## Tag Archives: rouche

### 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 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 $k$.