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Our next goal is to examine so-called “max/min” problems for coplex-valued functions of complex variables.Since order makes no sense for complex numbers, we will investigate max/min problems for the absolute value of a complex-valued function.For the corresponding question for real-valued functions of real variables, we have as our basic resultthe First Derivative Test. Indeed, when searching for the poinhts where a differentiable real-valued function f on an interval [ a , b ] attains its extreme values, we consider first the poinhts where it attains a local max or min.Of course, to find the absolute minimum and maximum, we must also check the values of the function at the endpoints.

Our next goal is to examine so-called “max/min” problems for coplex-valued functions of complex variables.Since order makes no sense for complex numbers, we will investigate max/min problems for the absolute value of a complex-valued function.For the corresponding question for real-valued functions of real variables, we have as our basic resultthe First Derivative Test ( [link] ). Indeed, when searching for the poinhts where a differentiable real-valued function f on an interval [ a , b ] attains its extreme values, we consider first the poinhts where it attains a local max or min, to which purpose end [link] is useful. Of course, to find the absolute minimum and maximum, we must also check the values of the function at the endpoints.

An analog of [link] holds in the complex case, but in fact a much different result is really valid. Indeed, it is nearly impossible for the absolute value ofa differentiable function of a complex variable to attain a local maximum or minimum.

Let f be a continuous function on a piecewise smooth geometric set S , and assume that f is differentiable on the interior S 0 of S . Suppose c is a point in S 0 at which the real-valued function | f | attains a local maximum. That is, there exists an ϵ > 0 such that | f ( c ) | | f ( z ) | for all z satisfying | z - c | < ϵ . Then f is a constant function on S ; i.e., f ( z ) = f ( c ) for all z S . In other words, the only differentiable functions of a complex variable, whose absolute value attains a local maximum on the interior of a geometric set, are constant functions on that set.

If f ( c ) = 0 , then f ( z ) = 0 for all z B ϵ ( c ) . Hence, by the Identity Theorem ( [link] ), f ( z ) would equal 0 for all z S . so, we may as well assume that f ( c ) 0 . Let r be any positive number for which the closed disk B ¯ r ( c ) is contained in B ϵ ( c ) . We claim first that there exists a point z on the boundary C r of the disk B ¯ r ( c ) for which | f ( z ) | = | f ( c ) | . Of course, | f ( z | | f ( c ) | for all z on this boundary by assumption. By way of contradiction, suppose that | f ( ζ ) | < | f ( c ) | for all ζ on the boundary C r of the disk. Write M for the maximum value of the function | f | on the compact set C r . Then, by our assumption, M < | f ( c ) | . Now, we use the Cauchy Integral Formula:

| f ( c ) | = | 1 2 π i C r f ( ζ ) ζ - c d ζ | = 1 2 π | 0 2 π f ( c + r e i t ) r e i t i r e i t d t | 1 2 π 0 2 π | f ( c + r e i t ) | d t 1 2 π 0 2 π M d t = M < | f ( c ) | ,

and this is a contradiction.

Now for each natural number n for which 1 / n < ϵ , let z n be a point for which | z n - c | = 1 / n and | f ( z n ) | = | f ( c ) | . We claim that the derivative f ' ( z n ) of f at z n = 0 for all n . What we know is that the real-valued function F ( x , y ) = | f ( x + i y ) | 2 = ( u ( x , y ) 2 + ( v ( x , y ) ) 2 attains a local maximum value at z n = ( x n , y n ) . Hence, by [link] , both partial derivatives of F must be 0 at ( x n , y n ) . That is

2 u ( x n , y n ) t i a l u t i a l x ( x n , y n ) + 2 v ( x n , y n ) t i a l v t i a l x ( x n , y n ) = 0

and

2 u ( x n , y n ) t i a l u t i a l y ( x n , y n ) + 2 v ( x n , y n ) t i a l v t i a l y ( x n , y n ) = 0 .

Hence the two vectors

V 1 = ( t i a l u t i a l x ( x n , y n ) , t i a l v t i a l x ( x n , y n ) )

and

V 2 = ( t i a l u t i a l y ( x n , y n ) , t i a l v t i a l y ( x n , y n ) )

are both perpendicular to the vector V 3 = ( u ( x n , y n ) , v ( x n , y n ) ) . But V 3 0 , because V 3 = | f ( z n ) | = | f ( c ) | > 0 , and hence V 1 and V 2 are linearly dependent. But this implies that f ' ( z n ) = 0 , according to [link] .

Since c = lim z n , and f ' is analytic on S 0 , it follows from the Identity Theorem that there exists an r > 0 such that f ' ( z ) = 0 for all z B r ( c ) . But this implies that f is a constant f ( z ) = f ( c ) for all z B r ( c ) . And thenm, again using the Identity Theorem, this implies that f ( z ) = f ( c ) for all z S , which completes the proof.

REMARK Of course, the preceding proof contains in it the verification that if | f | attains a maximum at a point c where it is differentiable, then f ' ( c ) = 0 . This is the analog for functions of a complex variable of [link] . But, [link] certainly asserts a lot more than that. In fact, it says that it is impossible for the absolute value of a nonconstant differentiable function of a complexvariable to attain a local maximum. Here is the coup d'grâs:

Maximum modulus principle

Let f be a continuous, nonconstant, complex-valued function on a piecewise smooth geometric set S , and suppose that f is differentiable on the interior S 0 of S . Let M be the maximum value of the continuous, real-valued function | f | on S , and let z be a point in S for which | f ( z ) | = M . Then, z does not belong to the interior S 0 of S ; it belongs to the boundary of S . In other words, | f | attains its maximum value only on the boundary of S .

  1. Prove the preceding corollary.
  2. Let f be an analytic function on an open set U , and let c U be a point at which | f | achieves a local minimum; i.e., there exists an ϵ > 0 such that | f ( c ) | | f ( z ) | for all z B ϵ ( c ) . Show that, if f ( c ) 0 , then f is constant on B ϵ ( c ) . Show by example that, if f ( c ) = 0 , then f need not be a constant on B ϵ ( c ) .
  3. Prove the “Minimum Modulus Principle:” Let f be a nonzero, continuous, nonconstant, function on a piecewise smooth geometric set S , and let m be the minimum value of the function | f | on S . If z is a point of S at which this minimum value is atgtained, then z belongs to the boundary C S of S .

Questions & Answers

what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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