Zeros of Meromorphic function

Abstract

We introduce and develop the root locus method in mathematics. And we study the distribution of zeros of meromorphic functions by root locus method.

Full text

arXiv:2210.10366v1 [math.CV] 19 Oct 2022
ZEROS OF MEROMORPHIC FUNCTION
LANDE MA AND ZHAOKUN MA
Abstract. We introduce and develop the root locus method in mathematics.
And we study the distribution of zeros of meromorphic functions by root locus
method.
Introduction
The meromorphic functions are the basic functions with general meanings in
mathematics. Meromorphic functions include all complex functions in the engi-
neering. The complex functions that need to be studied in the engineering must
have poles. The poles mainly determine the dynamic characteristics of a process.
Complex functions without poles have no actual values in the engineering. The
main purpose of mathematical research is to provide tools for engineering and other
sciences. In pure mathematics, many functions like the Riemann zeta function and
gamma function, belong to the meromorphic functions. However, there are only
several properties concerning the general meromorphic functions. And, there is
almost no result of its derivative function. Regarding the properties of zeros of
derivative of general meromorphic functions, so far, except for the definition of the
zero of derivative, nothing has been discovered.
It is well known that the distribution of zeros of the derivative of meromorphic
functions has been studied[8, 9, 10], and several special meromorphic functions are
also studied due to their great influence, such as Riemann zeta function[11, 12, 13,
14, 15, 16]. By utilizing the results which we obtained, we get the necessary and
sufficient results about the zeros of derivatives of meromorphic functions.
Speiser showed that the Riemann hypothesis is equivalent to the absence of non-
trivial zeros of the derivative of the Riemann zeta-function left of the critical line[19].
Levinson and Montgomery proved the quantitative version of the Speiser’s result,
namely, that the Riemann zeta-function and its derivative have approximately the
same number of zeros left of the critical line[12]. We prove there don’t exist the zeros
of derivatives of Riemann zeta function and Dirichlet ate series on the critical line
or on the imaginary axis. We also prove that the results of the values of argument of
Riemann zeta function and Dirichlet ate series strict and monotonic decrease on the
critical line. Riemann hypothesis is equivalent to that argument of xi-function and
Riemann zeta function and Dirichlet ate series concerning the imaginary variable t
strictly and monotonically decrease at the left side of the critical line.
Date: April 19th, 2022.
2020 Mathematics Subject Classification. Primary 11M26,11M06 Secondary 30D30,93C05.
Key words and phrases. Root locus; Poles; Zeros; Meromorphic function;
1

2 LANDE MA AND ZHAOKUN MA
1. The results
In textbook of automatic control theory[7], the factor in the left 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 which obtains the real
number values. So, the root locus equations and the results of the root locus in
automatic control theory are all very special and limit. The proofs of results in
automatic control theory are not comprehensive and no accurate. For the pure
mathematical study, the factor which is in the left side of the root locus equation
is best substituted by a general meromorphic function. So, we need to give the
results of the root locus of the general meromorphic function.
If the number of poles is finite or infinite, and the number of zeros is finite
or infinite, namely the two finite or infinite products Qm
l=1(1 −s
zl)γlGlz (s) and
Qn
j=1(1 −s
pj)βjGjp (s) both converge, and Kis as: K=|Qn
j=1(1−s
pj)βjGjp(s)
G(s)Qm
l=1(1−s
zl)γlGlz(s)|.
The Kis the reciprocal of the modulus of the meromorphic function G(s)Qm
l=1(1 −
s
zl)γlGlz (s)/Qn
j=1(1 −s
pj)βjGjp (s), let, W(s) = G(s)Qm
l=1(1−s
zl)γlGlz(s)
Qn
j=1(1−s
pj)βjGjp(s)=u(σ, t) +
iv(σ, t), W(s) is a meromorphic function with one complex variable in the extended
complex plane C∪ {∞}. So, after Kand the meromorphic function W(s) are mul-
tiplied, the product result is the unit complex value of the meromorphic function.
Definition 1.1. In the extended complex plane C∪ {∞}, the equation (1.1) is
called as the root locus equation.
(1.1) KG(s)Qm
l=1(1 −s
zl)γlGlz (s)
Qn
j=1(1 −s
pj)βjGjp (s)=a+ib
In which, a+ib is the unit complex number value of meromorphic function W(s).
