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According to the maximum modulus principle, the highest possible value of the modulus of a function defined on a bounded domain can only occur on the domain boundary.
- It is also known as the maximum modulus theorem for complicated analytic functions.
- The function is stable if the function modulus has an absolute maximum value within the domain.
- It is an important concept used in complex analysis.
- Subsequently, the maximum modulus principle describes the characteristics of an analytic function that is local maximum across a domain.
- If the function is not stable, the maximum value can only be reached on the border.
- Phragmén–Lindelöf principle, which is an extension to unbounded domains, is an important application of the maximum modulus principle.
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Key Terms: Maximum Modulus Principle, Analytic Functions, Bounded Domain, Continuity, Complex Analysis, Absolute Maximum Value, Domain
Statement of Maximum Modulus Principle
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According to the maximum modulus principle, if f is a form of holomorphic function, then in that case, modulus function f will not return a local maximum value that is not within the domain of f.
- It can be assumed that G ⊂ C here can be considered the series of the complicated number, which can be bounded and connected with the open series.
- The greatest value of |f(z)| arises on G, not within G, if f is an analytic function determined on G, assuming f is a constant function.
- Stated differently, if M represents the highest value of |f(z)| across and within G, then |f(z)| < M for each point z inside G until f is stable.
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Maximum Modulus Principle Proof
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According to the theorem's hypothesis, f is a continuous function both inside and outside of G, which means that f is an analytic function both inside and outside of G then G is a bounded complicated domain.
⇒ At some point on or within G, the function f reaches its maximum value M; therefore assert that this point only exists on G and not within G.
= Assume that this maximum value M of f happens at any location z = a and inside G, if at all possible.
= Consequently, for any z in G, max|f(z)| = |f(a)| = M and |f(z)| ≤ M.
Let us consider a circle
\(\Gamma\) with center an inside G. f has to be present at a point z = b, an adjacent point of a within \(\Gamma\) , such that |f(b)| ≤ M since f is constant throughout G.= Considering that M - ε = |f(b)|, where ε > 0.
⇒ According to the concept of continuity, since f is continuous at b, there exists δ > 0 such that for any arbitrarily selected ε > 0.
⇒ Each instance |z – b| < δ, ||f(z)| – |f(b)|| < ε/2 ……...(1)
= Considering that |f(z)| - |f(b)| ≤ ||f(z)| - |f(b)|| < ε/2
Thus, |f(z)| - |f(b)| < ε/2
Alternatively, M – ε + ε/2 = M – ε/2 = |f(z)| < |f(b)| + ε/2
As a result, |f(z)| < M – ε/2 ……….. (2)
which means that γ: |z – b| < δ, for every point z within the circle γ having centre b and radius δ.
⇒ Construct a second circle, \(\Gamma\), starting at a and going through b. The circle γ contains the arc PQ of the circle \(\Gamma\)', which means
⇒ When z is on arc PQ, |f(z)| < M – ε/2.
= However, on PQ's primary arc, |f(z)| ≤ M
Let us assume that the circle \(\Gamma\): |b – a| = r. Cauchy's Integral Formula gives us
\(f(a)=\frac{1}{2\pi i} \int_{\Gamma}\frac{f(z)}{z-a}dz\)
⇒ We have \(\Gamma\) we have z – a = \(re^{\Gamma \theta}\), meaning that
\(f(a)=\frac{1}{2\pi i} \int^{2\pi}_{0}\frac{f(a+re^{i\theta})}{re^{i\theta}}re^{i\theta}d\theta\)
\(\frac{1}{2\pi i}\int^{2\pi}_0f(a+re^{i\theta})d\theta\)
⇒ If c is the angle that the arc PQ at center a contains, then
\(f(a)=\frac{1}{2\pi i} \int^{\alpha}_{0}f(a+re^{i\theta})d\theta+\frac{1}{2\pi i}\int^{2\pi}_{\alpha}f(a+re^{i\theta})d\theta\)
Thus,
\(|f(a)|\leq\frac{1}{2\pi i} \int^{\alpha}_{0}|f(a+re^{i\theta})|d\theta+\frac{1}{2\pi i}\int^{2\pi}_{\alpha}|f(a+re^{i\theta}|)d\theta\)
\(< \frac{1}{2\pi i}\int^{\alpha}_0|M-\frac{e}{2}|d\theta+\frac{1}{2\pi i} \int^{2 \pi}_{\alpha}|M|d\theta \)
\(<\frac{\alpha}{2\pi i}(M-\frac{e}{2})+\frac{M}{2\pi i}(2\pi-\alpha)\)
\(=M-\frac{\alpha \epsilon}{2 \pi}\)
Thus the theorem shows that |f(a)| = M < M – \(\frac{\alpha \epsilon}{2 \pi}\) is incompatible. As a result, it is incorrect that we assumed f would reach its maximum value inside G. Therefore, rather than inside G, f must reach its greatest value on G.
