# The mean value problem of Smale’s problems

Lande Ma School of Mathematical Sciences, Tongji University, Shanghai, 200092,
China<EMAIL_ADDRESS>and  ZhaoKun Ma  YanZhou College, ShanDong Radio and
TV University, YanZhou, ShanDong 272100 China<EMAIL_ADDRESS>
(Date: December 13th, 2021.)

###### Abstract.

We show two results of mean value problem, Smale’s mean value problem is
comprehensively solved in this paper.

###### Key words and phrases:

Smale’s problems; mean value problem; critical point; root locus; zeros;

###### 2020 Mathematics Subject Classification:

Primary 30C15,30C10 Secondary 30D30,93C05

## Introduction

In 1998, Smale proposed some mathematics problems which need to be solved[1]
to a request from Vladimir Arnold, to propose a list of problems for the 21st
century. The mean value problem on the list is still open in full generality.
Early in 1981, Smale posed the mean value problem of complex polynomial[2]:
Given a complex polynomial $f$ of degree $d\geq 2$ and a complex number $s$,
is there a critical point $\theta$ of $f$(ie, $f^{{}^{\prime}}(\theta)=0$)
such that $\frac{\lvert f(s)-f(\theta)\rvert}{\lvert s-\theta\rvert}\leq
c\lvert f^{{}^{\prime}}(s)\rvert$, $c=1$. Mathematicians have obtained some
partial results[3, 4, 5].

In this paper, through the root locus method established by us. We prove two
results: Given a complex polynomial $f$ of degree $d\geq 2$ and a complex
number $s$. For every critical point $\theta_{i}$ (ie,
$f^{{}^{\prime}}(\theta_{i})=0$), $i=1,2,\cdots,n-1$, there exists its
adjacent domain $\Omega_{i}$. For all points in the extended complex plane
except the adjacent domain $\Omega=\\{\Omega_{i}\\}$, $i=1,2,\cdots,n-1$, such
that $\frac{\lvert f(s)-f(\theta_{i})\rvert}{\lvert
s-\theta_{i}\rvert}\leq\lvert f^{{}^{\prime}}(s)\rvert$.

Namely, the inequality in Smale’s mean value problem is true.

Given a complex polynomial $f$ of degree $d\geq 2$ and a complex number $s$.
For every critical point $\theta_{i}$ (ie, $f^{{}^{\prime}}(\theta_{i})=0$),
$i=1,2,\cdots,n-1$, there exists its adjacent domain $\Omega_{i}$. For all
points in the adjacent domain $\Omega=\\{\Omega_{i}\\}$, $i=1,2,\cdots,n-1$,
such that $\frac{\lvert f(s)-f(\theta_{i})\rvert}{\lvert
s-\theta_{i}\rvert}>\lvert f^{{}^{\prime}}(s)\rvert$.

Namely, the inequality in Smale’s mean value problem is true, when symbol in
inequality is converse.

##  1\. Root Locus Of the meromorphic function $W(s)$ and Their Properties

In textbooks of _automatic control theory_ , the factor at the left side of
the root locus equation is the rational fraction function of the constant
coefficient. The root locus equation is only concerning two degree of 0 and
180 degree numbers which obtains the real number values[6]. So, the root locus
equations and the results of the root locus in _automatic control theory_ are
all very special and limitted. The proofs of results in _automatic control
theory_ are not comprehensive and accurate.

We need to study the more general root locus equations of the the meromorphic
function $W(s)$. Let
$K=|\frac{\prod^{n}_{j=1}(1-\frac{s}{p_{j}})^{\beta_{j}}GP_{j}(s)}{G(s)\prod^{m}_{l=1}(1-\frac{s}{z_{l}})^{\gamma_{l}}GZ_{l}(s)}|$.
The $K$ is the reciprocal of the modulus of meromorphic function
$W(s)=u(\sigma,t)+iv(\sigma,t)$. After $K$ and meromorphic function $W(s)$ are
multiplied, the product result is the unit complex value of meromorphic
function $W(s)$.

######  Definition 1.1.

In the extended complex plane $\mathbb{C}\cup\\{\infty\\}$, the equation (1.1)
is called as _the root locus equation_.

(1.1)
$KG(s)\frac{\prod^{m}_{l=1}(1-\frac{s}{z_{l}})^{\gamma_{l}}G_{lz}(s)}{\prod^{n}_{j=1}(1-\frac{s}{p_{j}})^{\beta_{j}}G_{jp}(s)}=a+ib$

In which, $a+ib$ is the unit complex number value of meromorphic function
$W(s)$. $\alpha=2q\pi+\arg(\frac{b}{a})$. The zeros $z_{l}$ and poles $p_{j}$
are points in $\mathbb{C}\cup\\{\infty\\}$, and may be no conjugate.

######  Lemma 1.2.

After the coincident and finite zeros and poles of Eq(1.1) are cancelled, for
any finite zero of Eq(1.1), its $K$ value is $K=+\infty$; for any finite pole
of Eq(1.1), its $K$ value is $K=0$.

###### Proof.

The Eq(1.1) can be transformed to the following characteristic equation.

$KG(s)\prod^{m}_{l=1}(1-\frac{s}{z_{l}})^{\gamma_{l}}G_{lz}(s)-(a+ib)\prod^{n}_{j=1}(1-\frac{s}{p_{j}})^{\beta_{j}}G_{jp}(s)=0$.
Obviously, if $K=0$, all roots of the pole factor
$\prod^{n}_{j=1}(1-\frac{s}{p_{j}})^{\beta_{j}}G_{jp}(s)$ are the roots of
characteristic equation. Conversely, all the roots of characteristic equation
if $K=0$ are all of the roots of the pole factor. So, at poles of Eq(1.1),
$K=0$.

The Eq(1.1) can be transformed to its another characteristic equation.
$G(s)\prod^{m}_{l=1}$$(1-\frac{s}{z_{l}})^{\gamma_{l}}G_{lz}(s)-\frac{a+ib}{K}\prod^{n}_{j=1}(1-\frac{s}{p_{j}})^{\beta_{j}}G_{jp}(s)=0$.
If $K=+\infty$, all roots of the zero factor
$\prod^{m}_{l=1}(1-\frac{s}{z_{l}})^{\gamma_{l}}G_{lz}(s)$ are the roots of
the characteristic equation. Conversely, all roots of the characteristic
equation when $K=+\infty$ are the roots of the zero factor. So, at zeros of
Eq(1.1), $K=+\infty$. ∎

According to the expression of $K$ of the root locus equation (1.1) and
according to Lemma 1.2, it is obvious that the $K$ values are continuous
concerning the complex variable $s$ in $\mathbb{C}\cup\\{\infty\\}$. So, we
can prove Theorem 1.3 simply.

######  Theorem 1.3.

Let $s\in\mathbb{C}\cup\\{\infty\\}$, the $K$ value of
$s\in\mathbb{C}\cup\\{\infty\\}$ takes all of the non-negative real number
value from 0 to the $+\infty$.

In complex analysis, the phase angle is namely argument[7]. In _automatic
control theory_ , it is called as the phase angle. Here, we need to use name
of the phase angle. In this paper, $K$ is called as the gain.

Let $\Delta=\\{s=\sigma+it\in\mathbb{C}\cup\\{\infty\\}$, $s$ is not zero or
pole of $W(s)$ and $Eq(1.1)\\}$.

######  Lemma 1.4.

For any point $s\in\Delta$, the phase angle of meromorphic function $W(s)$ can
be written as the $\varphi(\sigma,t)$ function:
$\varphi(\sigma,t)=\arg(\frac{G_{y}(\sigma,t)}{G_{x}(\sigma,t)})+\sum^{m}_{l=1}(\gamma_{l}\arg(\frac{t-t_{l}}{\sigma-\sigma_{l}})-\gamma_{l}\arg(\frac{t_{l}}{\sigma_{l}})+\arg(\frac{G_{zy}(\sigma,t)}{G_{zx}(\sigma,t)}))-\sum^{n}_{j=1}(\beta_{j}\arg(\frac{t-t_{j}}{\sigma-\sigma_{j}})-\beta_{j}\arg(\frac{t_{j}}{\sigma_{j}})+\arg(\frac{G_{py}(\sigma,t)}{G_{px}(\sigma,t)}))$.

Let $s=(\sigma,t)\in\Delta$, then the phase angle condition equation of
Eq(1.1) on $s$ is:

(1.2) $\varphi(\sigma,t)=2q\pi+arg{(b/a)}$

In which, $q$ is an integer number.

######  Lemma 1.5.

Let Eq(1.1) be the root locus equation of $2q\pi+\alpha$ degree, for any point
$s=(\sigma,t)\in\Delta$, if the point $s$ satisfies the phase angle condition
equation (1.2) of $2q\pi+\alpha$ degree, then it must be on the root locus of
$2q\pi+\alpha$ degree of Eq(1.1).

