta$, with respect to which the center of gravity of $z_1, z_2, \ldots , z_n$ is precisely $\zeta$.

(2) The multiple zeros of $f(z)$, distinct from $\zeta$, each with multiplicity reduced by $1$.

(3) $\zeta$ itself only if it is a zero of $f(z)$ and in that case with the same multiplicity.

(4) The point $\infty$ if it is at least a double zero of $f(z)$; moreover also if it is not a zero of $f(z)$ but if $\zeta$ is the ordinary center of gravity of $f(z)$, i.e. the one constructed with respect to $\infty$.

A circular domain, whose boundary contains $\zeta$ as well as a point of the derived system with respect to $\zeta$, itself contains points of the original system.

If a circular domain contains the points $z_1, z_2, \ldots , z_n$ and does not contain the point $\zeta$, then it contains also the derived system of $z_1, z_2, \ldots , z_n$ with respect to $\zeta$.

The derived system of the points $z_1, z_2, \ldots , z_n$ lies in the smallest circular-arc polygon, circularly convex with respect to $\zeta$, that contains $z_1, z_2, \ldots , z_n$. (Generalization of Problem 31 of Part Three.)

If $f(z)$ is of degree $n$, then $(A_{\zeta })^{n-1}f(z)$ is of the first degree. Calculate the only zero of $(A_{\zeta })^{n-1}f(z)$!

Let

be an arbitrary polynomial of degree $n$ and let $\zeta_1, \zeta_2, \ldots , \zeta_n$ be arbitrary points in the complex plane. The expression

is a symmetric multilinear function of the variables $\zeta_1, \zeta_2, \ldots , \zeta_n$, which we can write as follows:

Here $\Sigma_0, \Sigma_1, \Sigma_2, \ldots , \Sigma_{n-1}, \Sigma_n$ are the elementary symmetric functions of $\zeta_1, \zeta_2, \ldots , \zeta_n$:

If the last $k$ of the numbers $\zeta_1, \zeta_2, \ldots , \zeta_n$ (and only these) lie at infinity, then we set $\Sigma_0=\Sigma_1=\cdots =\Sigma_{k-1}=0$, $\Sigma_k=1$, $\Sigma_{k+1}=\zeta_1+\zeta_2+\cdots +\zeta_{n-k}$, $\Sigma_{k+2}=\zeta_1\zeta_2+\zeta_1\zeta_3+\cdots +\zeta_{n-k-1}\zeta_{n-k}, \ldots , \Sigma_n=\zeta_1\zeta_2\ldots \zeta_{n-k}$.

Let

be an arbitrary polynomial of degree $n$ with zeros $z_1, z_2, \ldots , z_n$ and

an arbitrary polynomial of degree $n$ with zeros $\zeta_1, \zeta_2, \ldots , \zeta_n$. Calculate $A(\zeta_1, \zeta_2, \ldots , \zeta_n)f(z)$ and $A(z_1, z_2, \ldots , z_n)g(z)$.

If the two polynomials are related in such a manner that

or, equivalently [solution 139], $A(z_1, z_2, \ldots , z_n)g(z)=0$, then the two polynomials $f(z)$ and $g(z)$ are termed apolar. The term suggests the vanishing of the $n$th polar $A_{\zeta_1}A_{\zeta_2}\ldots A_{\zeta_n}f(z)$. The condition of apolarity is

[139.] The two systems $z_1, z_2, \ldots , z_n$ and $\zeta_1, \zeta_2, \ldots , \zeta_n$ are also termed mutually apolar.

What is the geometrical significance of the apolarity of $z_1, \zeta_1$, and that of $z_1, z_2$ and $\zeta_1, \zeta_2$?

Which are the systems $\zeta_1, \zeta_2, \ldots , \zeta_n$, that are apolar to the system consisting of the roots of the equation $z^n-1=0$?

Let $z_1, z_2, \ldots , z_n$ be an arbitrary system of points. There are $n$ systems $\zeta ,\zeta ,\ldots ,\zeta$ of coincident points that are apolar to the given system. Which are they?

The polynomial of degree $n$

is apolar to any system of numbers whose sum is $n$ and product is $0$.

Let $f(z)$ be an arbitrary polynomial of degree $n$, all of whose zeros lie in the circular domain $C$. If $\zeta_1, \zeta_2, \ldots , \zeta_k$, $k<n$, are arbitrary points lying outside $C$, then all the zeros of the polynomial of degree $n-k$, $A_{\zeta_1}A_{\zeta_2}\ldots A_{\zeta_k}f(z)$ lie in $C$.

Let the two polynomials $f(z)$ and $g(z)$ of degree $n$ be apolar to each other. Every circular domain that contains all the zeros of one polynomial also contains at least one zero of the other.

Let the two polynomials $f(z)$ and $g(z)$ of degree $n$ be apolar to each other. The two smallest convex polygons that contain all the zeros of $f(z)$ and $g(z)$ respectively have at least one common point. More generally, the two smallest circular-arc polygons that are circularly convex with respect to the same point and that contain all the zeros of $f(z)$ and $g(z)$ respectively, also have at least one common point.

The polynomial

always has one zero in the circle $\left|z\right| \leq 2$.

The polynomial

always has one zero in the circle $\left|z-1\right| \leq 1$.

The polynomial of $k+1$ terms

always has one zero in the circle

and hence in the circle

(Generalization of 147.)

If a polynomial of degree $n$ assumes the same value at two points $a$ and $b$, $a \neq b$, then its derivative has at least one zero in the circular disc with center at the midpoint of the line segment $ab$ and radius $(\left|a-b\right|/2)\cot (\pi /n)$. (Analogue of Rolle's theorem in the complex domain.)

Let

be a polynomial of degree $n$ all of whose zeros lie in a circular domain $C$. Also let

be a polynomial of degree $n$ with zeros $\beta_1, \beta_2, \ldots , \beta_n$. Every zero $\gamma$ of the polynomial

formed by ``composition'' of $f(z)$ and $g(z)$ then has the form

where $\nu$ denotes a suitably chosen index and $k$ a suitably chosen point in $C$. (Here we must set $\infty \cdot \infty = \infty$ and $0 \cdot \infty =$ indeterminate.)

Let the zeros of the polynomials $f(z)$ and $g(z)$ of degree $n$ all lie in the unit circle $\left|z\right| \leq 1$, and the zeros of at least one of the two polynomials in fact in $\left|z\right|<1$. Then also all the zeros of the polynomial of degree $n$ formed from $f(z)$ and $g(z)$ by composition [151] lie in the interior of the unit circle $\left|z\right|<1$.

Let the polynomial $f(z)$ of degree $n$ have all its zeros in a convex domain $\mathfrak {C}$ that contains the origin. Let the polynomial $g(z)$ of the same degree have only real zeros which lie in the interval $-1, 0$. The polynomial $h(z)$ of degree $n$ derived from the two polynomials by composition [151] then also has all its zeros in $\mathfrak {C}$.

If all the zeros of the polynomial

of degree $n$ lie in the real interval $-a, a$, and those of the polynomial