Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

12.1: Principle of the Argument

( \newcommand{\kernel}{\mathrm{null}\,}\)

Setup

γ a simple closed curve, oriented in a counterclockwise direction. f(z) analytic on and inside γ, except for (possibly) some finite poles inside (not on) γ and some zeros inside (not on) γ.

Let p1, ...,pm be the poles of f inside γ.
Let z1, ...,zn be the zeros of f inside γ.
Write mult(zk) = the multiplicity of the zero at zk. Likewise write mult(pk) = the order of the pole at pk.

We start with a theorem that will lead to the argument principle.

Theorem 12.1.1

With the above setup

γf(z)f(z) dz=2πi(mult(zk)mult(pk)).

Proof

To prove this theorem we need to understand the poles and residues of f(z)/f(z). With this in mind, suppose f(z) has a zero of order m at z0. The Taylor series for f(z) near z0 is

f(z)=(zz0)mg(z)

where g(z) is analytic and never 0 on a small neighborhood of z0. This implies

f(z)f(z)=m(zz0)m1g(z)+(zz0)mg(z)(zz0)mg(z)=mzz0+g(z)g(z)

Since g(z) is never 0, g(z)/g(z) is analytic near z0. This implies that z0 is a simple pole of f(z)/f(z) and

Res(f(z)f(z),z0)=mmult(z0).

Likewise, if z0 is a pole of order m then the Laurent series for f(z) near z0 is

f(z)=(zz0)mg(z)

where g(z) is analytic and never 0 on a small neighborhood of z0. Thus,

f(z)f(z)=m(zz0)m1g(z)+(zz0)mg(z)(zz0)mg(z)=mzz0+g(z)g(z)

Again we have that z0 is a simple pole of f(z)/f(z) and

Res(f(z)f(z),z0)=m=mult(z0).

The theorem now follows immediately from the Residue Theorem:

γf(z)f(z) dz=2πi sum of the residues=2πi(mult(zk)mult(pk)).

Definition

We write Zf,γ for the sum of multiplicities of the zeros of f inside γ. Likewise for Pf,γ. So the Theorem 12.2.1 says,

γff dz=2πi(Zf,γPf,γ).

Definition: Winding Number

We have an intuition for what this means. We define it formally via Cauchy’s formula. If γ is a closed curve then its winding number (or index) about z0 is defined as

Ind(γ,z0)=12πiγ1zz0 dz.

Mapping Curves: fγ

One of the key notions in this topic is mapping one curve to another. That is, if z=γ(t) is a curve and w=f(z) is a function, then w=fγ(t)=f(γ(t)) is another curve. We say f maps γ to fγ. We have done this frequently in the past, but it is important enough to us now, so that we will stop here and give a few examples. This is a key concept in the argument principle and you should make sure you are very comfortable with it.

Example 12.1.1

Let γ(t)=eit with 0t2π (the unit circle). Let f(z)=z2. Describe the curve fγ.

Solution

Clearly fγ(t)=e2it traverses the unit circle twice as t goes from 0 to 2π.

Example 12.1.2

Let γ(t)=it with <t< (the y-axis). Let f(z)=1/(z+1). Describe the curve fγ(t).

Solution

f(z) is a fractional linear transformation and maps the line given by γ to the circle through the origin centered at 1∕2. By checking at a few points:

f(i)=1i+1=1+i2, f(0)=1, f(i)=1i+1=1i2, f()=0.

We see that the circle is traversed in a clockwise manner as t goes from to .

屏幕快照 2020-09-14 下午3.11.36.png
The curve z=γ(t)=it is mapped to w=fγ(t))=1/(it+1).

Argument Principle

You will also see this called the principle of the argument.

Theorem 12.1.2 Argument principle

For f and γ with the same setup as above

γf(z)f(z) dz=2πiInd(fγ,0)=2πi(Zf,γPf,γ)

Proof

Theorem 12.2.1 showed that

γf(z)f(z) dz=2πi(Zf,γPf,γ)

So we need to show is that the integral also equals the winding number given. This is simply the change of variables w=f(z). With this change of variables the countour z=γ(t) becomes w=fγ(t) and dw=f(z) dz so

γf(z)f(z) dz=fγdww=2πiInd(fγ,0)

The last equality in the above equation comes from the definition of winding number.

Note that by assumption γ does not go through any zeros of f, so w=f(γ(t)) is never zero and 1/w in the integral is not a problem.

Here is an easy corollary to the argument principle that will be useful to us later.

Corollary

Assume that fγ does not go through −1, i.e. there are no zeros of 1+f(z) on γ then

γff+1=2πiInd(fγ,1)=2πi(Z1+f,γPf,γ).

Proof

Applying the argument principle in Equation 12.2.11 to the function 1+f(z), we get

γ(1+f)f(z)1+f(z) dz=2πiInd(1+fγ,0)=2πi(Z1+f,γP1+f,γ)

Now, we can compare each of the terms in this equation to those in Equation 12.2.14:

γ(1+f)f(z)1+f(z) dz=γff(z)1+f(z) dz  (because (1+f)=f)Ind(1+fγ,0)=Ind(fγ,1)  (1+f winds around 0  winds around -1)Z1+f,γ=Z1+f,γ  (same in both equation))P1+f,γ=Pf,γ  (poles of f=poles of 1+f)

Example 12.1.3

Let f(z)=z2+z Find the winding number of fγ around 0 for each of the following curves.