###### Proof.

Assuming that a point $s_{1}=(\sigma_{1},t_{1})\in\Delta$ satisfies the phase
angle condition equation (1.2),
$\varphi(\sigma_{1},t_{1})=2q\pi+arg{(b_{1}/a_{1})}$. The phase angle of
$W(s)$ is $\varphi(\sigma,t)$, this phase angle expression is same as the
expression of the left side of Eq(1.2).

Because the point $s_{1}$ satisfies the phase angle condition equation (1.2),
when the point $s_{1}$ is substituted into the function $W(s)$, $W(s)$ obtains
a complex value, according to the phase angle expression of Eq(1.2) of the
point $s_{1}$ and the condition which a point $s_{1}$ satisfies the phase
angle condition equation (1.2), this complex value can be written as:
$K_{1}^{*}(a_{1}+ib_{1})$, and $\alpha_{1}=\arg{(b_{1}/a_{1})}$, $K_{1}^{*}$
is the modulus of function $W(s)$ of point $s_{1}$. $K_{1}^{*}$ is a non-zero
positive real value. $\alpha_{1}$ is the phase angle of the function $W(s)$ of
the point $s_{1}$, and it is the value of the right side of Eq(1.2).

If we bring the point $s_{1}$ into the gain expression
$|\frac{\prod^{n}_{j=1}(1-\frac{s}{p_{j}})^{\beta_{j}}G_{jp}(s)}{G(s)\prod^{m}_{l=1}(1-\frac{s}{z_{l}})^{\gamma_{l}}G_{lz}(s)}|$,
according to the definition of the gain, we can obtain an unique gain $K_{1}$,
$K_{1}=|\frac{\prod^{n}_{j=1}(1-\frac{s_{1}}{p_{j}})^{\beta_{j}}G_{jp}(s_{1})}{G(s_{1})\prod^{m}_{l=1}(1-\frac{s_{1}}{z_{l}})^{\gamma_{l}}G_{lz}(s_{1})}|$,
the gain $K_{1}$ is a reciprocal of the modulus $K_{1}^{*}$ of $W(s)$ of the
point $s_{1}$. Further obtain $K_{1}^{*}$. $K_{1}^{*}=1/K_{1}$. For that the
gain $K_{1}$ multiplied the expression
$G(s_{1})\prod^{m}_{l=1}(1-\frac{s_{1}}{z_{l}})^{\gamma_{l}}G_{lz}(s_{1})/\prod^{n}_{j=1}(1-\frac{s_{1}}{p_{j}})^{\beta_{j}}G_{jp}(s_{1})$,
we have,
$K_{1}G(s_{1})\prod^{m}_{l=1}(1-\frac{s_{1}}{z_{l}})^{\gamma_{l}}G_{lz}(s_{1})/\prod^{n}_{j=1}(1-\frac{s_{1}}{p_{j}})^{\beta_{j}}G_{jp}(s_{1})=a_{1}+ib_{1}$.
This equation is a concrete situation which a point $s_{1}$ satisfies Eq(1.1).
The previous results prove that the point $s_{1}$ is a root of Eq(1.1).

So, the point $s_{1}$ satisfies Eq(1.1). Therefore, it is proved that the
point $s_{1}$ is on the root locus of Eq(1.1), and which its gain is $K_{1}$
and its degree is $2q\pi+\alpha_{1}$. Hence, if the point $s_{1}$ satisfies
Eq(1.2), then the point $s_{1}$ is on the root locus of Eq(1.1), and which its
gain is $K_{1}$ and its degree is $2q\pi+\alpha_{1}$. ∎

######  Lemma 1.6.

Let Eq(1.1) be the root locus equation of $2q\pi+\alpha$ degree, for any point
$s=(\sigma,t)\in\Delta$, if the point $s$ is on the root locus of
$2q\pi+\alpha$ degree of Eq(1.1), then it must satisfy the phase angle
condition equation of $2q\pi+\alpha$ degree of Eq(1.2).

###### Proof.

Assume that the point $s_{2}=(\sigma_{2},t_{2})\in\Delta$ is an arbitrary
point on the root locus of Eq(1.1). The points which satisfy Eq(1.1) are all
on the root locus of Eq(1.1). So, the point $s_{2}$ surely satisfies Eq(1.1).
When $s_{2}$ is substituted into Eq(1.1),
$K_{2}G(s_{2})\prod^{m}_{l=1}(1-\frac{s_{2}}{z_{l}})^{\gamma_{l}}G_{lz}(s_{2})/\prod^{n}_{j=1}(1-\frac{s_{2}}{p_{j}})^{\beta_{j}}G_{jp}(s_{2})=a_{2}+ib_{2}$.
According to Theorem 1.3, here, the gain $K_{2}$ is a positive real number,
its phase angle of $K_{2}$ is $2q_{1}\pi$, the phase angle of the right side
of the equation in this paragraph on the point $s_{2}$ is
$2q_{2}\pi+\arg{(b_{2}/a_{2})}=2q_{2}\pi+\alpha_{2}$.

The factor of the left side of the equation in last paragraph on the point
$s_{2}$ can be looked as two factors. One is $K_{2}$, another is
$G(s_{2})\prod^{m}_{l=1}(1-\frac{s_{2}}{z_{l}})^{\gamma_{l}}G_{lz}(s_{2})/\prod^{n}_{j=1}(1-\frac{s_{2}}{p_{j}})^{\beta_{j}}G_{jp}(s_{2})$.
The phase angle of the factor of the left side of the equation in last
paragraph on the point $s_{2}$ is equal to the summation of two phase angles
of two factors of $K_{2}$ and
$G(s_{2})\prod^{m}_{l=1}(1-\frac{s_{2}}{z_{l}})^{\gamma_{l}}G_{lz}(s_{2})/\prod^{n}_{j=1}(1-\frac{s_{2}}{p_{j}})^{\beta_{j}}G_{jp}(s_{2})$.
The phase angle of the factor of the left side of the equation in last
paragraph on the point $s_{2}$ is equal to the phase angle of the right side
of the equation in the last paragraph on the point $s_{2}$. The phase angle
$2q_{2}\pi+\alpha_{2}$ subtracts the phase angle $2q_{1}\pi$ is equal to
$2q_{2}\pi+\alpha_{2}-2q_{1}\pi=2q\pi+\alpha_{2}$.

The difference of the phase angle of the factor $a_{2}+ib_{2}$ and the factor
$K_{2}$ is: $2q\pi+\alpha_{2}$. So , the phase angle of expression
$G(s_{2})\prod^{m}_{l=1}(1-\frac{s_{2}}{z_{l}})^{\gamma_{l}}G_{lz}(s_{2})/\prod^{n}_{j=1}(1-\frac{s_{2}}{p_{j}})^{\beta_{j}}G_{jp}(s_{2})$
is $2q\pi+\alpha_{2}=2q\pi+\arg{(b_{2}/a_{2})}$. We can obtain: the phase
angle of $W(s)$ is
$\varphi(\sigma,t)=\arg(\frac{G_{y}(\sigma,t)}{G_{x}(\sigma,t)})+\sum^{m}_{l=1}(\gamma_{l}\arg(\frac{t-t_{l}}{\sigma-\sigma_{l}})-\gamma_{l}\arg(\frac{t_{l}}{\sigma_{l}})+\arg(\frac{G_{zy}(\sigma,t)}{G_{zx}(\sigma,t)}))-\sum^{n}_{j=1}(\beta_{j}\arg(\frac{t-t_{j}}{\sigma-\sigma_{j}})-\beta_{j}\arg(\frac{t_{j}}{\sigma_{j}})+\arg(\frac{G_{py}(\sigma,t)}{G_{px}(\sigma,t)}))$.
So, according to the previous proof, we can obtain:
$\varphi(\sigma_{2},t_{2})=2q\pi+arg{(b_{2}/a_{2})}$, the point $s_{2}$
satisfies the phase angle condition equation (1.2) . Hence, if the point
$s_{2}$ is on the root locus of Eq(1.1), which its gain is $K_{2}$ and its
degree is $2q\pi+\alpha_{2}$, and satisfies Eq(1.1), then point $s_{2}$
satisfies Eq(1.2). ∎

Lemma 1.5 gives the sufficient condition result of Theorem 1.7. Lemma 1.6
gives the necessary condition result of Theorem 1.7. So, sum up Lemma 1.5 and
Lemma 1.6, we can obtain the following theorem.

######  Theorem 1.7.

Let Eq(1.1) be $2q\pi+\alpha$ degree and $s=(\sigma,t)\in\Delta$ be an
arbitrary point. A necessary and sufficient condition for that the point $s$
is on the root locus of $2q\pi+\alpha$ degree of Eq(1.1) is that point $s$
whether or not satisfies the phase angle condition equation (1.2).