The argument of the unit complex number value is as: α= 2qπ+ arg( b
a). The factor
at the left side of the root locus equation (1.1) is meromorphic function W(s).
The zeros zland poles pjare points in C∪ {∞}, and may be no conjugate. The
exponents γland βjare all non-negative real numbers, and they are the exponents
of zeros zland poles pjof Eq(1.1).
Lemma 1.2. After the coincident and finite zeros and poles of Eq(1.1) are can-
celled, for any finite zero of Eq(1.1), its Kvalue is K= +∞; for any finite pole of
Eq(1.1), its Kvalue is K= 0.
Proof. The Eq(1.1) can be transformed to the following characteristic equation.
KG(s)Qm
l=1(1 −s
zl)γlGlz (s)−(a+ib)Qn
j=1(1 −s
pj)βjGjp (s) = 0. Obviously, if
K= 0, all roots of the pole factor Qn
j=1(1 −s
pj)βjGjp (s) are the roots of charac-
teristic equation. Conversely, all roots of characteristic equation if K= 0 are all
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)Qm
l=1(1−
s
zl)γlGlz (s)−a+ib
KQn
j=1(1 −s
pj)βjGjp (s) = 0. If K= +∞, all roots of the zero fac-
tor Qm
l=1(1 −s
zl)γlGlz (s) are the roots of the characteristic equation. Conversely,
all roots of the characteristic equation when K= +∞are the roots of the zero
factor. So, at zeros of Eq(1.1), K= +∞.

ZEROS OF MEROMORPHIC FUNCTION 3
According to the expression of Kof the root locus equation (1.1) and according
to Lemma 1.2, it is obvious that the Kvalues are continuous concerning the complex
variable sin C∪ {∞}. So, we can prove Theorem 1.3 simply.
Theorem 1.3. Let s∈C∪ {∞}, the Kvalue of s∈C∪ {∞} takes all of the
non-negative real number value from 0 to the +∞.
For the argument[8] in complex analysis, in automatic control theory, it is called
as the phase angle. Kis called as the gain K.
Let ∆ = {s=σ+it ∈C∪ {∞},sis not zero or pole of W(s) and Eq(1.1)}.
Lemma 1.4. For any point s∈∆, the phase angle of meromorphic function W(s)
can be written as the ϕ(σ, t)function: ϕ(σ, t) = arg(Gy(σ,t)
Gx(σ,t)) + Pm
l=1(γlarg( t−tl
σ−σl)−
γlarg( tl
σl) + arg( Gz y(σ,t)
Gzx(σ,t))) −Pn
j=1(βjarg( t−tj
σ−σj)−βjarg( tj
σj) + arg( Gpy (σ,t)
Gpx(σ,t))).
Let s= (σ, t)∈∆, then the phase angle condition equation of Eq(1.1) on sis:
(1.2) ϕ(σ, t) = 2qπ +arg(b/a)
In which, qis an integer number.
Lemma 1.5. Let Eq(1.1) be the root locus equation of 2qπ +αdegree, for any point
s= (σ, t)∈∆, if the point ssatisfies the phase angle condition equation (1.2) of
2qπ +αdegree, then it must be on the root locus of 2qπ +αdegree of Eq(1.1).
Proof. Assuming that a point s1= (σ1, t1)∈∆ satisfies the phase angle condition
equation (1.2), ϕ(σ1, t1) = 2qπ +arg(b1/a1). The phase angle of W(s) is ϕ(σ, t),
this phase angle expression is same as the expression of the left side of Eq(1.2).
Because the point s1satisfies Eq(1.2), when the point s1is 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 s1and the condition which a point s1satisfies
the phase angle condition equation (1.2), this complex value can be written as:
K∗
1(a1+ib1), and α1= arg (b1/a1), K∗
1is the modulus of function W(s) of point
s1.K∗
1is a non-zero positive real value. α1is the phase angle of the function W(s)
of the point s1, and it is the value of the right side of Eq(1.2).
