Proof that 1-1 analytic functions have nonzero derivative

Question

I recently saw a lecturer prove the following theorem (assuming the result that every analytic function is locally 1-1 whenever its derivative is nonzero): Let $\Omega \subset \mathbb{C}$ be open, and let $f : \Omega \to \mathbb{C}$ be 1-1 and analytic on $\Omega$. Then $f'(z_0) \not = 0$ for every $z_0 \in \Omega$.

I got the basic idea behind the proof: we assume for contradiction that $f'(z_0) = 0$, and, assuming without loss of generality that $z_0 = f(z_0) =0$, we have (from the power-series expansion) that $f(z) = z^kg(z)$ for some analytic $g$ in some disk at the origin (i.e., $z_0$) and some $k \ge 2$. Since $z^k$ is not 1-1 in any such disk (because there are multiple roots of unity), then $f$ isn't either.

However, the proof he gave was rather awkward and technical- it involved defining three different axillary functions, even though the idea was simple, and I've since forgotten how it exactly worked. In any case, I'm convinced there's a better way.

The problem is that I'm having trouble turning the idea into a real proof- I know that it obviously follows if $g$ is 1-1, but I'm also pretty sure that that is too strong an assumption. Am I missing something, or does the argument just have to be more complicated?

Answer 1

I like proving this theorem via its contrapositive rather than by contradiction (though the computations are essentially the same).

Suppose $f:\Omega\to\mathbb{C}$ is analytic with $f'(z_0)=0$. The goal is to show that every disc about the origin contains distinct $z_1,z_2$ with $f(z_1)=f(z_2)$. We may assume that $z_0=f(z_0)=f'(z_0)=0$ (using $f(z+z_0)-f(z_0)$ if necessary, as translation doesn't affect injectivity). Since $f$ is analytic at $z=0$ and $f'(0)=f(0)=0$, $f$ has a power series expansion $$f(z)=a_k z^k + a_{k+1} z^{k+1} + \dots$$ where $k>1$. Pulling out a $z^k$ gives $$f(z)=z^k (a_k + a_{k+1}z + \dots) = z^k g(z)$$ where $g$ is analytic with $g(0)\neq0$. Since $g$ is nonzero on a sufficiently small disc centered at the origin, we can define an appropriate branch its log so that its k^th root is well-defined. Call this function $h$, so that $h$ is analytic with $h(z)^k=g(z)$ near the origin. Hence $$f(z)=\left(zh(z)\right)^k.$$ Note that $\phi(z)=zh(z)$ is analytic (near $z=0$). Therefore, for any $\epsilon>0$ (sufficiently small), $\phi(D(0,\epsilon))$ is open (by the Open Mapping Theorem) and hence contains a disc $D(0,2\delta)$. In particular, there exist $z_1,z_2\in D(0,\epsilon)$ with $\phi(z_1)=\delta$ and $\phi(z_2)=\delta \exp\left(\frac{2\pi i}{k}\right)$. Therefore $$f(z_2)=\delta^k \exp\left(\frac{2\pi i}{k}\right)^k = \delta^k = f(z_1)$$ as desired.

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The proof does look a bit clunky, especially with all the auxilliary functions. However, it's actually fairly simple and the extra functions are really just to show why each step is valid. In fact, the gist of the proof is:

1. Show that $f$ is the k^th power of some analytic function $\phi$

2. Show that you can always find $z_1,z_2$ where $\phi(z_1)$ and $\phi(z_2)$ lie on the same circle and their arguments differ by $\frac{2\pi}{k}$, so that their k-th powers are equal.

Answer 2

Suppose $f$ is analytic at $z_0$ and non-constant, but $f'(z_0) = 0$. Then the order of the zero of $f(z) - f(z_0)$ at $z_0$ is some integer $k > 1$. Take some circle $\Gamma$ around $z_0$ so $f$ is analytic on and inside $\Gamma$ and there are no other zeros of $f(z) - f(z_0)$ or $f'(z)$ on or inside $\Gamma$. Now the sum of the orders of the zeros of $f(z) - p$ inside $\Gamma$ is $\dfrac{1}{2\pi i}\int_\Gamma \frac{f'(z)}{f(z) - p}\, dz.$, which is equal to $k$ for $p$ in a neighbourhood of $f(z_0)$. But since $f'(z)$ has no other zeros inside $\Gamma$, those zeros are simple, i.e. there are $k$ distinct solutions to $f(z) = p$ inside $\Gamma$.

Answer 3

A proof using Rouche's Theorem: if $f$ is analytic on $\Omega$ and $f'(z_0)=0$, then we can assume $f$ isn't a constant, so $z_0$ is an isolated zero of $f'$. Let $m\ge 2$ be the order of the zero at $z_0$ of $g(z)=f(z)-f(z_0)$. Let $B_1$ be an open ball about $z_0$ included in $\Omega$ in which $g$ and $f'$ have no other zeroes apart from $z_0$ and let $\delta>0$ be so that $|g(z)|>\delta$ for all $z$ on the boundary of $B_1$. If $h$ is the constant map $z\mapsto -\delta$ on $\Omega$, then by Rouche's theorem $g+h$ has the same number of zeroes as $g$ does inside $B_1$. As the zeroes of $g+h$ are simple ($(g+h)'(z)=f'(z)\ne 0$ for any $z\ne z_0$ inside $B_1$), it follows $f(z)=f(z_0)+\delta$ for $m$-many $z$ inside $B_1$.