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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.

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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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