Tag Archives: line segment

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

\mathbf{Problem:} Assume that U is an open subset of the complex plane \mathbb{C} and f:U\to \mathbb{C} is a holomorphic function. Assume, in addition, that f is one-to-one, that is f(z)\neq f(z^\prime) whenever z\neq z^\prime. Prove that the derivative f^\prime(z) is different from zero for every z\in U.

\mathbf{Proof:} Suppose \exists\, z_0=0, such that f^\prime(z_0)=0 and f(z) is one-to-one, then f can not be a constant and hence zeros of f^\prime(z) is isolated.
WLG assume z_0=0, f(0)=f^\prime(0)=0 and \mathbb{D}=\{z\in \mathbb{C}||z|<1\}\subset U. Also assume f^\prime(z)\neq 0 on \mathbb{D}\backslash\{0\}. Let \epsilon=\inf\limits_{\partial \mathbb{D}}|f(z)|>0, define g(z)=f(z)-\frac \epsilon 2. Then |g(z)-f(z)|=\frac \epsilon 2<f(z) for \forall x\in \partial \mathbb{D}. By Rouche’s theorem, f(z) and g(z) has the same number of zeros in \mathbb{D}. Since f has at least two zeros in \mathbb{D}, f(z)=\frac \epsilon 2 has at least two preimages. But f^\prime(z)\neq 0 in \mathbb{D}\backslash\{0\} implies these preimages are different from each other. This contradicts the injective property of f.

\mathbf{Remark:}

For the converse of this problem
1. If f^\prime(z)\neq 0, \forall z\in U, then we can only conclude f'(z) is locally injective. Consider f(z)=z^2 on U=\mathbb{C}\backslash\{0\}, then f^\prime(z)\neq 0 on U. But f is not globally injective.
2. Even if the domain is simply connected, f is still not necessarily globally injective. The following simple proof is wrong: suppose z\neq z^\prime and f(z)=f(z^\prime), then there exists a point \xi\in U on the line segment of z and z^\prime(because U is simply connected) such that

0=f(z)-f(z^\prime)=f^\prime(\xi)(z-z^\prime).

So f^\prime(\xi)=0. Contradiction.
This is because the mean value theorem only holds for real-value function, not for complex-value function.
3. Consider f(z)=e^z, f'(z) is never zero, but \displaystyle e^{2k\pi i}=1 for infinitely many k.

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