argument.htm Argument Principle
Argument Principle


Zeroes and Poles
For the moment, we shall consider a function f(z) analytic in the punctured disk
⋅
D
 
z0,R 
 = {z |0 < |z−z0| ≤ R}.
Then
f(z)
= ∞
∑
n = −∞ 
 an (z−z0)n,
an
= 1
2πi
 ⌠
(⎜)
⌡


Cz0,r 
 f(ζ) (ζ−z0)−n − 1  dζ.
•
If f(z) = ∑n = 0∞ an (z−z0)n, f(z) may be extended by defining f(z0) = a0, and the resulting function is analytic in |z−z0| ≤ R.

•
If f(z) = ∑n = N∞ an (z−z0)n, N ≥ 0, aN ≠ 0, f(z) is said to have a zero of order N at z=z0. Near z=z0,
f(z) = (z−z0)N ·g(z),
where g(z) is analytic in |z−z0| ≤ R, g(z0) ≠ 0.

•
If f(z) = ∑n = − M∞ an (z−z0)n, M ≥ 0, a−M ≠ 0, f(z) is said to have a pole of order M at z=z0. Near z=z0,
f(z) = (z−z0)−M ·g(z),
where g(z) is analytic in |z−z0| ≤ R, g(z0) ≠ 0.

•
If f(z) = ∑n = − ∞∞ an (z−z0)n, an ≠ 0 for infinitely many negative n, then f(z) is said to have an essential singularity at z=z0.

•
The coefficient of (z−z0)−1 is called the residue of f(z) at z=z0, and is written
Res(f,z=z0) = Resf(z)|z=z0 = 1
2πi
 ⌠
(⎜)
⌡


Cz0,r 
 f(ζ)  dζ.

•
Let f(z) be analytic in the punctured disk
⋅
D
 
z0,R 
 = {z |0 < |z−z0| ≤ R}.
Then for r small and positive,
⌠
(⎜)
⌡


Cz0,r 
 f(ζ)  dζ = 2 πi Resf(z)|z=z0.

•
Let f(z) be analytic in the punctured disk
⋅
D
 
z0,R 
 = {z |0 < |z−z0| ≤ R}.
Suppose that f(z) has a zero of order N > 0, at z=z0.
For z near z0, f(z) = (z−z0)N g(z), g(z) analytic, and g(z0) ≠ 0. It follows that
f′(z)
f(z)
= N(z − z0)N−1 g(z) + (z−z0)N g′(z)
(z−z0)N g(z)
=N (z − z0)−1 + analytic,
so that
Res⎛
⎝ 		f′
f
 ,z = z0⎞
⎠
= N
= order of zero at z=z0
Then for r small and positive,
1
2πi
 ⌠
(⎜)
⌡


Cz0,r 
 f′(ζ)
f(ζ)
  dζ = N.

There is another interpretation of the number N. For the moment let fN(z) = (z − z0)N. Follow the argfN(z) as Cz0,r is traversed in the counterclockwise direction. The change in argument of (z − z0)N, denoted by ∆Cz0,r argfN(z) is exactly 2π N. This is the first statement of the Argument Principle :
1
2 π
 ∆Cz0,r argfN(z)
= N
= order of zero.
In the case f(z) has a zero of order N at z=z0, we expect that an antiderivative of the function [(f′(z))/f(z)] is log(f(z)). This is the case locally, at least if we are near enough to a point z1 on Cz0,r. As the path Cz0,r is traversed counterclockwise, the logarithm of f(z) may be defined locally in a continuous manner, but when we make one full revolution around the circle returning to z1, the argument of f(z) may have changed by a multiple of 2π. We have that
⌠
(⎜)
⌡


Cz0,r 
 f′(ζ)
f(ζ)
  dζ
= i ·∆Cz0,r argf(z)
= 2 πi ·N.
= 2 πi ·order of zero at z=z0.
Thus for the small circle Cz0,r,
N
= 1
2π
 ∆Cz0,r arg f(z)
= 1
2πi
 ⌠
(⎜)
⌡


Cz0,r 
 f′(ζ)
f(ζ)
  dζ

•
Let f(z) be analytic in the punctured disk
⋅
D
 
z0,R 
 = {z |0 < |z−z0| ≤ R}.
Suppose that f(z) has a pole of order M > 0, at z=z0.
Then for r small and positive,
1
2πi
 ⌠
(⎜)
⌡


Cz0,r 
 f′(ζ)
f(ζ)
  dζ = − M.
Mimicking the discussion above for zeroes, we obtain for the small circle Cz0,r
− M
= 1
2π
 ∆Cz0,r arg f(z)
= 1
2πi
 ⌠
(⎜)
⌡


Cz0,r 
 f′(ζ)
f(ζ)
  dζ


We are now ready to state the Argument Principle .

Theorem. Let C be a simple closed path. Suppose that f(z) is analytic and nonzero on C and meromorphic inside C.
•
List the zeroes of f inside C as z1, …,zk with multiplicities N1, …,Nk, and let
ZC = N1 + …+ Nk.
•
List the poles of f inside C as w1, …,wj with orders N1, …,Nk, an let
PC = M1 + …+ Mj.
Then
ZC − PC
= 1
2π
 ∆C arg f(z)
= 1
2πi
 ⌠
(⎜)
⌡


C 
 f′(ζ)
f(ζ)
  dζ.

Proof. calculate the integral two ways. First take a local antiderivative log(f(z)) to obtain
1
2πi
 ⌠
(⎜)
⌡


C 
 f′(ζ)
f(ζ)
  dζ
= 1
2π
 ∆C arg f(z).
Second take small circles around each zi and wi and the usual cuts from C to to the circles. In this way, obtain
1
2πi
 ⌠
(⎜)
⌡


C 
 f′(ζ)
f(ζ)
  dζ
= j
∑
i=1 
 1
2πi
 ⌠
(⎜)
⌡


Czi,r 
 f′(ζ)
f(ζ)
  dζ + k
∑
i=1 
 1
2πi
 ⌠
(⎜)
⌡


Cwi,r 
 f′(ζ)
f(ζ)
  dζ
= j
∑
i=1 
 Ni − k
∑
i=1 
 Mi
= ZC − PC.



Corollary. Let C be a simple closed path. Suppose that f(z) is analytic and nonzero on C and analytic inside C.
•
List the zeroes of f inside C as z1, …,zk with multiplicities N1, …,Nk, an let
ZC = N1 + …+ Nk.
Then
ZC
= 1
2π
 ∆C arg f(z)
= 1
2πi
 ⌠
(⎜)
⌡


C 
 f′(ζ)
f(ζ)
  dζ.
Briefly stated: Let C be a simple closed path. Suppose that f(z) is analytic and nonzero on C and analytic inside C. Then
1
2π
 ∆C arg f(z)
= number of zeroes inside C - counting multiplicities.


Indices and Winding Numbers


Let C be a simple closed path. Suppose that f(z) is analytic and nonzero on C and meromorphic inside C. Then w = f(z) = f(z(t)) is a closed path (not necessarily simple). Call this path f(C). As w traverses f(c), the number of times the argument of w changes by a multiple of 2π is called the index or winding number of the path f(C). The Argument Principle says that the winding number of f(C) is ZC − PC.


File translated from TEX by TTH, version 4.03.
On 16 Nov 2013, 20:49.