## Derivative of holomorphic function and its injective property or one-to-one propety

Assume that is an open subset of the complex plane and is a holomorphic function. Assume, in addition, that is one-to-one, that is whenever . Prove that the derivative is different from zero for every .

Suppose , such that and is one-to-one, then can not be a constant and hence zeros of is isolated.

WLG assume , and . Also assume on . Let , define . Then for . By Rouche’s theorem, and has the same number of zeros in . Since has at least two zeros in , has at least two preimages. But in implies these preimages are different from each other. This contradicts the injective property of .

For the converse of this problem

1. If , then we can only conclude is locally injective. Consider on , then on . But is not globally injective.

2. Even if the domain is simply connected, is still not necessarily globally injective. The following simple proof is wrong: *suppose and , then there exists a point on the line segment of and (because is simply connected) such that*

*So . Contradiction.*

This is because the mean value theorem only holds for real-value function, not for complex-value function.

3. Consider , is never zero, but for infinitely many .

### Like this:

Like Loading...

*Related*

or leave a trackback:

Trackback URL.

## Comments

you typed one-to-to instead of one-to-one in line 2 of your problem

thank you for your note.

Why does f have at least two zeros in D?

because f(0)=f'(0)=0