If we bring the point s1into the gain expression |Qn
j=1(1−s
pj)βjGjp (s)
G(s)Qm
l=1(1−s
zl)γlGlz(s)|, we can
obtain an unique gain K1,K1=|Qn
j=1(1−s1
pj)βjGjp(s1)
G(s1)Qm
l=1(1−s1
zl)γlGlz(s1)|, the gain K1is a recipro-
cal of the modulus K∗
1of W(s) of the point s1. Further obtain K∗
1.K∗
1= 1/K1. For
that the gain K1multiplied the expression G(s1)Qm
l=1(1 −s1
zl)γlGlz (s1)/Qn
j=1(1 −
s1
pj)βjGjp (s1), we have, K1G(s1)Qm
l=1(1 −s1
zl)γlGlz (s1)/Qn
j=1(1 −s1
pj)βjGjp (s1) =
a1+ib1. This equation is a concrete situation which a point s1satisfies Eq(1.1).
The previous results prove that the point s1is a root of Eq(1.1).
So, the point s1satisfies Eq(1.1). Therefore, it is proved that the point s1is
on the root locus of Eq(1.1), and which its gain is K1and its degree is 2qπ +α1.
Hence, if the point s1satisfies Eq(1.2), then the point s1is on the root locus of
Eq(1.1), and which its gain is K1and its degree is 2qπ +α1.
Lemma 1.6. Let Eq(1.1) be the root locus equation of 2qπ +αdegree, for any point
s= (σ, t)∈∆, if the point sis on the root locus of 2qπ +αdegree of Eq(1.1), then
it must satisfy the phase angle condition equation of 2qπ +αdegree of Eq(1.2).

4 LANDE MA AND ZHAOKUN MA
Proof. Assume that the point s2= (σ2, t2)∈∆ 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 s2surely satisfies Eq(1.1). When s2is substituted into
Eq(1.1), K2G(s2)Qm
l=1(1 −s2
zl)γlGlz (s2)/Qn
j=1(1 −s2
pj)βjGjp (s2) = a2+ib2. Ac-
cording to Lemma 1.3, here, the gain K2is a positive real number, its phase angle
of K2is 2q1π, the phase angle of the right side of the equation in this paragraph
on the point s2is 2q2π+ arg (b2/a2) = 2q2π+α2.
The factor of the left side of the equation in last paragraph on the point s2can be
looked as two factors. One is K2, another is G(s2)Qm
l=1(1 −s2
zl)γlGlz (s2)/Qn
j=1(1 −
s2
pj)βjGjp (s2). The phase angle of the factor of the left side of the equation in last
paragraph on the point s2is equal to the summation of two phase angles of two
factors of K2and G(s2)Qm
l=1(1 −s2
zl)γlGlz(s2)/Qn
j=1(1 −s2
pj)βjGjp (s2). The phase
angle of the factor of the left side of the equation in last paragraph on the point s2
is equal to the phase angle of the right side of the equation in the last paragraph
on the point s2. The phase angle 2q2π+α2subtracts the phase angle 2q1πis equal
to 2q2π+α2−2q1π= 2qπ +α2.
The difference of the phase angle of the factor a2+ib2and the factor K2is:
2qπ +α2. So , the phase angle of expression G(s2)Qm
l=1(1 −s2
zl)γlGlz (s2)/Qn
j=1(1 −
s2
pj)βjGjp (s2) is 2qπ +α2= 2qπ + arg (b2/a2). We can obtain: the phase angle of
W(s) is ϕ(σ, t) = arg(Gy(σ,t)
Gx(σ,t)) + Pm
l=1(γlarg( t−tl
σ−σl)−γlarg( tl
σl) + arg( Gz y (σ,t)
Gzx(σ,t))) −
Pn
j=1(βjarg( t−tj
σ−σj)−βjarg( tj
σj) + arg( Gpy (σ,t)
Gpx(σ,t))). So, according to the previous
proof, we can obtain: ϕ(σ2, t2) = 2qπ +arg(b2/a2), the point s2satisfies the phase
angle condition equation (1.2) . Hence, if the point s2is on the root locus of Eq(1.1),
which its gain is K2and its degree is 2qπ +α2, and satisfies Eq(1.1), then point s2
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 Theorem 1.7.
Theorem 1.7. Let Eq(1.1) be 2qπ +αdegree and s= (σ, t)∈∆be an arbitrary
point. A necessary and sufficient condition for that the point sis on the root locus
of 2qπ +αdegree of Eq(1.1) is that point swhether or not satisfies the phase angle
condition equation (1.2).