  1. γ1 = circle of radius 2.
  2. γ2 = circle of radius 1/2.
  3. γ3 = circle of radius 1.
Solution

f(z) has zeros at 0, −1. It has no poles.

So, f has no poles and two zeros inside γ1. The argument principle says Ind(fγ1,0)=Zf,γ1Pf,γ=2

Likewise f has no poles and one zero inside γ2, so Ind(fγ2,0)=10=1

For γ3 a zero of f is on the curve, i.e. f(1)=0, so the argument principle doesn’t apply. The image of γ3 is shown in the figure below – it goes through 0.

屏幕快照 2020-09-15 下午2.14.35.png
The image of 3 different circles under f(z)=z2+z.

Rouché’s Theorem

Theorem 12.1.3: Rouché’s theorem

Make the following assumptions:

  • γ is a simple closed curve
  • f,h are analytic functions on and inside γ, except for some finite poles.
  • There are no poles of f and h on γ.
  • |h|<|f| everywhere on γ.

Then

Ind(fγ,0)=Ind((f+h)γ,0).

That is,

Zf,γPf,γ=Zf+h,γPf+h,γ

Proof

In class we gave a heuristic proof involving a person walking a dog around fγ on a leash of length hγ. Here is the analytic proof.

The argument principle requires the function to have no zeros or poles on γ. So we first show that this is true of f,f+h,(f+h)/f. The argument is goes as follows.

Zeros: The fact that 0|h|<|f| on γ implies f has no zeros on γ. It also implies f+h has no zeros on γ, since the value of h is never big enough to cancel that of f. Since f and f+h have no zeros, neither does (f+h)/f.

Poles: By assumption f and h have no poles on γ, so f+h has no poles there. Since f has no zeors on γ, (f+h)/f has no poles there.

Now we can apply the argument principle to f and f+h

12πiγff dz=Ind(fγ,0)=Zf,γPf,γ.

12πiγ(f+h)f+h dz=Ind((f+h)γ,0)=Zf+h,γPf+h,γ.

Next, by assumption |hf|<1, so (hf)γ is inside the unit circle. This means that 1+hf=f+hf maps γ to the inside of the unit disk centered at 1. (You should draw a figure for this.) This implies that

Ind((f+hf)γ,0)=0.

Let g=f+hf. The above says Ind(gγ,0)=0. So, γgg dz=0. (We showed above that g has no zeros or poles on γ.)

Now, it's easy to compute that gg=(f+h)f+hff. So, using

Ind(gγ,0)=γgg dz=γ(f+h)f+h dzγff dz=0Ind((f+h)γ,0)=Ind(fγ,0).

Now equations 12.2.19 and 12.2.20 tell us Zf,γPf,γ=Zf+h,γPf+h,γ, i.e. we have proved Rouchés theorem.

Corollary

Under the same hypotheses, If h and f are analytic (no poles) then

Zf,γ=Zf+h,γ.

Proof

Since the functions are analytic Pf,γ and Pf+h,γ are both 0. So Equation 12.2.18 shows Zf=Zf+h. QED

We think of h as a small perturbation of f.

Example 12.1.4

Show all 5 zeros of z5+3z+1 are inside the curve C2:|z|=2.

Solution

Let f(z)=z5 and h(z)=3z+1. Clearly all 5 roots of f (really one root with multiplicity 5) are inside C2. Also clearly, |h|<7<32=|f| on C2. The corollary to Rouchés theorem says all 5 roots of f+h=z5+3z+1 must also be inside the curve.

Example 12.1.5

Show z+3+2ez has one root in the left half-plane.

Solution

Let f(z)=z+3, h(z)=2ez. Consider the contour from iR to iR along the y-axis and then the left semicircle of radius R back to iR. That is, the contour C1+CR shown below.

屏幕快照 2020-09-15 下午2.41.51.png

To apply the corollary to Rouchés theorem we need to check that (for R large) |h|<|f| on C1+CR. On C1, z=iy, so

|f(z)|=|3+iy|3,   |h(z)|=2|eiy|=2.

So |h|<|f| on C1.

On CR, z=x+iy with x<0 and |z|=R. So,

|f(z)|>R3 for R large, |h(z)|=2|ex+iy|=2ex<2 (since x<0).

So |h|<|f| on CR.

The only zero of f ia at z=3, which lies inside the contour.

Therefore, by the Corollary to Rouchés theorem, f+h has the same number of roots as f inside the contour, that is 1. Now let R go to infinity and we see that f+h has only one root in the entire half-plane.

Theorem 12.1.4: Fundamental Theorem of Algebra

Rouchés theorem can be used to prove the fundamental theorem of algebra as follows.

Proof

Let

P(z)=zn+an1zn1+ ...+a0

be an nth order polynomial. Let f(z)=zn and h=Pf. Choose an R such that R>max(1,n|an1|,...,n|a0|). Then on |z|=R we have

|h||an1|Rn1+|an2|Rn2+ ...+|a0|RnRn1+RnRn2+ ...+Rn<Rn.

On |z|=R we have |f(z)|=Rn, so we have shown |h|<|f| on the curve. Thus, the corollary to Rouchés theorem says f+h and f have the same number of zeros inside |z|=R. Since we know f has exactly n zeros inside the curve the same is true for the polynomial f+h. Now let R go to infinity, we’ve shown that f+h has exactly n zeros in the entire plane.

Note

The proof gives a simple bound on the size of the zeros: they are all have magnitude less than or equal to max(1,n|an1|,...,n|a0|).


This page titled 12.1: Principle of the Argument is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?