In other words, if R(z) is a rational function, then there exist polynomial P (z) and Q(z) in C[z] with Q(z) 6= 0 such that R(z

1. Argument Principle

A rational function is a quotient of a complex polynomial by a nonzero complex poly- nomial¹. In other words, if R(z) is a rational function, then there exist polynomial P (z) and Q(z) in C[z] with Q(z) 6= 0 such that R(z) = P (z)/Q(z). When deg P (z) > deg Q(z), division algorithm implies P (z) = Q(z)g(z) + r(z) for some g(z) r(z) in C[z] such that deg r(z) < deg Q(z) or r(z) = 0. Hence

R(z) = g(z) + r(z) Q(z).

Using division algorithm, we will simply consider rational function R(z) = P (z)/Q(z) with deg P (z) < deg Q(z). Denote n = deg P (z) and m = deg Q(z). Also assume that P (z), Q(z) are relatively prime.

Rational functions define meromorphic functions on C. Assume that p1, · · · , pn and q₁, · · · , q_m are zeros of P (z) and Q(z) respectively. Then R(z) is holomorphic on U = C \ {q1, · · · , qm} and

R⁰(z)

R(z) = P⁰(z)

P (z) −Q⁰(z)

Q(z), z ∈ U.

Let D be a closed disk containing {p₁, · · · , p_n} ∪ {q₁, · · · , q_m} whose boundary C = ∂D is an oriented circle with positive orientation. Then R⁰(z)/R(z) is defined on C and

Z

C

R⁰(z) R(z)dz =

Z

C

P⁰(z) P (z)dz −

Z

C

Q⁰(z) Q(z)dz.

The function P⁰(z)/P (z) has the following expansion P⁰(z)

P (z) = 1 z − p1

+ 1

z − p2

+ · · · + 1 z − pn

.

Let C_i be the small circle with positive orientation center at p_i of radius δ > 0 such that C_i are disjoint. Using the property of path integration, we obtain

I

C

P⁰(z) P (z)dz =

n

X

i=1

Z

Ci

dz z − pi

= 2πin.

Similarly, we have

I

C

Q⁰(z) Q(z)dz =

m

X

j=1

Z

C_j⁰

dz

z − q_i = 2πim.

where C_j⁰ is the small circle with positive orientation center at qi of radius δ⁰ > 0 for 1 ≤ j ≤ m such that C_j⁰ are disjoint. This implies that

I

C

R⁰(z)

R(z)dz = 2πi(n − m).

Let f (z) be a nonzero meromorphic function on a region (open connected set) Ω. Let D be a closed disk in Ω whose boundary C is an oriented circle with positive orientation. Let Z_C(f ) and P_C(f ) be the set of zeros and the set of poles of f (z) lying in the interior D^◦ of D. Since D is a compact, f can only have finitely many zeros and poles in D. Both of ZC(f ) and Z_P(f ) are finite sets. Let us denote Z_C(f ) = {p₁, · · · , p_r} and P_C(f ) = {q₁, · · · , q_s}

1We only consider rational function over C.

1

2

where pi and qj are all distinct. Denote the order of pi by ni and the order of qj by mj. We construct a rational function R(z) by

R(z) = Q_r

i=1(z − p_i)ⁿⁱ Qs

j=1(z − q_j)^mj.

Then the zeros and poles of R(z) lie inside D^◦. Define a new function g(z) = f (z)/R(z).

Then p_i are removable singularities of g and hence g can be extended to a meromorphic function on Ω such that g is holomorphic in D^◦ and g is nonzero on D. Write f (z) = R(z)g(z). Let U be any open subset of Ω such that f is holomorphic. For z ∈ U,

f⁰(z)

f (z) = R⁰(z)

R(z) +g⁰(z) g(z).