Definition 1.8. Let Ξ be a point set in C∪ {∞}, that are consist of all of points
on the path of roots of Eq(1.1) traced out in C∪ {∞} as 2qπ +α= 2qπ + arg( b
a) is
a constant. That path of Eq(1.1) in C∪ {∞} is called as the root locus of Eq(1.1).
Namely, the set Ξ is the root locus of the 2qπ +αdegree.
Definition 1.9. When the phase angle of meromorphic function W(s) at the left
side of the root locus equation (1.1) is 2qπ+αdegree, we call the root locus equation
(1.1) as the 2qπ +αdegree root locus equation.
In C∪{∞}, the phase angle of meromorphic function W(s) at the left side of the
root locus equation (1.1) is the degree 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 2qπ +αdegree, we call the root locus of the
root locus equation (1.1) as 2qπ +αdegree root locus.

ZEROS OF MEROMORPHIC FUNCTION 5
Lemma 1.11. For all finite zeros of Eq(1.1), they are on the root locus of all
degrees of Eq(1.1) from 2qπ degree to 2qπ + 2γlπdegree.
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)Qm
l=1(1−z p
zl)γlGlz(z p)
Qn
j=1(1−z p
pj)βjGjp (zp)= 0. This equation can also be expressed as:
G(zp)Qm
l=1(1−z p
zl)γlGlz(z p)
Qn
j=1(1−z p
pj)βjGjp (zp)=Kzp eiθzp = 0. In this equation, Kzp is the modulus of
the function W(s) and θzp is the phase angle of the function W(s). So, Kzp = 0,
and if θzp = 2qπ +α,Kzp ei(2qπ+α)= 0 ∗(cos(θzp ) + isin(θzp )) = 0 is true, and θz p
is an arbitrary degree from 2qπ degree to 2qπ + 2γlπdegree.
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(θz p) + isin(θzp ) is the non-zero and
non-infinity, the modulus of cos(θzp ) + isin(θz p) is 1. θzp represents an arbitrary
degree, which shows: no matter what value θzp is, there is Kzpei(2qπ+α)= 0 ∗
(cos(θzp) + isin(θzp )) = 0.
This proves: For the finite zero zp of Eq(1.1), when it is substituted into the
meromorphic function W(s), the phase angle of the complex value of the meromor-
phic function W(s) can be the arbitrary 2qπ +αdegree. 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 2qπ +αis an arbitrary degree, when it obtains
all of degrees, this shows the finite zeros of Eq(1.1) are simultaneously on the root
locus of all of degrees of Eq(1.1) from 2qπ degree to 2qπ + 2γlπdegree. 
Lemma 1.12. For all finite poles of Eq(1.1), they are on the root locus of all
degrees of Eq(1.1) from 2qπ degree to 2qπ + 2βjπdegree.
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)Qm
l=1(1−pz
zl)γlGlz(pz )
Qn
j=1(1−pz
pj)βjGjp (pz)=∞.
In complex analysis, 0 can be written as: 0 ∗eiθ = 0 ∗(co s θ+isin θ). The points
of non-zero and non-infinity finite values can also be written as: k∗eiθ =k∗(cos θ+
isin θ). The infinity can also be written as: (+∞)∗eiθ = (+∞)∗(cos θ+isin θ).
In which, (cos θ+isin θ) is a non-zero and non-infinity, and its modulus is 1 of the
unit complex number value. For (+∞)∗eiθ = (+∞)∗(cos θ+isin θ), no matter
what value θ= 2qπ +αobtains on the unit circle in C∪ {∞}, the values of this
expression all obtain the infinity.
The above equation can also be expressed as: Gpz =G(pz)Qm
l=1(1−pz
zl)γlGlz(pz )
Qn
j=1(1−pz
pj)βjGjp (pz)=
Kpzeiθp z =∞. In this equation, Kpz is the modulus of the meromorphic function
W(s) and θpz is the phase angle of the meromorphic function W(s). So, Kpz = +∞,
and if θpz = 2qπ +α,Kpz ∗eiθpz = (+∞)∗(cos θpz +isin θpz ) = ∞is true, and
θpz = 2qπ +αis an arbitrary degree from 2qπ degree to 2qπ + 2βjπdegree.
