Proving that an Analytic Function with vanishing Derivative at a Point is Not One-to-One

Understanding the behavior of analytic functions, particularly when their derivative vanishes at a point, is crucial in complex analysis. In this article, we will delve into proving that if an analytic function has a zero derivative at a specific point, it cannot be one-to-one in any small neighborhood of that point. This proof involves several key concepts, including the open mapping theorem and the power series representation of analytic functions.

Introduction to the Problem

Consider an analytic function ( f: D to mathbb{C} ) where ( D subset mathbb{C} ) is a domain and ( f'(z_0) 0 ) for some ( z_0 in D ). Our goal is to show that ( f ) is not one-to-one near ( z_0 ), meaning that within any small disc around ( z_0 ), there exist distinct points ( z_1 ) and ( z_2 ) such that ( f(z_1) f(z_2) ).

Transformation and Assumptions

To simplify our proof, we can assume without loss of generality that ( z_0 0 ) and ( f(0) f'(0) 0 ). This can be achieved through translation and scaling. Let ( g(z) f(z z_0) - f(z_0) ). If ( g(0) 0 ) and ( g'(0) f'(z_0) ), then we can further translate using ( f(z) g(z - z_0) f(z_0) ). Hence, we can assume ( f(0) f'(0) 0 ) and work with ( f ) directly.

Power Series Representation

Since ( f ) is analytic at ( z 0 ), it can be represented by its power series around ( z 0 ):

[ f(z) a_k z^k a_{k 1} z^{k 1} cdots ]

where ( k geq 1 ) and ( a_k eq 0 ).

We can factor this as:

[ f(z) z^k left( a_k a_{k 1} z a_{k 2} z^2 cdots right) ]

For simplicity, denote ( g(z) a_k a_{k 1} z a_{k 2} z^2 cdots ) and ( g(0) eq 0 ).

Defining and Analyzing ( g )

Since ( g(0) eq 0 ), on a sufficiently small disc around the origin, ( g ) is non-zero. We can define an appropriate branch of the ( k )-th root of ( g ), denoted by ( h ), such that ( g(z) h(z)^k ) near the origin. Hence,

[ f(z) z^k h(z) ]

Now, ( h(z) ) is an analytic function near the origin with ( h(0) eq 0 ).

Constructing a Disk and Applying the Open Mapping Theorem

Let ( D_0(epsilon) { z in mathbb{C} : |z| 0 ).

Applying the Argument Principle

Consider the function ( r(z) z h(z)^k ). Since ( h(z) ) is non-zero in ( D_0(epsilon) ) and analytic, ( r(z) ) is also analytic near ( 0 ). Let ( delta e^{2pi i / k} ) be a point in the image of ( r ) on the boundary of ( D_0(epsilon) ). By the argument principle, for any ( epsilon > 0 ) sufficiently small, there exist ( z_1, z_2 in D_0(epsilon) ) such that

[ r(z_1) delta quad text{and} quad r(z_2) delta e^{2pi i / k} ]

Thus,

[ f(z_1) (z_1 h(z_1))^k (delta)^k delta e^{2pi i / k} (z_2 h(z_2))^k f(z_2) ]

This shows that ( f ) is not one-to-one in ( D_0(epsilon) ).

Conclusion

In conclusion, we have shown that if an analytic function ( f ) has a zero derivative at a point ( z_0 ), it cannot be one-to-one in any sufficiently small neighborhood of ( z_0 ). The proof relies on the properties of analytic functions, the open mapping theorem, and the power series representation of analytic functions.