Since g is nonzero on D^◦, g⁰(z)/g(z) is holomorphic in D^◦. By the previous argument, 1

2πi I

C

f⁰(z)

f (z)dz = 1 2πi

I

C

R⁰(z)

R(z)dz + 1 2πi

I

C

g⁰(z) g(z)dz

= 1 2πi

I

C

R⁰(z) R(z)dz

= (n1+ · · · + nr) − (m1+ · · · + ms)

Let us denote #Z_C(f ) = n₁ + · · · + n_r and #P_C(f ) = m₁+ · · · + m_s. Then #Z_C(f ) and

#PC(f ) are numbers of zeros and numbers of poles in D^◦ counting multiplicities.

Theorem 1.1. (Argument Principle) Let f be a meromorphic function on a region Ω. Let D be a closed disk whose boundary C is a oriented circle with positive orientation. Let

#Z_C(f ) and #P_C(f ) be numbers of zeros and numbers of poles of f in D^◦with multiplicities counted. Then

1 2πi

I

C

f⁰(z)

f (z)dz = #ZC(f ) − #PC(f ).

If f is a holomorphic function on a region Ω, we would like to study the number of solutions to the equation

(1.1) f (z) = ω

in the closed disk D in Ω; here we require that ∂D contains no solution of (1.1). When ω = 0, the numbers of solution to (1.1) can be calculated using argument principle. In fact, if we denote F (z) = f (z) − ω, then (1.1) is equivalent to F (z) = 0. Using argument principle, the number of solutions to F (z) = 0 in D is given by the count our integral

1 2πi

I

C

F⁰(z)

F (z)dz = 1 2πi

I

C

f⁰(z) f (z) − ωdz.

2. Generalization of Argument Principle

Let h(z) be a holomorphic function on a region Ω and R(z) be a rational function whose zeros and poles are contained in a closed disk D in Ω. Assume R(z) is nonzero on C and C is equipped with the positive orientation. Then we would like to compute

1 2πi

I

C

h(z)R⁰(z) R(z)dz.

3

Let R(z) and P (z) and Q(z) be as the previous section. Using Cauchy’s integral formula, 1

2πi I

C

h(z)P⁰(z) P (z)dz =

n

X

i=1

1 2πi

Z

C

h(z) (z − pi)dz

=

n

X

i=1

h(p_i).

When R(z) = P (z)/Q(z), we have 1

2πi I

C

h(z)R⁰(z)

R(z)dz = 1 2πi

I

C

h(z)P⁰(z)

P (z)dz + 1 2πi

I

C

h(z)Q⁰(z) Q(z)dz

=

n

X

i=1

h(p_i) −

m

X

j=1

h(q_j).

Let f (z) be a meromorphic function on Ω. Write f (z) = R(z)g(z) where R(z) is a rational function and g(z) is a meromorphic function on Ω so that g(z) is holomorphic on D^◦ and nonzero on D. Let U be any open subset of Ω so that f (z) is holomorphic on U. For any z ∈ U

h(z)f⁰(z)

f (z) = h(z)R⁰(z)

R(z) + h(z)g⁰(z) g(z). Since h(z)g⁰(z)/g(z) is holomorphic on D^◦, we obtain

1 2πi

I

C

h(z)f⁰(z)

f (z)dz = 1 2πi

I

C

h(z)R⁰(z)

R(z)dz + 1 2πi

I

C

h(z)g⁰(z) g(z)dz

= 1 2πi

I

C

h(z)R⁰(z) R(z)dz

=

n

X

i=1

h(p_i) −

m

X

j=1

h(q_j).

Theorem 2.1. Let h be a holomorphic function on a region Ω and f be a meromorphic function on Ω. Let D be a closed disk whose boundary C does not contain any of zeros or poles of f. Let p₁, · · · , p_n be zeros of f in D and q₁, · · · , q_m be poles of f in D. Then

1 2πi

I

C

h(z)f⁰(z) f (z)dz =

n

X

i=1

h(p_i) −

m

X

j=1

h(q_j).

A univalent function on an open set U is a holomorphic injection f : U → C. For each ω ∈ f (U ), the solution to f (z) = ω is unique. Let us denote the solution by z(ω). Choose a closed disk D such that the point z(ω) is contained in the interior of D. We obtain

z(ω) = 1 2πi

I

C

z f⁰(z) f (z) − ωdz.

Here C = ∂D.