For the point pz that lets the left side of Eq(1.1) obtain ∞, no matter what
phase angle it is, since its modulus is ∞,cos(θpz ) + isin(θpz ) is the non-zero and
non-infinity, the modulus of cos(θpz ) + isin(θpz ) is 1. θpz = 2qπ +αrepresents an
arbitrary degree, which shows: no matter what value θpz = 2qπ +αis, there is
Kpzeiθp z = (+∞)∗(cos(θpz ) + isin(θpz)) = ∞.

6 LANDE MA AND ZHAOKUN MA
For the finite pole pz of Eq(1.1), when it is substituted into the meromorphic
function W(s), the phase angle of the complex value of the meromorphic function
W(s) can be 2qπ +αdegree. 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 2qπ +αis an arbitrary degree from 2qπ degree to 2qπ + 2βjπdegree, when
it obtains all of degrees, this shows the finite poles of Eq(1.1) are simultaneously on
the root locus of all of degrees of Eq(1.1) from 2qπ degree to 2q π +2βjπdegree. 
Lemma 1.13. All of the root locus of arbitrary 2qπ +αdegree of Eq(1.1) are
originated from poles of Eq(1.1), and are finally received by zeros of Eq(1.1).
Proof. The arbitrary 2qπ +αdegree root locus of Eq(1.1) are the curves in C∪
{∞}. So, they all need to have their own origination points and receiving points.
According to the relationship between the points in C∪ {∞} and the root locus
of Eq(1.1), the points in C∪ {∞} 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
2qπ +αdegree root locus of Eq(1.1), and the last type is the infinity points in
C∪ {∞}.
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 C∪ {∞} is on a root loci of a certain
degree of Eq(1.1). And a general finite point in C∪ {∞} only has the phase angles
of only one degree 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 C∪ {∞} let the meromorphic function W(s) obtain
a non-zero finite value A. The gain values of the meromorphic function W(s) on
the infinite point in C∪ {∞} are the non-zero finite value |A|. The phase angles
of the meromorphic function W(s) are all equal to 2qπ +arg(A) degree. So, these
infinite points in C∪ {∞} are also the ordinary points of the finite values with the
gain |A|on the 2qπ +arg(A) degree root locus, they can not originate or receive
the root locus.
When the infinite points in C∪ {∞} are the infinite zeros or poles of Eq(1.1).
On the infinite zero or pole, the gain value is K= +∞or K= 0 respectively.
The finite poles and zeros of Eq(1.1) are on the root locus of all of degrees of
Eq(1.1), and the infinite number of the different degrees 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= +∞. 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 zeros of Eq(1.1) are the receiving points of the root locus. 
Definition 1.14. In C∪ {∞}, if an angle φis between the positive direction of
the real axis and the tangent line of the root loci of arbitrary 2qπ +αdegree that
is originated from a finite pole, then angle φis called as angle of origination of the
root loci of 2qπ +αdegree of Eq(1.1).
Definition 1.15. In C∪ {∞}, if an angle ϕis between the positive direction of
the real axis and the tangent line of the root loci of arbitrary 2qπ +αdegree that
receives at a finite zero, then angle ϕis called as angle of receiving of the root loci
of 2qπ +αdegree of Eq(1.1).

ZEROS OF MEROMORPHIC FUNCTION 7
In which, when the rotation is counterclockwise, the obtained angle is the positive
angle, when the rotation is clockwise the obtained angle is the negative angle.
Theorem 1.16. In C∪ {∞}, let Pkbe a βkrepeated finite poles of Eq(1.1). The
angle of origination from Pkon the root locus of arbitrary 2qπ +αdegree of Eq(1.1)
is:
θPk=1
βk(2qπ−α+arg(G(Pk))+βkarg(−pk)−arg(Gkp (Pk))+
m
P
l=1
(γlarg(Pk−zl)−
γlarg(−zl) + arg(Glz (Pk)))−
n
P
j=1,j6=k
(βjarg(Pk−pj)−βjarg(−pj)+arg(Gjp (Pk))).
In which, βkis a real number.
