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By way of contradiction, suppose B s / 2 ( f ( c ) is not contained in f ( U ) , , and let w B s / 2 ( f ( c ) ) be a complex number that is not in f ( U ) . We have that | w - f ( c ) | < s / 2 , which implies that | w - f ( z ) | > s / 2 for all z C r . Consider the function g defined on the closed disk B ¯ r ( c ) by g ( z ) = 1 / ( w - f ( z ) ) . Then g is continuous on the closed disk B ¯ r ( c ) and differentiable on B r ( c ) . Moreover, g is not a constant function, for if it were, f would also be a constant function on B r ( c ) and therefore, by the Identity Theorem, constant on all of U , whichg is not the case by hypothesis. Hence, by the Maximum Modulus Principle,the maximum value of | g | only occurs on the boundary C r of this disk. That is, there exists a point z ' C r such that | g ( z ) | < | g ( z ' ) | for all z B r ( c ) . But then

2 s = 1 s / 2 < 1 | w - f ( c ) | < 1 | w - f ( z ' ) | 1 s ,

which gives the desired contradiction. Therefore, the entire disk B s / 2 ( f ( c ) ) belongs to f ( U ) , and hence the point f ( c ) belongs to the interior of the set f ( U ) . Since this holds for any point c U , it follows that f ( U ) is open, as desired.

Now we can give the version of the Inverse Function Theorem for complex variables.

Let S be a piecewise smooth geometric set, and suppose f : S C is continuously differentiable at a point c = a + b i , and assume that f ' ( c ) 0 . Then:

  1. There exists an r > 0 , such that B ¯ r ( c ) S , for which f is one-to-one on B ¯ r ( c ) .
  2.   f ( c ) belongs to the interior of f ( S ) .
  3. If g denotes the restriction of the function f to B r ( c ) , then g is one-to-one, g - 1 is differentiable at the point f ( c ) , and g - 1 ' ( f ( c ) = 1 / f ' ( c ) .

Arguing by contradiction, suppose that f is not one-to-one on any disk B ¯ r ( c ) . Then, for each natural number n , there must exist two points z n = x n + i y n and z n ' = x n ' + i y n ' such that | z n - c | < 1 / n , | z n ' - c | < 1 / n , and f ( z n ) = f ( z n ' ) . If we write f = u + i v , then we would have that u ( x n , y n ) - u ( x n ' , y n ' ) = 0 for all n . So, by part (c) of [link] , there must exist for each n a point ( x ^ n , y ^ n ) , such that ( x ^ n , y ^ n ) is on the line segment joining z n and z n ' , and for which

0 = u ( x n , y n ) - u ( x n ' , y n ' ) = t i a l u t i a l x ( x ^ n , y ^ n ) ( x n - x n ' ) + t i a l u t i a l y ( x ^ n , y ^ n ) ( y n - y n ' ) .

Similarly, applying the same kind of reasoning to v , there must exist points ( x ˜ n , y ˜ n ) on the segment joining z n to z n ' such that

0 = t i a l v t i a l x ( x ˜ n , y ˜ n ) ( x n - x n ' ) + t i a l v t i a l y ( x ˜ n , y ˜ n ) ( y n - y n ' ) .

If we define vectors U n and V n by

U n = ( t i a l u t i a l x ( x ^ n , y ^ n ) , t i a l u t i a l y ( x ^ n , y ^ n ) )

and

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

then we have that both U n and V n are perpendicular to the nonzero vector ( ( x n - x n ' ) , ( y n - y n ' ) ) . Therefore, U n and V n are linearly dependent, whence

det ( ( tial u tial x ( ( x ^ n , y ^ n ) tial u tial y ( x ^ n , y ^ n ) tial v tial x ( x ˜ n , y ˜ n ) tial v tial y ( x ˜ n , y ˜ n ) ) ) = 0 .

Now, since both { x ^ n + i y ^ n } and { x ˜ n + i y ˜ n } converge to the point c = a + i b , and the partial derivatives of u and v are continuous at c , we deduce that

det ( ( tial u tial x ( a , b ) tial u tial y ( a , b ) tial v tial x ( a , b ) tial v tial y ( a , b ) ) ) = 0 .

Now, from [link] , this would imply that f ' ( c ) = 0 , and this is a contradiction. Hence, there must exist an r > 0 for which f is one-to-one on B ¯ r ( c ) , and this proves part (1).

Because f is one-to-one on B r ( c ) , f is obviously not a constant function. So, by the Open Mapping Theorem, the point f ( c ) belongs to the interior of the range of f , and this proves part (2).

Now write g for the restriction of f to the disk B r ( c ) . Then g is one-to-one. According to part (2) of [link] , we can prove that g - 1 is differentiable at f ( c ) by showing that

lim z f ( c ) g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) = 1 f ' ( c ) .

That is, we need to show that, given an ϵ > 0 , there exists a δ > 0 such that if 0 < | z - f ( c ) | < δ then

| g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) - 1 f ' ( c ) | < ϵ .

First of all, because the function 1 / w is continuous at the point f ' ( c ) , there exists an ϵ ' > 0 such that if | w - f ' ( c ) | < ϵ ' , then

| 1 w - 1 f ' ( c ) | < ϵ .

Next, because f is differentiable at c , there exists a δ ' > 0 such that if 0 < | y - c | < δ ' then

| f ( y ) - f ( c ) y - c - f ' ( c ) | < ϵ ' .

Now, by [link] , g - 1 is continuous at the point f ( c ) , and therefore there exists a δ > 0 such that if | z - f ( c ) | < δ then

| g - 1 ( z ) - g - 1 ( f ( c ) | < δ ' .

So, if | z - f ( c ) | < δ , then

| g - 1 ( z ) - c | = | g - 1 ( z ) - g - 1 ( f ( c ) ) | < δ ' .

But then,

| f ( g - 1 ( z ) ) - f ( c ) g - 1 ( z ) - c - f ' ( c ) | < ϵ ' ,

from which it follows that

| g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) - 1 f ' ( c ) | < ϵ ,

as desired.

Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
ya I also want to know the raman spectra
Bhagvanji
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
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
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
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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
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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