######  Definition 1.8.

Let $\Xi$ be a point set in $\mathbb{C}\cup\\{\infty\\}$, that are consist of
all of points on the path of roots of Eq(1.1) traced out in
$\mathbb{C}\cup\\{\infty\\}$ as $2q\pi+\alpha=2q\pi+\arg(\frac{b}{a})$ is a
constant. That path of Eq(1.1) in $\mathbb{C}\cup\\{\infty\\}$ is called as
_the root locus of Eq(1.1)_. Namely, the set $\Xi$ is _the root locus of the
$2q\pi+\alpha$ degree_.

The subset of $\Xi$ is part or entire root locus except zeros and poles. And
the empty set $\emptyset$ is the case when zero and pole are coincident, the
case has no root locus. Let $\tau$ be the collection of subsets of $\Xi$, the
ordered pair $(\Xi,\tau)$ satisfying the following properties: $\emptyset$ and
$\Xi$ itself are both open in $\tau$, the intersection of any two open sets is
open in $\tau$, and the union of every collection of open sets is open in
$\tau$. So, the collection $\tau$ is a topology on $\Xi$, and the ordered pair
$(\Xi,\tau)$ is a topological space.

The meromorphic function $W(s)$ is the factor at Eq(1.1)’s left side. So, the
$\varphi(\sigma,t)$ function which is in Lemma 1.4 can be used to compute the
degree of the root locus of Eq(1.1). The factor at the right side of Eq(1.1)
is the computing value.

######  Definition 1.9.

When the phase angle of meromorphic function $W(s)$ at the left side of the
root locus equation (1.1) is $2q\pi+\alpha$ degree, we call the root locus
equation (1.1) as _the $2q\pi+\alpha$ degree root locus equation_.

In $\mathbb{C}\cup\\{\infty\\}$, the degree number of the phase angle of
meromorphic function $W(s)$ at the left side of the root locus equation (1.1)
is the degree number of the root locus of the root locus equation (1.1). So,
we can give a definition of the degree of the root locus of Eq(1.1).

######  Definition 1.10.

When the phase angle of meromorphic function $W(s)$ at the left side of the
root locus equation (1.1) is $2q\pi+\alpha$ degree, we call the root locus of
the root locus equation (1.1) as _$2q\pi+\alpha$ degree root locus_.

######  Lemma 1.11.

For all finite zeros of Eq(1.1), they are on the root locus of all degree
numbers of Eq(1.1) from $2q\pi$ degree to $2q\pi+2\gamma_{l}\pi$ degree.

###### Proof.

Assuming that the point $zp$ is an arbitrary finite zeros of Eq(1.1). When the
point $zp$ is substituted into meromorphic function $W(s)$ at the left side of
Eq(1.1), we have:
$G(zp)\frac{\prod^{m}_{l=1}(1-\frac{zp}{z_{l}})^{\gamma_{l}}G_{lz}(zp)}{\prod^{n}_{j=1}(1-\frac{zp}{p_{j}})^{\beta_{j}}G_{jp}(zp)}=0$.
This equation can also be expressed as:
$G(zp)\frac{\prod^{m}_{l=1}(1-\frac{zp}{z_{l}})^{\gamma_{l}}G_{lz}(zp)}{\prod^{n}_{j=1}(1-\frac{zp}{p_{j}})^{\beta_{j}}G_{jp}(zp)}=K_{zp}e^{i\theta_{zp}}=0$.
In this equation, $K_{zp}$ is the modulus of the function $W(s)$ and
$\theta_{zp}$ is the phase angle of the function $W(s)$. So, $K_{zp}=0$, and
$\theta_{zp}=2q\pi+\alpha$,
$K_{zp}e^{i(2q\pi+\alpha)}=0*(cos(\theta_{zp})+isin(\theta_{zp}))=0$ is true,
and $\theta_{zp}$ is an arbitrary degree number from $2q\pi$ degree to
$2q\pi+2\gamma_{l}\pi$ degree.

For the point $zp$ that lets the left side of Eq(1.1) obtain 0, no matter what
phase angle it is, since its modulus is 0.
$cos(\theta_{zp})+isin(\theta_{zp})$ is the non-zero and non-infinity, the
modulus of the factor $cos(\theta_{zp})+isin(\theta_{zp})$ is 1. $\theta_{zp}$
represents an arbitrary degree number, which shows: no matter what value
$\theta_{zp}$ is, there is
$K_{zp}e^{i(2q\pi+\alpha)}=0*(\cos(\theta_{zp})+i\sin(\theta_{zp}))=0$.

This proves: For the finite zero $zp$ of Eq(1.1), when it is substituted into
the meromorphic function $W(s)$, for the phase angle of the arbitrary
$2q\pi+\alpha$ degree number of the meromorphic function $W(s)$, its values
are all equal to 0. According to Definition 1.10, the phase angle of the
meromorphic function $W(s)$ is namely the phase angle of the root locus of
Eq(1.1). Because $2q\pi+\alpha$ is an arbitrary degree number, when it obtains
all of degree numbers, this shows the finite zeros of Eq(1.1) are
simultaneously on the root locus of all of degree numbers of Eq(1.1) from
$2q\pi$ degree to $2q\pi+2\gamma_{l}\pi$ degree. ∎

######  Lemma 1.12.

For all finite poles of Eq(1.1), they are on the root locus of all degree
numbers of Eq(1.1) from $2q\pi$ degree to $2q\pi+2\beta_{j}\pi$ degree.

###### Proof.

Assuming that the point $pz$ is an arbitrary finite pole of Eq(1.1). When the
point $pz$ is substituted into the meromorphic function $W(s)$, we can obtain
a value
$Gpz=G(pz)\frac{\prod^{m}_{l=1}(1-\frac{pz}{z_{l}})^{\gamma_{l}}G_{lz}(pz)}{\prod^{n}_{j=1}(1-\frac{pz}{p_{j}})^{\beta_{j}}G_{jp}(pz)}=\infty$.

In complex analysis, 0 can be written as:
$0*e^{i\theta}=0*(\cos\theta+i\sin\theta)$. The points of non-zero and non-
infinity finite values can also be written as:
$k*e^{i\theta}=k*(\cos\theta+i\sin\theta)$. The infinity can also be written
as: $(+\infty)*e^{i\theta}=(+\infty)*(\cos\theta+i\sin\theta)$. In which,
$(\cos\theta+i\sin\theta)$ is a non-zero and non-infinity, and its modulus is
1 of the unit complex number value. For
$(+\infty)*e^{i\theta}=(+\infty)*(\cos\theta+i\sin\theta)$, no matter what
value $\theta=2q\pi+\alpha$ obtains on the unit circle in
$\mathbb{C}\cup\\{\infty\\}$, the values of this expression all obtain the
infinity.

The above equation can also be expressed as:
$Gpz=G(pz)\frac{\prod^{m}_{l=1}(1-\frac{pz}{z_{l}})^{\gamma_{l}}G_{lz}(pz)}{\prod^{n}_{j=1}(1-\frac{pz}{p_{j}})^{\beta_{j}}G_{jp}(pz)}=K_{pz}e^{i\theta_{pz}}=\infty$.
In this equation, $K_{pz}$ is the modulus of the meromorphic function $W(s)$
and $\theta_{pz}$ is the phase angle of the meromorphic function $W(s)$. So,
$K_{pz}=+\infty$, and $\theta_{pz}=2q\pi+\alpha$,
$K_{pz}*e^{i\theta}=(+\infty)*(\cos\theta+i\sin\theta)=\infty$ is true, and
$\theta_{pz}=2q\pi+\alpha$ is an arbitrary degree number from $2q\pi$ degree
to $2q\pi+2\beta_{j}\pi$ degree.

For the finite pole $pz$ of Eq(1.1), when it is substituted into the
meromorphic function $W(s)$, for the phase angle of the arbitrary
$2q\pi+\alpha$ degree number of the meromorphic function $W(s)$, its values
are all equal to infinty. According to Definition 1.10, the phase angle of the
meromorphic function $W(s)$ is namely the phase angle of the root locus of
Eq(1.1). Because $2q\pi+\alpha$ is an arbitrary degree number from $2q\pi$
degree to $2q\pi+2\beta_{j}\pi$ degree, when it obtains all of degree numbers,
this shows the finite poles of Eq(1.1) are simultaneously on the root locus of
all of degree numbers of Eq(1.1) from $2q\pi$ degree to $2q\pi+2\beta_{j}\pi$
degree. ∎

######  Lemma 1.13.

All of the root locus of arbitrary $2q\pi+\alpha$ degree number of Eq(1.1) are
originated from poles of Eq(1.1), and are finally received at zeros of
Eq(1.1).

###### Proof.

The arbitrary $2q\pi+\alpha$ degree number root locus of Eq(1.1) are the
curves in $\mathbb{C}\cup\\{\infty\\}$. So, they all need to have their own
origination points and receiving points. According to the relationship between
the points in $\mathbb{C}\cup\\{\infty\\}$ and the root locus of Eq(1.1), the
points in $\mathbb{C}\cup\\{\infty\\}$ can be divided into four types, one is
the poles of Eq(1.1), one is the zeros of Eq(1.1), the other is the general
finite points on the $2q\pi+\alpha$ degree root locus of Eq(1.1), and the last
type is the infinity points in $\mathbb{C}\cup\\{\infty\\}$.