Maximum Modulus Principle Examples
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The example of maximum modulus principle is as follows:
Example: When f(z) = z2 – 3z + 2, find the highest value of |f(z)| in |z| ≤ 1? Ans: Given that f is a polynomial function on the complex plane C, it is continuous both on |z| ≤ 1 and within it. Thus, to obtain the maximum of |f(z)|, we can apply the maximum modulus principle to the provided function f. As per the principle of maximum modulus, the maximum value is reached when |z| < 1, meaning that -1 ≤ z ≤ 1. = However, since 1 and 2 are the zeros of the polynomial function f, the maximum cannot occur at z = 1. = Maximum thus occurs when z = -1. = |f(-1)| equals |(-1)|1 + 3 + 2| = |6| = 6; 2 – 3(-1) + 2| = Consequently, the greatest value of |f(z)| is 6. |
Things to Remember
- A non-constant analytic function on a connected open set G ⊂ C cannot have a local maximum point in G, as per the maximum modulus principle.
- The open map theorem, which says that a non constant holomorphic function maps open ranges to open collections, can be seen as a specific example of the maximum modulus principle.
- An adequately small open neighborhood of a cannot have an open image if |f| reaches a local maximum at a. F is constant as a result.
- Unless f is constant, |f(z)| < M for each point z within G if M is the greatest value of |f(z)| on and within G.
- The Borel–Carathéodory theorem, an application of the maximum modulus principle, is used to express an analytic function in terms of its real part.
- Another example of a theorem including Hadamard three-lines theorem shows the behaviour of holomorphic functions between two other parallel lines in the complex plane.
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Sample Question
Ques. What is the principle of maximal modulus? (2 marks)
Ans. The absolute value of a non-constant analytic function on a connected open set G ⊂ C cannot have a local maximum point in G, according to the maximum modulus principle for complex analytic functions.
Ques. When f(z) = z4 + z2 + 1, find the highest value of |f(z)| in |z| ≤ 1? (3 marks)
Ans. Given that f is a polynomial function on the complex plane C, it is continuous both on |z| ≤ 1 and within it. Thus, to obtain the maximum of |f(z)|, we can apply the maximum modulus principle to the provided function f.
= The maximum modulus principle states that maximum value happens on |z| ≤ 1, that is, -1 ≤ z ≤ 1.
= Consequently, |f(-1)| = |f(1)| = |(1)4 + (1)2 + 1| = 3 because f is an even function.
= Consequently, the highest value hof |f(z)| is 3.
Ques. Is the value of the maximum modulus principle unbounded? (2 marks)
Ans. Yes, bounded domain problems are covered by the maximum modulus concept. The principle known as the Phragmen-Lindelöf variant is unbounded. The problem appears not in the domain's boundedness, which is sometimes enough to guarantee function boundedness, but rather in the function's boundedness.
Ques. What does the maximum modulus principle require? (2 marks)
Ans. The function must be analytic within the specified domain in order to comply with the maximum modulus concept. An adequately small open neighborhood of a cannot have an open image if |f| reaches a local maximum at a. F is constant as a result.
Ques. In what way is the minimum modulus principle demonstrated by the maximal modulus principle theorem? (2 marks)
Ans. The minimal modulus principle states that an analytic function defined on a bounded domain has its absolute minimum value on the domain border. The maximum modulus principle for 1/f can be used to demonstrate the minimum modulus principle.
Ques. What uses does the maximum modulus principle have? (2 marks)
Ans. Numerous significant theorems in complex analysis, including Schwarz's Lemma, Hadamard's three-line theorem, the Borel-Caratheodory theorem, and the fundamental theorem of algebra, are proved using the maximum modulus principle.
Ques. How can the highest modulus of a complicated function be determined? (2 marks)
Ans. The highest Modulus Theorem in complex analysis states that the highest value of \f (z) | in R must occur on the boundary of R if a non-constant function/(z) is continuous on a closed bounded region R and analytic at every interior point of R.
Ques. What is the harmonic function maximum modulus principle? (2 marks)
Ans. According to Maximum Modulus Theorem for Harmonic Functions, in the event where D is a bounded domain, u is continuous on D, harmonic in D, and u ≤ M on ∂D, then u ≤ M on D. In other words, u reaches its maximum on the ∂D border.
Ques. For x = -5 and x = 10, find the value of the modulus function |x|? (2 marks)
Ans. |x| = |-5| = 5 if x = -5.
= |x| = |10| = 10 if x = 10.
Thus, |x| = 10 for x = 10 and |x| = 5 for x = -5.
Ques. Using the definition of the modulus function, solve |x + 3| = 8? (3 marks)
Ans. Given that the value of the modulus function is always non-negative, there are two possible scenarios:
When x + 3 is greater than 0 and x + 3 is less than 0, the formula is |x + 3| = x + 3.
First case: If x + 3 > 0, then
|x + 3| equals three times x.
⇒ 3x + = 8
x = 8 - 3 = 5
In case number two, x + 3 < 0.
⇒ |x + 3| equals -(x + 3)
⇒ -(x + 3) = 8
⇒ x - 3 = 8
⇒ x = -3 - 8 = -11
Hence, the solution is x = 5, -11.
Ques. Address the inequality where |x - 1| < 3? (2 marks)
Ans. In this instance, 3 > 0. Therefore, based on the modulus function's qualities,
⇒ 3 < x - 1 < 3
By adding one on each side,
⇒ 3 + 1 < x < 3 + 1
⇒ -2 < x < 4.
The above inequality has a solution of -2 < x < 4.
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