$f$ an injective holomorphic function on a domain then $f'\ne 0$ on the domain.

Question

Problem: Let $f$ be a holomorphic function on a domain $D$ ( let's say a connected open subset of $\mathbb C$) and it is one-one. Show that $f'(z)\ne0$ for all $z\in D$

My approach: Let us suppose $f'(z_0)=0$ for some $z_0\in D$ and also let $f(z_0)=w_0$. Since $f$ is injective $f'$ is not identically zero .

Therefore $\exists r>0$ such that $f(z)\ne w_0$ and $f'(z)\ne 0$ for all $z\in B(z_0,r):=\{z\in D:0<|z-z_0|<r\}$.

We can have a sequence $\{w_n\}_{n=1}^\infty$ such that $w_n\to w_0$ and $w_n\ne w_0$ for all $n$

Then let us define $g(z)=f(z)-w_0$ on $B(0,\frac{r}{2})$. And sequence of holomorphic function $\{g_n\}$ on $B(0,\frac{r}{2})$ such that $g_n(z)=f(z)-w_n$.

Observe that $g$ has a zero of order $m\ge2$ at $z_0$.

So $g_n\to g$ uniformly on $B(0,\frac{r}{2})$. Here we are in a position to apply Hurwitz Theorem.

Therefore $\exists n_0$ such that $g_n$ and $g$ have the same number of zeroes on $B(0,\frac{r}{2})$ counted with multiplicity (i.e. $m\ge 2$ here) for all $n\ge{n_0}$.

So $g_{n_0}$ has $m$ zeroes counted with multiplicity. But observe that there are atleast two distinct zeroes of $g_{n_0}$ in $B(0,\frac{r}{2})$ (the zeroes may have different multiplicities).

Otherwise, if we have $g_{n_0}(z_1)=0$ with multiplicity $m\ge 2$ then $g_{n_0}'(z_1)=f'(z_1)=0$, which is not possible as $z_1\in B(0,\frac{r}{2})$.

Let the distinct roots be $z_1$ and $z_2$ then $g_{n_0}(z_1)=g_{n_0}(z_2)=0\implies f(z_1)=f(z_2)=w_{n_0}$ .

This contradicts the fact that $f$ is one-one.

Let me know if it is right. If wrong, provide me with a better approach. If any simple solution exists please let me know.

Thank You.

Answer 1

Suppose $f'(z_0)=0$ for some $z_0\in D$. The holomorphic function $h:D\to \mathbb{C},h(z)=f(z)-f(z_0),$ has a zero in $z_0$ of order 2 or higher. For sufficiently small $\epsilon>0$, there would exist an $\delta>0$ such that for all $a$ with $\mid a-f(z_0)\mid<\delta$ the equation $f(z)=a$ has 2 or more solutions in the disk $\mid z-z_0\mid <\epsilon$, which would contradict the fact that $f$ is one-to-one.

For a proof, see Theorem 11, p. 131, Complex Analysis by Ahlfors.

Answer 2

I think your proof is correct. Here is another way that uses that a nonvanishing holomorphic function on a disc has a logarithm. Let $P$ be a point where $f'(P) = 0$, and without loss of generality, $P = 0$ and $f(P) = 0$ (compose and precompose $f$ with biholomorphisms). Then near $0$, for some $n\ge 2$, we can write with $a_n \ne 0$ \begin{align*} \zeta = f(\tilde z) &= \sum_{k\ge n}a_k\tilde z^k\\ &= \tilde z^n\big(\sum_{k\ge 0}a_{k+n}\tilde z^k\big)\\ &\equiv \tilde z^n F(\tilde z). \end{align*} The function $F(\tilde z)$ is nonvanishing on a disc containing $0$, so it has a logarithm (in particular it has an $n$th root), so put $h(\tilde z) = F(\tilde z)^{1/n}$ and write $$ \zeta = \tilde z^nh(\tilde z)^n = (\tilde z\,h(\tilde z))^n. $$ The map $\tilde z\, h(\tilde z)$ is a local biholomorphism, so put $z = \tilde z\,h(\tilde z)$. This shows that $f$ is locally equivalent to $\zeta = z^n$, with $n\ge 2$, so in particular, $f$ is at least $n$-to-$1$ near $0$, and is not injective.