Proof. Selecting a point s1on the root locus of 2qπ +αdegree, the point s1should
be infinitely approaching to βkrepeated finite pole Pkthat needs to compute its
angle of origination. Because point s1is infinitely approaching to the finite pole
Pk, the phase angle βkarg(s1−pj) of vectors between the point s1and poles (pj,
j6=k. Namely, except pole Pk.) all can be substituted by angles βkarg(Pk−pj),
j6=k. The phase angle γkarg(s1−zl) of vectors between the point s1and all of
the finite zeros zlof Eq(1.1) all can be substituted by angles γkarg(Pk−zl).
The point s1is on the root locus of 2qπ +αdegree, because this point is infinitely
approaching to the finite pole Pk, the point s1is infinitely approaching to the
tangent line of the root loci of arbitrary 2qπ +αdegree that originates from the
finite pole Pk, so, the vector between point s1and the finite pole Pkis infinitely
approaching to the tangent line of the root loci of arbitrary 2qπ +αdegree that
originates from the finite pole Pk. So, according to the Definition 1.14, we can
let the phase angle of vectors between point s1and the finite pole Pkbe infinitely
approaching to angle of origination θPk.
The point s1must satisfy the phase angle condition equation, arg(G(s1)) +
m
P
l=1
(γlarg(s1−zl)−γlarg(−zl)+ arg(Glz (s1)))−
n
P
j=1
(βjarg(s1−pj)−βjarg(−pj)+
arg(Gj p (s1))) = 2qπ +α. So, when the finite pole Pksubstitutes point s1, we
can obtained. arg(G(Pk)) +
m
P
l=1
(γlarg(Pk−zl)−γlarg(−zl) + arg(Glz (Pk))) −
n
P
j=1,j6=k
(βjarg(Pk−pj)−βjarg(−pj) + arg(Gjp (Pk))) −βkθPk+βkarg(−pk)−
arg(Gkp (Pk)) = 2qπ +α.
After terms are moved, obtain: βkθPk=arg(G(Pk)) +
m
P
l=1
(γlarg(Pk−zl)−
γlarg(−zl)+arg(Glz (Pk)))−
n
P
j=1,j6=k
(βjarg(Pk−pj)−βjarg(−pj)+arg(Gjp(Pk)))+
βkarg(−pk)−arg(Gkp (Pk)) −2qπ −α
The 2qπ and −2qπ are equivalent, so, 2qπ is substituted by −2qπ, after trans-
positions, we can obtain the formula in Theorem 1.16. 
Theorem 1.17. In C∪ {∞}, let Zkbe a γkrepeated zeros of Eq(1.1). The angle
of receiving by Zkon the root locus of arbitrary 2qπ +αdegree of Eq(1.1) is:
θZk=1
γk(2qπ+α−arg(G(Zk))+γkarg(−zk)−arg(Gkz (Zk))−
m
P
l=1,l6=k
(γlarg(Zk−zl)−
γlarg(−zl) + arg(Glz (Zk))) +
n
P
j=1
(βjarg(Zk−pj)−βjarg(−pj) + arg(Gjp (Zk))).

8 LANDE MA AND ZHAOKUN MA
In which, γkis a real number.
Proof. Selecting a point s1on the root locus of 2qπ +αdegree. The point s1is
infinitely approaching to the γkrepeated zero Zkthat needs to compute its angle
of receiving. Because the point s1is infinitely approaching to zero Zk. The phase
angle βkarg(s1−pj) of vectors between point s1and all poles pjof Eq(1.1) all
can be substituted by angles βkarg(Zk−pj). The phase angle γkarg(s1−zl) of
vectors between point s1and all zeros (zl,l6=k, except zero Zk) of Eq(1.1) all can
be substituted by angles γkarg(Zk−zl).
The point s1is on the root locus of 2qπ +αdegree, because this point is infinitely
approaching to the finite zero Zk, the point s1is infinitely approaching to the
tangent line of the root loci of arbitrary 2qπ +αdegree that is received by the
finite zero Zk, so, the vector between point s1and the finite zero Zkis infinitely
approaching to the tangent line of the root loci of arbitrary 2qπ +αdegree that
is received by the finite zero Zk. So, according to the Definition 1.15, we can let
the phase angle of vectors between point s1and the finite zero Zkbe infinitely
approaching to angle of receiving θZk.