Except the finite and infinite zeros of Eq(1.1), and except the finite and
infinite poles of Eq(1.1), a general finite point in
$\mathbb{C}\cup\\{\infty\\}$ is on a root loci of a certain degree number of
Eq(1.1). And a general finite point in $\mathbb{C}\cup\\{\infty\\}$ only has
the phase angles of only one degree number and non-zero finite gain value. It
can not be the originating point and receiving point of the root locus of
Eq(1.1).

When the infinite point in $\mathbb{C}\cup\\{\infty\\}$ let the meromorphic
function $W(s)$ obtain a constant $A$. The gain values of the infinite point
in $\mathbb{C}\cup\\{\infty\\}$ are the constant $\lvert A\rvert$. The phase
angles of the meromorphic function $W(s)$ are all equal to $2q\pi+arg{(A)}$
degree. So, these infinite points in $\mathbb{C}\cup\\{\infty\\}$ are also the
ordinary points of the finite values with the gain $\lvert A\rvert$ on the
$2q\pi+\arg(a)$ degree root locus. They can not originate or receive the root
locus.

When the infinite points in $\mathbb{C}\cup\\{\infty\\}$ are the infinite
zeros or poles of Eq(1.1). On the infinite zero or pole, the gain value is
$K=+\infty$ or $K=0$ respectively.

The finite poles and zeros of Eq(1.1) are on the root locus of all of degree
numbers of Eq(1.1), and the infinite number of the different degree numbers
root locus are at the same one point, so, they satisfy the condition that the
root locus can be originated or received. On the finite and infinite poles of
Eq(1.1), there is $K=0$. On the finite and infinite zeros of Eq(1.1), there is
$K=+\infty$. Thus, we can let the finite and infinite poles of Eq(1.1) as the
origination points of the root locus. The finite and infinite zero points of
Eq(1.1) are the receiving points of the root locus. ∎

######  Definition 1.14.

_The origination points_ of the root locus of the arbitrary $2q\pi+\alpha$
degree of Eq(1.1) are the points on the $2q\pi+\alpha$ degree root locus that
their corresponding gain values equal to zero in $\mathbb{C}\cup\\{\infty\\}$,
$K=0$.

######  Definition 1.15.

_The receiving points_ of the root locus of the arbitrary $2q\pi+\alpha$
degree of Eq(1.1) are the points on the $2q\pi+\alpha$ degree root locus that
their corresponding gain values equal to infinity in
$\mathbb{C}\cup\\{\infty\\}$, $K=+\infty$.

######  Lemma 1.16.

(1). For any infinite point of $\mathbb{C}\cup\\{\infty\\}$, if it satisfies

$K=\lim\limits_{\lvert
s\rvert\to+\infty}\frac{\prod^{n}_{j=1}|1-\frac{s}{p_{j}}|^{\beta_{j}}|G_{jp}(s)|}{{|G(s)|\prod^{m}_{l=1}|1-\frac{s}{z_{l}}|^{\gamma_{l}}|G_{lz}(s)|}}=0$,
then it is the infinite pole of the root locus of Eq(1.1). These infinite
points must be the origination points of $2q\pi+\alpha$ degrees root locus of
Eq(1.1) in $\mathbb{C}\cup\\{\infty\\}$.

(2). For any infinite point of $\mathbb{C}\cup\\{\infty\\}$, if it satisfies

$K=\lim\limits_{\lvert
s\rvert\to+\infty}\frac{\prod^{n}_{j=1}|1-\frac{s}{p_{j}}|^{\beta_{j}}|G_{jp}(s)|}{{|G(s)|\prod^{m}_{l=1}|1-\frac{s}{z_{l}}|^{\gamma_{l}}|G_{lz}(s)|}}=A$,
in which $A\neq 0$ is a finite value, then it is a general point that its gain
is a finite value. These infinite points are the general points on the root
locus of Eq(1.1) in $\mathbb{C}\cup\\{\infty\\}$.

(3). For any infinite point of $\mathbb{C}\cup\\{\infty\\}$, if it satisfies

$K=\lim\limits_{\lvert
s\rvert\to+\infty}\frac{\prod^{n}_{j=1}|1-\frac{s}{p_{j}}|^{\beta_{j}}|G_{jp}(s)|}{{|G(s)|\prod^{m}_{l=1}|1-\frac{s}{z_{l}}|^{\gamma_{l}}|G_{lz}(s)|}}=+\infty$,
then the infinite point is the infinite zero of the root locus of Eq(1.1).
These infinite points must be the receiving points of $2q\pi+\alpha$ degrees
root locus of Eq(1.1) in $\mathbb{C}\cup\\{\infty\\}$.

###### Proof.

(i). For any infinite points of $\mathbb{C}\cup\\{\infty\\}$, if they satisfy

$K=\lim\limits_{\lvert
s\rvert\to+\infty}\frac{\prod^{n}_{j=1}|1-\frac{s}{p_{j}}|^{\beta_{j}}|G_{jp}(s)|}{{|G(s)|\prod^{m}_{l=1}|1-\frac{s}{z_{l}}|^{\gamma_{l}}|G_{lz}(s)|}}=0$,
according to the definition of the origination points of the arbitrary degree
root locus of Eq(1.1), the infinity points on the root locus of the arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) are also origination points of the arbitrary
$2q\pi+\alpha$ degree root locus of Eq(1.1).

(ii). For any infinite points of $\mathbb{C}\cup\\{\infty\\}$, if they satisfy

$K=\lim\limits_{\lvert
s\rvert\to+\infty}\frac{\prod^{n}_{j=1}|1-\frac{s}{p_{j}}|^{\beta_{j}}|G_{jp}(s)|}{{|G(s)|\prod^{m}_{l=1}|1-\frac{s}{z_{l}}|^{\gamma_{l}}|G_{lz}(s)|}}=A$,
in which $A\neq 0$ is a finite value. So, all of the infinity points in
$\mathbb{C}\cup\\{\infty\\}$ that their gains are the finite values are all
general points. The infinity points in $\mathbb{C}\cup\\{\infty\\}$ are the
general points that their gains are the finite values $A$.

(iii). For any infinite points of $\mathbb{C}\cup\\{\infty\\}$, if they
satisfy

$K=\lim\limits_{\lvert
s\rvert\to+\infty}\frac{\prod^{n}_{j=1}|1-\frac{s}{p_{j}}|^{\beta_{j}}|G_{jp}(s)|}{{|G(s)|\prod^{m}_{l=1}|1-\frac{s}{z_{l}}|^{\gamma_{l}}|G_{lz}(s)|}}=+\infty$.
According to the definition of the receiving points of the arbitrary degree
root locus of Eq(1.1), the infinity points on the root locus of the arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) are also receiving points of the arbitrary
$2q\pi+\alpha$ degree root locus of Eq(1.1). ∎

Sum up Lemma 1.13 and Lemma 1.16, we can obtain Theorem 1.17.

######  Theorem 1.17.

In $\mathbb{C}\cup\\{\infty\\}$, all root locus of the arbitrary
$2q\pi+\alpha$ degree number of Eq(1.1) are originated from their finite poles
or the infinite poles. And these root locus are received by their finite zeros
or the infinite zeros.

##  2\. Some properties of root locus of $W(s)$

In this section, we show the expressions of the argument[7] of meromorphic
function $W(s)$ in $\mathbb{C}\cup\\{\infty\\}$.

(1). When $u(\sigma,t)>0$, $v(\sigma,t)\geq 0$.
$\varphi(\sigma,t)=arg(\frac{v(\sigma,t)}{u(\sigma,t)})=\arctan(\frac{v(\sigma,t)}{u(\sigma,t)})$.

(2). When $u(\sigma,t)<0$, $v(\sigma,t)\geq 0$.
$\varphi(\sigma,t)=arg(\frac{v(\sigma,t)}{u(\sigma,t)})=\pi-\arctan(\frac{v(\sigma,t)}{-u(\sigma,t)})$.

(3). When $u(\sigma,t)<0$, $v(\sigma,t)<0$.
$\varphi(\sigma,t)=arg(\frac{v(\sigma,t)}{u(\sigma,t)})=\pi+\arctan(\frac{v(\sigma,t)}{u(\sigma,t)})$.

(4). When $u(\sigma,t)>0$, $v(\sigma,t)<0$.
$\varphi(\sigma,t)=arg(\frac{v(\sigma,t)}{u(\sigma,t)})=-\arctan(\frac{-v(\sigma,t)}{u(\sigma,t)})$.

