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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?

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It may be hard to answer whether there's a less complicated argument if we don't know what the argument was. One proof is in Theorem 7.4 of J.B. Conway's complex analysis text: books.google.com/… –  Jonas Meyer Apr 27 '11 at 1:03
    
Google isn't letting me see the page containing the proof. –  Calvin McPhail-Snyder Apr 27 '11 at 1:23
    
(Apparently, enter posts comments.) I found some notes that I took- the proof involved dividing both sides by $g(0)$ (which seems unnecessary), then defining a new function $\psi$ by $\psi(z) = g(z)/g(0)$, so that $f(z)/g(0) = z^k\psi(z)$. $\psi$ takes values in a disc away from $0$, so its log is well-defined. We can then put $\phi(z) = z \exp(\log(\psi(z)) /k)$, so that $\phi(z^k) = z^k \psi(z)$, and then it's easy to show that $f$ isn't 1-1, since $\phi$ is, as $\phi'(0) = 1$. That actually isn't as bad as I remembered- I think it's because he didn't assume $z_0 = f(z_0) = 0$. –  Calvin McPhail-Snyder Apr 27 '11 at 1:31
    
Isn't this just the inverse function theorem? –  gary Apr 27 '11 at 1:34
3  
@gary: No, the inverse function theorem implies the converse, that if the derivative is nonzero then the function is locally injective. –  Jonas Meyer Apr 27 '11 at 1:42

2 Answers 2

up vote 12 down vote accepted

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.


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.

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Thanks. I think the essential part is saying things like "$g$ 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", instead of proving it in detail. –  Calvin McPhail-Snyder Apr 28 '11 at 1:04

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 $\Gamma$.

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Robert Israel, which theorem did you use for counting the sum of the orders of zeros of $f(z)-p$? –  Yoav Bar Sinai Jan 14 at 8:45
    
I'm not aware that the theorem has an official name. It's just from computing the residue of $f'(z)/(f(z)-p)$ at $z = r$ where $f(z) - p = g(z) (z - r)^m$ and $g(r) \ne 0$. –  Robert Israel Jan 14 at 15:59
    
Thanks, I think that this is the argument principle. –  Yoav Bar Sinai Jan 18 at 14:30

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