Maximum Modulus Principle

The maximum modulus principle or maximum modulus theorem for complex analytic functions states that the maximum value of modulus of a function defined on a bounded domain may occur only on the boundary of the domain. If the modulus of the function has a maximum value inside the domain, then the function is constant.

Thus, the maximum modulus principle states the nature of the local maximum of an analytic function within a domain. The maximum value could only be attained on the boundary unless the function is constant.

Statement of Maximum Modulus Principle

Let G ⊂ C ( C is the set of complex numbers) be a bounded and connected open set. If f is an analytic function defined on G, then the maximum value of |f(z)| occurs on G, not inside G, unless f is a constant function.

In other words, if M is the maximum value of |f(z)| on and within G, then unless f is constant, |f(z)| < M for every point z within G.

Proof of Maximum Modulus Principle

By the hypothesis of the theorem, f is an analytic function within and on G, that is, f is differentiable within and on G, which implies f is continuous within and on G, G being a bounded complex domain.

The function f attains maximum value M at a point within or on G; we claim that this point only lies on G and not inside G.

Suppose, if possible, this maximum value M of f occurs at any point z = a inside G.

Then, max|f(z)| = |f(a)| = M and |f(z)| ≤ M for all z within G.

Let us take a circle 𝛤 within G centered at a. Since f is continuous within G, f must exist at a point z = b, a neighbourhood point of a within 𝛤 such that |f(b)| ≤ M.

Let |f(b)| = M – ε where ε > 0.

Since f is continuous at b, by the definition of continuity, for any arbitrary chosen ε > 0, there exist δ > 0 such that

||f(z)| – |f(b)|| < ε/2 whenever |z – b| < δ …………..(1)

Since |f(z)| – |f(b)| ≤ ||f(z)| – |f(b)|| < ε/2

⇒ |f(z)| – |f(b)| < ε/2

or, |f(z)| < |f(b)| + ε/2 = M – ε + ε/2 = M – ε/2

Thus, |f(z)| < M – ε/2 …………(2)

for all points z inside the circle γ with centre b and radius δ, that is, γ : | z – b| < δ.

Now draw another circle 𝛤’ with centre at a and passing through b. The arc PQ of the circle 𝛤’ lies within the circle γ so that

|f(z)| < M – ε/2 for z being on arc PQ

But on the major arc PQ |f(z)| ≤ M

Suppose the circle 𝛤’: |b – a| = r. By Cauchy’s Integral Formula, we have

On 𝛤’ we have z – a = re^i𝜃, so that

Let 𝜶 be the angle subtended by the arc PQ at center a, then

Therefore,

Thus, |f(a)| = M < M – 𝛼ε/2𝜋, which is a contradiction. Hence our assumption of f attaining maximum value inside G is wrong.

Hence, f must attain maximum value on G and not inside G.

Related Articles

• Complex Numbers
• Analytic Functions
• Limits and Continuity
• Differentiability
• Integration
• Complex Conjugate

Solved Examples on Maximum Modulus Principle

Example 1:

Find the maximum value of |f(z)| in |z| ≤ 1 when f(z) = z² – 3z + 2.

Solution:

Since f is a polynomial function throughout the complex plane C, hence continuous within and on |z| ≤ 1. Therefore, we can apply maximum modulus principle on the given function f to find maximum of |f(z)|.

According to the maximum modulus principle, the maximum value occurs on |z| ≤ 1, that -1 ≤ z ≤ 1.

But zeros of the polynomial function f are 1 and 2, so that maximum cannot occur at z = 1.

Hence, maximum occur at z = -1

|f(-1)| = |(-1)² – 3(-1) + 2| = |1 + 3 + 2| = |6| = 6

Thus, 6 is the maximum value of |f(z)|.

Example 2:

Find the maximum value of |f(z)| in |z| ≤ 1 when f(z) = z⁴ + z² + 1.

Solution:

Since f is a polynomial function throughout the complex plane C, hence continuous within and on |z| ≤ 1. Therefore, we can apply the maximum modulus principle on the given function f to find the maximum of |f(z)|.

According to the maximum modulus principle, maximum value occurs on |z| ≤ 1, that -1 ≤ z ≤ 1.

Since f is an even function, therefore, |f(-1)| = |f(1)| = |(1)⁴ + (1)² + 1| = 3

Thus, 3 is the maximum value of |f(z)|.