We need to create a sub-set $\Delta_{u}$ of set $\Delta$. Let
$\Delta_{u}=\\{s=(\sigma+it)\in\mathbb{C}\cup\\{\infty\\}$, $s$ is not zero or
pole of $W(s)$, and there is $u(\sigma,t)\neq 0\\}$.

######  Lemma 2.1.

Let $s=(\sigma,t)\in\Delta_{u}$, then we have:
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)=\frac{v^{{}^{\prime}}_{\sigma}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{\sigma}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$,
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)=\frac{v^{{}^{\prime}}_{t}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{t}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$.

This lemma can be easily obtained by partial derivative calculation.

Let $W_{3}(s)=iW(s)$, so, $W_{3}(s)=iu(\sigma,t)-v(\sigma,t)$. The phase angle
of function $W_{3}(s)$ is
$\varphi_{3}(\sigma,t)=\frac{\pi}{2}+\varphi(\sigma,t)$.
$\frac{\partial{\varphi_{3}}}{\partial{\sigma}}(\sigma,t)=\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)$,
$\frac{\partial{\varphi_{3}}}{\partial{t}}(\sigma,t)=\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)$.
We need to create another sub-set $\Delta_{v}$ of set $\Delta$.
$\Delta_{v}=\\{s=(\sigma+it)\in\mathbb{C}\cup\\{\infty\\}$, $s$ is not zero or
pole of $W(s)$, and there is $v(\sigma,t)\neq 0\\}$. According to Lemma 2.1
and the result which we give in this segment, we can obtain Lemma 2.2.

######  Lemma 2.2.

Let $s=(\sigma,t)\in\Delta_{v}$, then we have:
$\frac{\partial{\varphi_{3}}}{\partial{\sigma}}(\sigma,t)=\frac{v^{{}^{\prime}}_{\sigma}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{\sigma}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$,
$\frac{\partial{\varphi_{3}}}{\partial{t}}(\sigma,t)=\frac{v^{{}^{\prime}}_{t}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{t}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$

We need to create another sub-set $\Delta_{vu}$ of set $\Delta_{v}$. Let
$\Delta_{vu}=\\{s=(\sigma+it)\in\mathbb{C}\cup\\{\infty\\}$, $s$ is not zero
or pole of $W(s)$, and there is $u(\sigma,t)=0\\}$. According to Lemma 2.2, we
can obtain Lemma 2.3.

######  Lemma 2.3.

Let $s=(\sigma,t)\in\Delta_{vu}$, then we have:
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)=\frac{v^{{}^{\prime}}_{\sigma}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{\sigma}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$,
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)=\frac{v^{{}^{\prime}}_{t}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{t}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$.

$\Delta_{u}\cap\Delta_{vu}=\emptyset$, $\Delta_{u}\subset\Delta$,
$\Delta_{vu}\subset\Delta$, and $\Delta=\Delta_{u}\cup\Delta_{vu}$. So, we can
sum up Lemma 2.1 and Lemma 2.3, we can obtain Theorem 2.4.

######  Theorem 2.4.

Let $s=(\sigma,t)\in\Delta$, then we have:
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)=\frac{v^{{}^{\prime}}_{\sigma}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{\sigma}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$,
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)=\frac{v^{{}^{\prime}}_{t}(\sigma,t)u(\sigma,t)-u^{{}^{\prime}}_{t}(\sigma,t)v(\sigma,t)}{u^{2}(\sigma,t)+v^{2}(\sigma,t)}$.

In many textbooks of the complex variable function of one variable, the
following results exist.
$\frac{dW(s)}{ds}=\frac{\partial{u(\sigma,t)}}{\partial{\sigma}}+i\frac{\partial{v(\sigma,t)}}{\partial{\sigma}}=\frac{\partial{u(\sigma,t)}}{\partial{t}}-i\frac{\partial{v(\sigma,t)}}{\partial{t}}$.
$\frac{\partial{u(\sigma,t)}}{\partial{\sigma}}=\frac{\partial{v(\sigma,t)}}{\partial{t}}$,
$\frac{\partial{v(\sigma,t)}}{\partial{\sigma}}=-\frac{\partial{u(\sigma,t)}}{\partial{t}}$.

######  Theorem 2.5.

For any point $s=(\sigma,t)\in\Delta$, we have
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)=Re({\frac{\frac{dW(s)}{ds}}{W(s)}})$,
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)=Im({\frac{\frac{dW(s)}{ds}}{W(s)}})$

###### Proof.

$\frac{\frac{dW(s)}{ds}}{W(s)}=\frac{\frac{\partial{u(\sigma,t)}}{\partial{\sigma}}+i\frac{\partial{v(\sigma,t)}}{\partial{\sigma}}}{u(\sigma,t)+iv(\sigma,t)}=\frac{\frac{\partial{u(\sigma,t)}}{\partial{\sigma}}u(\sigma,t)+\frac{\partial{v(\sigma,t)}}{\partial{\sigma}}v(\sigma,t)+i(\frac{\partial{v(\sigma,t)}}{\partial{\sigma}}u(\sigma,t)-\frac{\partial{u(\sigma,t)}}{\partial{\sigma}}v(\sigma,t))}{u(\sigma,t)^{2}+v(\sigma,t)^{2}}=\frac{\frac{\partial{v(\sigma,t)}}{\partial{t}}u(\sigma,t)-\frac{\partial{u(\sigma,t)}}{\partial{t}}v(\sigma,t)}{u(\sigma,t)^{2}+v(\sigma,t)^{2}}+i\frac{\frac{\partial{v(\sigma,t)}}{\partial{\sigma}}u(\sigma,t)-\frac{\partial{u(\sigma,t)}}{\partial{\sigma}}v(\sigma,t)}{u(\sigma,t)^{2}+v(\sigma,t)^{2}}$.

Obviously, according to Theorem 2.4, we can obtain Theorem 2.5. ∎

According to Theorem 2.5, obviously, we can obtain Theorem 2.6.

######  Theorem 2.6.

For any point $s=(\sigma,t)\in\Delta$, if
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)=0$, then
$Re({\frac{\frac{dW(s)}{ds}}{W(s)}})=0$. If
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)=0$, then
$Im({\frac{\frac{dW(s)}{ds}}{W(s)}})=0$. Conversely, if
$Re({\frac{\frac{dW(s)}{ds}}{W(s)}})=0$, then
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)=0$. If
$Im({\frac{\frac{dW(s)}{ds}}{W(s)}})=0$, then
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)=0$.

######  Theorem 2.7.

Let $s_{0}\in\Delta$, except the situation which the point $s_{0}$ is an
infinite point in $\mathbb{C}\cup\\{\infty\\}$. Then $s_{0}$ is the finite
zero of $W_{3}^{{}^{\prime}}(s)$ if and only if
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ and
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ hold
simultaneously.

###### Proof.

On the zeros $s_{0}$ of $W^{{}^{\prime}}(s)$,
${\frac{\frac{dW(s)}{ds}}{W(s)}}\mid{{}_{s=s_{0}}}=0$, except the situation
which the point $s_{0}$ is an infinite point in $\mathbb{C}\cup\\{\infty\\}$.
So, the two equations
$Re({\frac{\frac{dW(s)}{ds}}{W(s)}})\mid{{}_{s=s_{0}}}=0$ and
$Im({\frac{\frac{dW(s)}{ds}}{W(s)}})\mid{{}_{s=s_{0}}}=0$ are true
simultaneously, so, two partial derivative equations
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ and
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ are
true simultaneously.

Conversely, except zeros and poles of the meromorphic function $W(s)$, and
except the situation which the point $s_{0}$ is an infinite point in
$\mathbb{C}\cup\\{\infty\\}$. When two partial derivative equations
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ and
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ are
true simultaneously. Then, two equations
$Re({\frac{\frac{dW(s)}{ds}}{W(s)}})\mid{{}_{s=s_{0}}}=0$ and
$Im({\frac{\frac{dW(s)}{ds}}{W(s)}})\mid{{}_{s=s_{0}}}=0$ are true
simultaneously, so, the equation
${\frac{\frac{dW(s)}{ds}}{W(s)}}\mid{{}_{s=s_{0}}}=0$ is true. So, when two
partial derivative equations
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ and
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ are
true simultaneously. Such points are the non-repeated zeros of derivative of
the meromorphic function $W(s)$.

On the infinite points in $\mathbb{C}\cup\\{\infty\\}$,
$\frac{\frac{dW(s)}{ds}}{W(s)}\mid{{}_{s=s_{0}}}=0$, but, in most situations,
two equations
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ and
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$ can not
be true simultaneously. So, we need to exclude the infinite zeros of
$W^{{}^{\prime}}(s)$ in this lemma. ∎

Let $\Delta_{d}=\\{s=(\sigma+it)\in\mathbb{C}\cup\\{\infty\\}$, $s$ is not
zero or pole of $W(s)$, the point $s$ is not an infinite point in
$\mathbb{C}\cup\\{\infty\\}$. $s$ is not the finite zeros of derivative of the
function $W(s)\\}$.

