Argument Principle

The argument principle in complex analysis relates the difference between the number of zeros and the number of poles to the closed integral of the logarithmic derivative of an analytic function. The argument principle is a corollary of the Cauchy’s Residue theorem. The principle follows with certain assumptions, let f(z) be a function defined on and within a simple closed contour such that

• f(z) is a meromorphic function within and on the closed contour except for the isolated poles inside the contour.
• f(z) has no zeros and poles on the boundary of the closed contour.
• f(z) has finitely many zeros and isolated poles inside the closed contour.

Statement of Argument Principle

Let f be a meromorphic function defined in a domain D bounded by a simple closed contour C. If f has poles p₁, p₂, …, p_m and zeros z₁, z₂, …, z_n counted according to their multiplicity such that any of the poles or zeros does not lie on C, then

That is, the closed contour integral of the logarithmic derivative of the given function is equal to the 2𝜋i times of the difference between total number of zeros and poles of that function, counted according to their multiplicity.

Proof of Argument Principle

Given f is a meromorphic function within and on closed contour C, f has finitely many poles p₁, p₂, …, p_m and zeros z₁, z₂, …, z_n counted according to their multiplicity such that any of the poles or zeros does not lie on C. Since each zeros and poles are isolated it if always possible to find

R > 0 such that within the circle |z – z_i| = R (i = 1, 2, …, n) or |z – p_j| = R (j = 1, 2, …, m) we shall not find any other zero or poles, respectively.

Now, by the corollary of Cauchy’s Theorem, we can have

Let z = z₁ be a zero of f of order M, then we write

f(z) = ( z – z₁)^M g(z)

where g(z) is analytic and g(z₁) ≠ 0.

Taking logarithm on both sides, we get

log f(z) = M log ( z – z₁) + log g(z)

Differentiating with respect to z, we get

f'(z)/f(z) = M/(z – z₁) + g'(z)/g(z)

Since g(z) is analytic and so, g'(z) will also be analytic. Hence, g'(z)/g(z) is an analytic function at z₁.

Then, by Cauchy’s Theorem, “The integral of an analytic function over a simple closed contour vanishes”.

This will true for all such contours 𝛾 within C.

Also,

Putting z – z₁ = Re^i𝜃

Hence, form (1)

where N is the total number of zeros, counted according to their multiplicity.

Now, let z = p₁ be a pole of f(z) of order K, then we write

f(z) = (z – p₁)^{– K} h(z)

where h(z) is analytic and h(p₁) ≠ 0.

Taking logarithm on both sides, we get

log f(z) = – K log ( z – p₁) + log h(z)

Differentiating with respect to z, we get

f'(z)/f(z) = – K/(z – p₁) + h'(z)/h(z)

And as argued before, we have

From (1), we have

where P is the number of poles, counted according to their multiplicity.

Thus from (1), (2) and (3), we get

This proves the theorem.

Why is the Argument Principle named so?

According to the Argument principle,

where N and P are the total number of zeros and poles.

Now,

where f(z) = w.

Thus, it represents the overall change in argument of f(z) as z moves around C.

Related Articles

• Liouvilles Theorem
• Morera’s Theorem
• Maximum Modulus Principle
• Harmonic Functions
• Rouche’s Theorem

Solved Examples on Argument Principle

Example 1:

Find the integral of [(3z² – 2z + 1)/{(z² + 1)(z – 1)}] over |z| = 2.

Solution:

Let f(z) = (z² + 1)(z – 1)

Then f’(z) = 3z² – 2z + 1

Also, f has only zeros within |z| = 2 which are 1, i, – i

Thus, we can apply the Argument Principle

(1/2 𝜋 i)∫_C [f’(z)/f(z)] dz = N – P

where N = 3 and P = 0

∫_{|z| = 2} [(3z² – 2z + 1)/{(z² + 1)(z – 1)}] = 3 × 2 𝜋 i = 6𝜋 i

Example 2:

Verify the Argument Principle for f(z) = [(z² + a²)/(z² + b²)], 0 < a < 1 and 0 < b < 1 over the unit circle.

Solution:

Given f(z) = [(z² + a²)/(z² + b²)], 0 < a < 1 and 0 < b < 1, has zeros at z = ± ai and poles at z = ±bi

Inside the unit circle |z| = 1. Thus N – P = 2 – 2 = 0

Now, ‘

= 1 + 1 – 1 – 1 = 0

Hence, the argument principle is verified.