The point s1must satisfy the phase angle condition equation, arg(G(s1)) +
m
P
l=1
(γlarg(s1−zl)−γlarg(−zl)+ arg(Glz (s1)))−
n
P
j=1
(βjarg(s1−pj)−βjarg(−pj)+
arg(Gj p (s1))) = 2qπ+α. So, when the finite zero Zksubstitutes point s1, we can ob-
tained. arg(G(Zk)) +
m
P
l=1,l6=k
(γlarg(Zk−zl)−γlarg(−zl) + arg(Glz (Zk)))+ γkθZk−
γkarg(−zk) + arg(Gkz (Zk)) −
n
P
j=1
(βjarg(Zk−pj)−βjarg(−pj) + arg(Gjp (Zk))) =
2qπ +α.
After terms are moved, obtain: γkθZk=−arg(G(Zk)) −
m
P
l=1,l6=k
(γlarg(Zk−zl)−
γlarg(−zl) + arg(Glz (Zk))) +
n
P
j=1
(βjarg(Zk−pj)−βjarg(−pj) + arg(Gjp (Zk))) +
γkarg(−zk)−arg(Gkz (Zk)) + 2qπ +α. After transpositions, we can obtain the
formula in Theorem 1.17. 
Theorem 1.18. The difference of degrees of the root locus that a pole emits on two
opposite directions is α1−α2=βkπ. In which, α1is the degree that a pole emits
the root locus on one direction, α2is the degree that the pole emits the root locus
on the opposite direction, βkis the exponent of the pole.
Proof. According to Theorem 1.16, two angles of departure at the βkrepeated pole
Pkon two opposite directions are:
θPk=1
βk(2qπ−α1+arg(G(Pk))+βkarg(−pk)−arg(Gkp (Pk))+
m
P
l=1
(γlarg(Pk−zl)−
γlarg(−zl) + arg(Glz (Pk)))−
n
P
j=1,j6=k
(βjarg(Pk−pj)−βjarg(−pj)+arg(Gjp (Pk))).
θPk+π=1
βk(2qπ−α2+arg(G(Pk))+βkarg(−pk)−arg(Gkp (Pk))+
m
P
l=1
(γlarg(Pk−zl)−
γlarg(−zl) + arg(Glz (Pk)))−
n
P
j=1,j6=k
(βjarg(Pk−pj)−βjarg(−pj)+arg(Gjp (Pk))).

ZEROS OF MEROMORPHIC FUNCTION 9
Two sides of two equations separately subtract, and obtain: −π=1
βk(−α1+α2).
−πβk=−α1+α2. So, α1−α2=βkπ.
Theorem 1.19. The difference of degrees of the root locus that a zero receives on
two opposite directions is α2−α1=γkπ, In which, α1is the degree that a zero
receives the root locus on one direction, α2is the degree that the zero receives the
root locus on the opposite direction, γkis the exponent of zero.
Proof. According to Theorem 1.17, two angles of arrival at the γkrepeated zero Zk
on two opposite directions are:
θZk=1
γk(2qπ+α1−arg(G(Zk))+γkarg(−zk)−arg(Gkz (Zk))−
m
P
l=1,l6=k
(γlarg(Zk−zl)−
γlarg(−zl) + arg(Glz (Zk))) +
n
P
j=1
(βjarg(Zk−pj)−βjarg(−pj) + arg(Gjp (Zk))).
θZk+π=1
γk(2qπ+α2−arg(G(Zk))+γkarg(−zk)−arg(Gkz (Zk))−
m
P
l=1,l6=k
(γlarg(Zk−zl)−
γlarg(−zl) + arg(Glz (Zk))) +
n
P
j=1
(βjarg(Zk−pj)−βjarg(−pj) + arg(Gjp (Zk))).
Two sides of two equations separately subtract, and obtain: −π=1
γk(α1−α2).
−πγk=α1−α2. So, α2−α1=γkπ.
Theorem 1.20. For the root locus which are originated from an arbitrary finite
pole Pkof Eq(1.1) in C∪ {∞}, if the angle θPkof origination is a independent
variable, then, the degree values of the root locus strictly and monotonously increase
clockwise.