######  Theorem 2.8.

For any point $s=(\sigma,t)\in\Delta_{d}$, the partial derivative of the phase
angle of $W(s)$ concerning the variable $\sigma$ and variable $t$ cannot
simultaneously equal to 0.

###### Proof.

According to Theorem 2.7, if two equations
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)=0$ and
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)=0$ hold simultaneously,
then the point $s$ is surely a finite zero of the derivative
$W^{{}^{\prime}}(s)$ of function $W(s)$. So, except the finite zeros and poles
of function $W(s)$, except the infinite points in
$\mathbb{C}\cup\\{\infty\\}$, and also except the finite zeros of derivative
of the function $W(s)$, and on other finite points in the extended complex
plane $\mathbb{C}\cup\\{\infty\\}$, the partial derivative of the phase angle
of the function $W(s)$ concerning the variable $t$ and variable $\sigma$
cannot simultaneously equal to 0. ∎

######  Remark 2.9.

Let $s_{0}=(\sigma_{0}+it_{0})\in\Delta_{d}$, then, if
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$, we
have $\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}\neq
0$; or, if
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$, we have
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}\neq
0$.

When $2q\pi+arg{(b/a)}$ is a constant, $\varphi(\sigma,t)=2q\pi+arg{(b/a)}$,
it is a function equation which contain two real variables, and it is the
phase angle condition equation of Eq(1.1). According to Theorem 1.7, except
zeros and poles of Eq(1.1), if the roots of this function equation exist in
$\mathbb{C}\cup\\{\infty\\}$, the roots of Eq(1.1) also exist in
$\mathbb{C}\cup\\{\infty\\}$. If the roots of this function equation
constitute the root locus in $\mathbb{C}\cup\\{\infty\\}$, the roots of
Eq(1.1) also constitute the root locus in $\mathbb{C}\cup\\{\infty\\}$.

In order to prove the existence and continuity of the roots of the implicit
function equation $\varphi(\sigma,t)=2q\pi+arg{(b/a)}$, we need to cite the
implicit function theorem and involved concepts[7].

######  Definition 2.10.

Let $f:R^{n+m}\longrightarrow R^{m}$ be a continuously differentiable
function. We think of $R^{n+m}$ as the Cartesian product $R^{n}\times R^{m}$,
and we write a point of this product as
$(x,y)=(x_{1},\cdots,x_{n}.y_{1},\cdots,y_{m})$. Starting from the given
function $f$, our goal is to construct a function $g:R^{n}\longrightarrow
R^{m}$ whose graph $(x,g(x))$ is precisely the set of all $(x,y)$ such that
$f(x,y)=0$.

As noted above, this may not always be possible. We will therefore fix a point
$(a,b)=(a_{1},\cdots,a_{n}.b_{1},\cdots,b_{m})$ which satisfies $f(a,b)=0$,
and we will ask for a g that works near the point $(a,b)$. In other words, we
want an open set $U$ of $R^{n}$ containing $a$, an open set $V$ of $R^{m}$
containing $b$, and a function $g:U\longrightarrow V$ such that the graph of
$g$ satisfies the relation $f(x,y)=0$ on $U\times V$, and that no other points
within $U\times V$ do so. In symbols, $((x,g(x))|x\in U)$ = $((x,y)\in U\times
V|f(x,y)=0)$.

To state the implicit function theorem, we need the Jacobian matrix of $f$,
which is the matrix of the partial derivatives of $f$. Abbreviating
$(a_{1},\cdots,a_{n}.b_{1},\cdots,b_{m})$ to $(a,b)$, the Jacobian matrix is

$(Df)(a,b)=[\begin{array}[]{ccc}\frac{\partial{f_{1}}}{\partial{x_{1}}}(a,b)&\cdots&\frac{\partial{f_{1}}}{\partial{x_{n}}}(a,b)\\\
\vdots&\ddots&\vdots\\\
\frac{\partial{f_{m}}}{\partial{x_{1}}}(a,b)&\cdots&\frac{\partial{f_{m}}}{\partial{x_{n}}}(a,b)\end{array}|\begin{array}[]{ccc}\frac{\partial{f_{1}}}{\partial{y_{1}}}(a,b)&\cdots&\frac{\partial{f_{1}}}{\partial{y_{m}}}(a,b)\\\
\vdots&\ddots&\vdots\\\
\frac{\partial{f_{m}}}{\partial{y_{1}}}(a,b)&\cdots&\frac{\partial{f_{m}}}{\partial{y_{m}}}(a,b)\end{array}]$
$=[X|Y]$.

Where $X$ is the matrix of partial derivatives in the variables $x_{l}$ and
$Y$ is the matrix of partial derivatives in the variables $y_{j}$. The
implicit function theorem says that if $Y$ is an invertible matrix, then there
are $U$, $V$, and $g$ as desired. Writing all the hypotheses together gives
the following statement.

The implicit function theorem concerning variable $x$.

######  Theorem 2.11.

Let $f:R^{n+m}\longrightarrow R^{m}$ be a continuously differentiable
function, and let $R^{n+m}$ have coordinates $(x,y)$. Fix a point
$(a,b)=(a_{1},\cdots,a_{n}.b_{1},\cdots,b_{m})$ with $f(a,b)=c$, where $c\in
R^{m}$. If the Jacobian matrix $J_{f,y}(a,b)=[(\partial f_{l}/\partial
y_{j})(a,b)]$ is invertible, then there exists an open set $U$ containing $a$,
an open set $V$ containing $b$, and a unique continuously differentiable
function $g:U\longrightarrow V$ such that $((x,g(x))|x\in U)$ = $((x,y)\in
U\times V|f(x,y)=c)$.

The implicit function theorem concerning variable $y$.

######  Theorem 2.12.

Let $f:R^{n+m}\longrightarrow R^{n}$ be a continuously differentiable
function, and let $R^{n+m}$ have coordinates $(x,y)$. Fix a point
$(a,b)=(a_{1},\cdots,a_{n}.b_{1},\cdots,b_{m})$ with $f(a,b)=c$, where $c\in
R^{n}$. If the Jacobian matrix $J_{f,x}(a,b)=[(\partial f_{l}/\partial
x_{k})(a,b)]$ is invertible, then there exists an open set $V$ containing $b$,
an open set $U$ containing $a$, and a unique continuously differentiable
function $h:V\longrightarrow U$ such that $((h(y),y)|y\in V)$ = $((x,y)\in
U\times V|f(x,y)=c)$.

######  Lemma 2.13.

For the root locus of the arbitrary $2q\pi+\alpha$ degree of Eq(1.1). And the
locus which the points that satisfy the implicit function equation
$\varphi(\sigma,t)=2q\pi+arg{(b/a)}$ are generated in the two-dimensional real
plane. The two locus are totally the same, and their properties like
continuity are equivalent.

###### Proof.

When $2q\pi+\alpha$ is a constant, the root locus of the arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) are curves that are constituted by roots of
Eq(1.1). Theorem 1.7 proves that the root locus of the arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) can also be sufficiently and necessary
determined by the real function which contains two real variables, namely,
when $2q\pi+\alpha$ is a constant, the roots locus of the arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) can also be sufficiently and necessary
determined by the implicit function equation
$\varphi(\sigma,t)=2q\pi+arg{(b/a)}$.

So, essentially, Theorem 1.7 proves: if the extended complex plane
$\mathbb{C}\cup\\{\infty\\}$ is as a two-dimensional plane which let two real
variables $\sigma$, $t$ be as its two coordinates, then, the root locus of the
arbitrary $2q\pi+\alpha$ degree of Eq(1.1) are constituted by points which are
determined by two real variables $\sigma$, $t$ that satisfy the implicit
function equation $\varphi(\sigma,t)=2q\pi+arg{(b/a)}$ in two-dimensional real
plane.

Therefore, roots of Eq(1.1) in $\mathbb{C}\cup\\{\infty\\}$ and points which
are determined by two real variables that satisfy the implicit function
equation $\varphi(\sigma,t)=2q\pi+arg{(b/a)}$ in two-dimensional plane are
same. Namely, if the extended complex plane $\mathbb{C}\cup\\{\infty\\}$ and
two-dimensional real plane which are determined by two real coordinates are
considered as same one, when the root locus of the arbitrary $2q\pi+\alpha$
degree of Eq(1.1) constitute the locus in $\mathbb{C}\cup\\{\infty\\}$, then,
the points which are determined by two real coordinates that satisfy the
implicit function equation $\varphi(\sigma,t)=2q\pi+arg{(b/a)}$ also
constitute the locus in two-dimensional real plane, two kinds of locus are
fully same.

