Given a one-one analytic function $f$ with $f(a) = 0$. Prove that $f'(a) \neq 0$.

Question

Let $D$ be a domain in $\mathbb C$ (assume whatever you want about $D$ - what I have is simply connected). Let $f: D \to \mathbb C$ and $a \in D$ such that $f(a) = 0$. Suppose that $f$ is one-one. Prove that $f'(a) \neq 0$.

I need this to construct a certain Green's function. What we have is: for any $\epsilon$, there's a $\delta$ such that $|f(z) - (z-a)f'(a)| \le \epsilon |z - a|$ for every $z \in D(a,\delta)$. If we assume that $f'(a) = 0$, then we will have $|f(z)| \le \epsilon |z-a|$. I don't see where to go from here. Another possibility is to write out $f$ as a power series under the assumption $f'(a) = 0$:

$$f(z) = \frac{f^{(2)}(a)}{2!} (z-a)^2 + \cdots$$

also clueless here. Any idea?

Answer 1

The value of $f$ at $a$ doesn't matter since we can always consider $z\mapsto f(z) - f(a)$ without affecting injectivity or derivatives. Now our goal is to show that if $f$ has vanishing derivative, then it cannot be globally one-to-one. In fact, we will show something stronger. For convenience, I'll take $a = 0$. Since $f$ is not identically 0, there is some minimal $N \geq 2$ with the property that $f^{(N)}(0) \neq 0$. Locally around $0$, write \begin{align*} f(z) &= f(0) + f'(0)z + \sum_{n=2}^\infty a_nz^n \\ &= z^N g(z) \end{align*} where $g(z)$ is a holomorphic function satisfying $g(0) \neq 0$ (by the minimality assumption on $N$). Now, there exists a locally holomorphic $h$ with $h(0) \neq 0$ and $g(z) = (h(z))^N$, so that $$ f(z) = [zh(z)]^N. $$ This in turn implies that $f$ is locally $N$-to-1; since $N \geq 2$ we are done.

N.B. The expression for $f$ is called the normal form for nonconstant analytic functions and is unfortunately not stressed enough in basic literature. It more generally takes the form $$f(z) = f(z_0) + [(z-z_0) h(z)]^N, \qquad h(z_0) \neq 0$$ and is derived in an identical fashion to this proof.

Answer 2

If $f'(a) = 0$, I claim that $f$ is not one-to-one in a neighbourhood of $a$. We may assume $f$ is not constant in a neighbourhood of $a$.

Consider a positively oriented circle $\Gamma$ around $a$ such that $D$ contains $\Gamma$ and the disk inside $\Gamma$. For $w$ inside $\Gamma$,

$$ N(w) = \dfrac{1}{2\pi i} \oint_\Gamma \dfrac{f'(z)\; dz}{f(z)-w} $$

is the number of zeros of $f(z)-w$ inside $\Gamma$, counted by multiplicity. Since $f'(a) = 0$, $N(f(a)) > 1$. But $N(w)$ is continuous (and thus constant) in a neighbourhood of $f(a)$, so $N(w) > 1$ there. The zeros of $f'$ form a discrete set, so for all but finitely many $w$ in this neighbourhood there are no zeros of $f(z)-w$ of multiplicity $>1$, thus this count must consist of at least two distinct zeros, implying $f$ is not one-to-one.