Proof. If the angle of origination of the α1degree root locus of Eq(1.1) on the fi-
nite pole Pkis: θPk1=1
βk(2qπ −α1+arg(G(Pk)) + βkarg(−pk)−arg(Gkp(Pk)) +
m
P
l=1
(γlarg(Pk−zl)−γlarg(−zl)+arg(Glz (Pk)))−
n
P
j=1,j6=k
(βjarg(Pk−pj)−βjarg(−pj)+
arg(Gj p (Pk))). If the angle of origination of the α2degree root locus of Eq(1.1) on
the finite pole is: θPk2=1
βk(2qπ −α2+arg(G(Pk)) + βkarg(−pk)−arg(Gkp (Pk)) +
m
P
l=1
(γlarg(Pk−zl)−γlarg(−zl)+arg(Glz (Pk)))−
n
P
j=1,j6=k
(βjarg(Pk−pj)−βjarg(−pj)+
arg(Gj p (Pk))). Two equations subtract, obtain: θPk1−θPk2=1
βk(α2−α1), so,
α2−α1=βk(θPk1−θPk2).
βkis the exponent of pole Pk, it is a real number. θPk1and θPk2are separately
the origination angle of the root locus of α1degree and α2degree. The angle
of origination is defined as a angle which is between the positive direction of the
real axis and the tangent line of the root loci that is originated from a finite pole.
Beginning at the positive real axis, rotating counterclockwise to the tangent line of
the root loci, a angle of origination can be obtained. So, if θPk1> θPk2,θPk1−θPk2>
0. The angle of origination of the root loci of α1degree is θPk1. The angle of
origination of the root loci of α2degree is θPk2.α2−α1>0, α2> α1is true. So,
it can prove: for two root locus which are originated on the two different angles of
origination from a same pole, the root loci of the larger degree is at the clockwise
direction of the root loci of the less degree.
If θPk1< θPk2,θPk1−θPk2<0. α2−α1<0, α2< α1is true. So, it can prove:
for two root locus which are originated on the two different angles of origination

10 LANDE MA AND ZHAOKUN MA
from a same pole, the root loci of the larger degree is at the clockwise direction of
the root loci of the less degree.
So, the degree values of the root locus that are originated from this pole Pk
strictly and monotonously increases clockwise. 
Theorem 1.21. For the root locus which are received by an arbitrary finite zero
Zkof Eq(1.1) in C∪ {∞}, if the angle θZkof receiving is a independent variable,
then its degree values strictly and monotonously decrease clockwise.
Let the difference of the degree numbers of a pair of the root locus which begin
at pole Pkon opposite direction be a positive, according to Theorem 1.18, we can
obtain Theorem 1.22. Let the difference of the degree numbers of a pair of the
root locus which begin at pole Zkon opposite direction be a positive, according to
Theorem 1.19, we can obtain Theorem 1.23.
Theorem 1.22. A necessary and sufficient condition for the βkrepeated pole Pk
to be simple is that the difference of the degree numbers of any one pair of the root
locus which begin at pole Pkon opposite direction is a positive integer times of π.
Theorem 1.23. A necessary and sufficient condition for the γkrepeated zero Zk
to be simple is that the difference of the degree number of any one pair of the root
locus which end at zero Zkon opposite direction is a positive integer times of π.
ξ(s) = 1
2s(s−1)Γ( s
2)π−s
2ζ(s) is a conjugate function, namely, if ξ(s) = u(σ+
it) + iv(σ+it), then, ξ(s) = u(σ+it)−iv(σ+it). ξ(s) = ξ(1 −s). So, ξ(1
2+it) =
ξ(1
2−it), namely, if ξ(1
2+it) = u(1
2+it) + iv(1
2+it), then, ξ(1
2+it) = ξ(1
2−
it) = u(1
2+it) + iv(1
2+it). By the property of the conjugate function, we have:
ξ(1
2−it) = u(1
2+it)−iv(1
2+it). So, v(1
2+it) = 0, on the critical line, xi-function
obtains the real number values.
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ZEROS OF MEROMORPHIC FUNCTION 11
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Current address: School of Mathematical Sciences, Tongji University, Shanghai, 200092, China
Email address:dzy200408@126.com
Current address: YanZhou College, ShanDong Radio and TV University, YanZhou, ShanDong
272100 China
Email address:dzy200408@sina.cn