Because two kinds of root locus are completely same, the properties like
existence and continuity of the two kinds of locus are equivalent. ∎

According to Lemma 2.13, the problems of the complex variable functions are
transformed into the problems of the real variable functions. So, we can use
theorems of the real variable functions and the definition of the continuity
to prove the existence and continuity of points which are determined by two
real variables that satisfy the implicit function equation
$\varphi(\sigma,t)=2q\pi+arg{(b/a)}$ in two-dimensional real plane.

######  Lemma 2.14.

Let $s_{0}=(\sigma_{0}+it_{0})\in\Delta_{d}$, $s_{0}$ satisfies
$\varphi(\sigma_{0},t_{0})=2q\pi+arg{(b_{0}/a_{0})}$. There exists an open set
$U$ containing $\sigma_{0}$, an open set $V$ containing $t_{0}$, and a unique
continuously differentiable function $g:U\longrightarrow V$ such that
$((\sigma,g(\sigma))|\sigma\in U)$ = $((\sigma,t)\in U\times
V|\varphi(\sigma,t)=2q\pi+arg{(b_{0}/a_{0})})$. Or, there exists an open set
$V$ containing $t_{0}$, an open set $U$ containing $\sigma_{0}$, and a unique
continuously differentiable function $h:V\longrightarrow U$ such that
$((h(t),t)|t\in V)$ = $((\sigma,t)\in U\times
V|\varphi(\sigma,t)=2q\pi+arg{(b_{0}/a_{0})})$.

###### Proof.

The function $\varphi(\sigma,t)$, $\varphi(\sigma,t):R^{2}\longrightarrow R$
is a continuously differentiable function, $R^{2}$ have two coordinates
$(\sigma,t)$.

For an arbitrary point $s_{0}=(\sigma_{0}+it_{0})\in\Delta$, it satisfies the
implicit function equation
$\varphi(\sigma_{0},t_{0})=2q\pi+arg{(b_{0}/a_{0})}$, where
$arg{(b_{0}/a_{0})}\in R$.

The point $s_{0}=(\sigma_{0}+it_{0})\in\Delta$ satisfies the condition of
Theorem 2.11, according to Theorem 2.8, we can obtain: if
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$, we
have $\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}\neq
0$; or, if
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}=0$, we have
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}\neq
0$.

The implicit function $\varphi(\sigma,t)=2q\pi+arg{(b/a)}$ satisfies the
implicit function theorem, so, according to the implicit function theorem
concerning variable $x$, we can obtain: If
$\frac{\partial{\varphi}}{\partial{t}}(\sigma,t)\mid{{}_{s=s_{0}}}\neq 0$.
There exists an open set $U$ containing $\sigma_{0}$, an open set $V$
containing $t_{0}$, and a unique continuously differentiable function
$g:U\longrightarrow V$ such that $((\sigma,g(\sigma))|\sigma\in U)$ =
$((\sigma,t)\in U\times V|\varphi(\sigma,t)=2q\pi+arg{(b_{0}/a_{0})})$.

According to the implicit function theorem concerning variable $y$. If
$\frac{\partial{\varphi}}{\partial{\sigma}}(\sigma,t)\mid{{}_{s=s_{0}}}\neq
0$. There exists an open set $V$ containing $t_{0}$, an open set $U$
containing $\sigma_{0}$, and a unique continuously differentiable function
$h:V\longrightarrow U$ such that $((h(t),t)|t\in V)$ = $((\sigma,t)\in U\times
V|\varphi(\sigma,t)=2q\pi+arg{(b_{0}/a_{0})})$. ∎

######  Lemma 2.15.

Let $s_{0}=(\sigma_{0}+it_{0})\in\Delta_{d}$, then the root locus of arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) are all continuous on the points $s_{0}$.

###### Proof.

According to the results of Lemma 2.14, we have: $t=g(\sigma)$ is a continuous
function inside an open set $U$. So, for any given $\varepsilon>0$, there
surely exists a $\delta_{\sigma 1}<\frac{\varepsilon}{\sqrt{(}2)}$, when
$|\sigma-\sigma_{0}|<\delta_{\sigma 1}<\frac{\varepsilon}{\sqrt{(}2)}$. Then,
$|t-t_{0}|=|g(\sigma)-g(\sigma_{0})|<\frac{\varepsilon}{\sqrt{(}2)}$ is true.
So, for any given $\varepsilon>0$, let $\delta_{\sigma 2}=\delta_{\sigma
1}<\frac{\varepsilon}{\sqrt{(}2)}$, when $|\sigma-\sigma_{0}|<\delta_{\sigma
1}<\frac{\varepsilon}{\sqrt{(}2)}$, then,
$|t-t_{0}|=|g(\sigma)-g(\sigma_{0})|<\frac{\varepsilon}{\sqrt{(}2)}$, and
obtain:
$|\sigma+it-\sigma_{0}-it_{0}|=|\sigma+ig(\sigma)-\sigma_{0}-ig(\sigma_{0})|=\sqrt{(\sigma-\sigma_{0})^{2}+(g(\sigma)-g(\sigma_{0}))^{2}}<\sqrt{\frac{\varepsilon^{2}}{2}+\frac{\varepsilon^{2}}{2}}=\varepsilon$.

According to Lemma 2.14, we have: $\sigma=h(t)$ is a continuous function
inside an open set $V$. So, for any given $\varepsilon>0$, there surely exists
a $\delta_{\sigma 1}<\frac{\varepsilon}{\sqrt{(}2)}$, when
$|t-t_{0}|<\delta_{\sigma 1}<\frac{\varepsilon}{\sqrt{(}2)}$, then,
$|\sigma-\sigma_{0}|=|h(t)-h(t_{0})|<\frac{\varepsilon}{\sqrt{(}2)}$ is true.
So, for any given $\varepsilon>0$, let $\delta_{\sigma 2}=\delta_{\sigma
1}<\frac{\varepsilon}{\sqrt{(}2)}$, when $|t-t_{0}|<\delta_{\sigma
1}<\frac{\varepsilon}{\sqrt{(}2)}$, then,
$|\sigma-\sigma_{0}|=|h(t)-h(t_{0})|<\frac{\varepsilon}{\sqrt{(}2)}$, and
obtain:
$|\sigma+it-\sigma_{0}-it_{0}|=|h(t)+it-h(t_{0})-it_{0}|=\sqrt{(t-t_{0})^{2}+(h(t)-h(t_{0}))^{2}}<\sqrt{\frac{\varepsilon^{2}}{2}+\frac{\varepsilon^{2}}{2}}=\varepsilon$

So, it is proved that the root locus of the arbitrary $2q\pi+\alpha$ degree
phase angle of Eq(1.1) are continuous on the point $(\sigma_{0},t_{0})$.

Because the point $(\sigma_{0},t_{0})$ is an arbitrary point except zeros and
poles of Eq(1.1), except the infinite points in $\mathbb{C}\cup\\{\infty\\}$
and except the derivative zeros of the function $W(s)$ in
$\mathbb{C}\cup\\{\infty\\}$. Based on that the root locus of the arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) are continuous on the point
$(\sigma_{0},t_{0})$, so, it is proved that the root locus of the arbitrary
$2q\pi+\alpha$ degree of Eq(1.1) can not have any break points except the
derivative zeros of the function $W(s)$ and zeros and poles of Eq(1.1). When
$s_{0}=(\sigma_{0}+it_{0})\in\Delta_{d}$, the root locus of arbitrary
$2q\pi+\alpha$ degree phase angle of Eq(1.1) must be continuous on the point
$s_{0}$. ∎

Because of the limition of the implicit function theorem, the zeros, poles and
the finite zeros of derivative of the function $W(s)$ are excluded in the
result of the continuity of the root locus. The root locus are the results in
$\mathbb{C}\cup\\{\infty\\}$. But, the proof of the continuity of the root
locus are achieved through utilizing the phase angle function which has two
real variables and real function values. In the previous results, except three
categories points which discretely distributed in
$\mathbb{C}\cup\\{\infty\\}$, every one root loci is continuous. We have used
the definition of the continuity to prove Lemma 2.15. Because the root locus
are all continuous around the three categories points, obviously, we can use
the continuity judging theorems about the real value function of the real
variables to prove that the root locus are surely continuous on the finite
zeros of derivative of the function $W(s)$. On the zeros and poles of Eq(1.1),
the root locus are surely the left continuous or the right continuous. The
proof method is the same as the proof of Lemma 2.15. Here, we needn’t to give
their proofs. And we directly give their results.

The situation of the infinite points is more complex, here, we don’t give the
proof of its result. Using the same method which has already been given in the
above segment, we can prove this problem. We only give the result which the
root locus are continuous on the infinite points in
$\mathbb{C}\cup\\{\infty\\}$.

######  Theorem 2.16.

Let $s=(\sigma+it)\in\mathbb{C}\cup\\{\infty\\}$, then, the root locus of
arbitrary $2q\pi+\alpha$ degree of Eq(1.1) are all continuous on the points
$s$ in $\mathbb{C}\cup\\{\infty\\}$.

##  3\. The results about the mean value problem

Specially, the meromorphic function $W(s)$ in Eq(1.1) can be a rational
function.
$W_{r}(s)=\frac{\prod^{m}_{l=1}(s-z_{l})}{\prod^{n}_{j=1}(s-p_{j})}$. In this
paper, let $n\geq m$. $n$ and $m$ are both natural number.

######  Lemma 3.1.

On every one root loci of Eq(3.1), there don’t exist two points which their
gain values are same.

(3.1) $KW_{r}(s)=a_{f}+ib_{f}$

###### Proof.

Because Eq(3.1) has $n$ poles. Every one pole only emits a root loci of
Eq(3.1) which its degree is $\arg(b_{f}/a_{f})$. Eq(3.1) emits $n$ root locus
which their degrees are $\arg(b_{f}/a_{f})$. According to Theorem 1.3, for
every one root loci of $n$ root locus which their degrees are
$\arg(b_{f}/a_{f})$, the gains of points on these root locus obtain gain
values from 0 on poles to the positive infinity on zeros. So, if there are two
points which their gain values are both $\frac{1}{\lvert FF\rvert}$ on a same
root loci of Eq(3.1), then, on every one root loci of other $n-1$ root locus,
there at least exists one point which its gain value is $\frac{1}{\lvert
FF\rvert}$. So, there are more than $n$ points which their phase angles are
all $\arg(b_{f}/a_{f})$. and gain values are all $\frac{1}{\lvert FF\rvert}$.

Namely, there are more than $n$ points which they satisfy the equation:
$\frac{1}{\lvert FF\rvert}W_{r}(s)=a_{f}+ib_{f}$. Further, $W_{r}(s)=\lvert
FF\rvert(a_{f}+ib_{f})$. Let $FF=\lvert FF\rvert(a_{f}+ib_{f})$. $FF$ is a
complex number. So, $W_{r}(s)=FF$.
$FF\prod^{n}_{j=1}(s-p_{j})-\prod^{m}_{l=1}(s-z_{l})=0$. It is $n$ order
polynomial equation. It has and only has $n$ roots. So, on every one root loci
of Eq(3.1), there exist two points which their gain values can be same, this
situation cannot be true. ∎

######  Theorem 3.2.

On every root loci of Eq(3.1), the gain of the root loci continuously and
strictly monotonically obtains values from 0 on poles of Eq(3.1) to the
positive infinity on zeros of Eq(3.1).

###### Proof.

If the gain $|K(s)|$ can not be monotonic on the root loci of Eq(3.1), then,
on the root loci , there surely exist two points $s_{1}$, $s_{2}$.
$|K(s_{1})|=|K(s_{2})|$. It is contradict to Lemma 3.1. According to Theorem
1.3, we can obtain Theorem 3.2. ∎

For the inequality in mean value problem

(3.2) $\frac{\lvert f(s)-f(\theta)\rvert}{\lvert s-\theta\rvert}\leq\lvert
f^{{}^{\prime}}(s)\rvert.$

Let $f(s)=\sum^{n}_{k=0}(a_{k}s^{k})$.
$f(\theta)=\sum^{n}_{k=0}(a_{k}{\theta}^{k})$.
$f(s)-f(\theta)=\sum^{n}_{k=0}(a_{k}s^{k})-\sum^{n}_{k=0}(a_{k}{\theta}^{k})=\sum^{n}_{k=1}(a_{k}(s^{k}-{\theta}^{k}))=\sum^{n}_{k=1}(a_{k}(s-\theta)\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})$.
$\frac{\lvert f(s)-f(\theta)\rvert}{\lvert
s-\theta\rvert}=\lvert\frac{f(s)-f(\theta)}{s-\theta}\rvert=\lvert\sum^{n}_{k=1}a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj}\rvert$.

Namely, $\frac{f(s)-f(\theta)}{s-\theta}$ is a polynomial of one variable
which the order is $n-1$. $f^{{}^{\prime}}(s)$ is the derivative polynomial of
the polynomial $f(s)$. Eq(3.2) can be equivalently transformed into the
inequality

(3.3)
$\lvert\frac{f^{{}^{\prime}}(s)}{\frac{f(s)-f(\theta)}{s-\theta}}\rvert\geq
1.$

The left side $\frac{f^{{}^{\prime}}(s)}{\frac{f(s)-f(\theta)}{s-\theta}}$ of
Eq(3.3) is a rational fraction which the polynomial $f^{{}^{\prime}}(s)$ is
divided by the polynomial $\frac{f(s)-f(\theta)}{s-\theta}$. The zeros of
Eq(3.3) are zeros of the polynomial $f^{{}^{\prime}}(s)$, and the zeros of
Eq(3.3) are zeros of derivative of the polynomial $f(s)$. The poles of of
Eq(3.3) are zeros of the polynomial
$\frac{f(s)-f(\theta)}{s-\theta}=$$\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})$.
Do the root locus equation:

(3.4)
$K\frac{f^{{}^{\prime}}(s)}{\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})}=a+ib.$

According to the results of the root locus in section 1 and section 2, we can
obtain: The root locus of Eq(3.4) begin at the zeros of the polynomial
$\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})$. And they end at
the zeros of $f^{{}^{\prime}}(s)$. From the zeros of the polynomial
$\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})$ to the zeros of
$f^{{}^{\prime}}(s)$, the gain
$K=\lvert\frac{\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})}{f^{{}^{\prime}}(s)}\rvert$
strictly and monotonically increase. The gain $K$ and the modulus of the
rational fraction
$\frac{f^{{}^{\prime}}(s)}{\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})}$
are the relationship of reciprocal. From the zeros of the polynomial
$\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})$ to the zeros of
$f^{{}^{\prime}}(s)$, the modulus of the rational fraction
$\lvert\frac{f^{{}^{\prime}}(s)}{\sum^{n}_{k=1}(a_{k}\sum^{k}_{jj=1}s^{jj-1}{\theta}^{k-jj})}\rvert$
strictly and monotonically decrease.

At every zero $\theta_{i}$ of $f^{{}^{\prime}}(s)$,
$f^{{}^{\prime}}(\theta_{i})=0$. So, around every zero of
$f^{{}^{\prime}}(s)$, there surely exists its adjacent domain $\Omega_{i}$,
$i=1,2,\cdots,n-1$. For all points in the adjacent domain of zero $\theta_{i}$
of $f^{{}^{\prime}}(s)$, they all satisfy:
$\lvert\frac{f^{{}^{\prime}}(s)}{\frac{f(s)-f(\theta_{i})}{s-\theta_{i}}}\rvert<1$,
namely, $\frac{\lvert f(s)-f(\theta_{i})\rvert}{\lvert
s-\theta_{i}\rvert}=\lvert\frac{f(s)-f(\theta_{i})}{s-\theta_{i}}\rvert>\lvert
f^{{}^{\prime}}(s)\rvert$.

But, for all points in the extended complex plane except all adjacent domains
$\Omega={\Omega_{i}}$, $i=1,2,\cdots n-1$, they all satisfy:
$\lvert\frac{f^{{}^{\prime}}(s)}{\frac{f(s)-f(\theta_{i})}{s-\theta_{i}}}\rvert\geq
1$, namely, $\frac{\lvert f(s)-f(\theta_{i})\rvert}{\lvert
s-\theta_{i}\rvert}=\lvert\frac{f(s)-f(\theta_{i})}{s-\theta_{i}}\rvert\leq\lvert
f^{{}^{\prime}}(s)\rvert$.

So, we have proved the following two theorems.

######  Theorem 3.3.

Given a complex polynomial $f$ of degree $d\geq 2$ and a complex number $s$.
For every critical point $\theta_{i}$ (ie, $f^{{}^{\prime}}(\theta_{i})=0$),
$i=1,2,\cdots,n-1$, there exists its adjacent domain $\Omega_{i}$. For all
points in the adjacent domain $\Omega=\\{\Omega_{i}\\}$, $i=1,2,\cdots,n-1$,
such that $\frac{\lvert f(s)-f(\theta_{i})\rvert}{\lvert
s-\theta_{i}\rvert}>\lvert f^{{}^{\prime}}(s)\rvert$.

######  Theorem 3.4.

Given a complex polynomial $f$ of degree $d\geq 2$ and a complex number $s$.
For every critical point $\theta_{i}$ (ie, $f^{{}^{\prime}}(\theta_{i})=0$),
$i=1,2,\cdots,n-1$, there exists its adjacent domain $\Omega_{i}$. For all
points in the extended complex plane except the adjacent domain
$\Omega=\\{\Omega_{i}\\}$, $i=1,2,\cdots,n-1$, such that $\frac{\lvert
f(s)-f(\theta_{i})\rvert}{\lvert s-\theta_{i}\rvert}\leq\lvert
f^{{}^{\prime}}(s)\rvert$.

Conflict of Interest: Authors declare that they have no conflict of interest.
This article does not contain any studies with human participants or animals
performed by any of the authors.

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