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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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    $\begingroup$ 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/… $\endgroup$ Apr 27, 2011 at 1:03
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    $\begingroup$ Google isn't letting me see the page containing the proof. $\endgroup$ Apr 27, 2011 at 1:23
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    $\begingroup$ (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 $z_0 = f(z_0) = 0$. $\endgroup$ Apr 27, 2011 at 1:31
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    $\begingroup$ @gary: No, the inverse function theorem implies the converse, that if the derivative is nonzero then the function is locally injective. $\endgroup$ Apr 27, 2011 at 1:42
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    $\begingroup$ I'd argue that although the idea is simple, any proof will have to be at least a little "awkward" or "technical" for the reason that the statement is analogous to other statements that are false. For example $f: \mathbb{R} \to \mathbb{R}$ given by $x \mapsto x^3$ is one-to-one and in many senses it's as nice a function from $\mathbb{R}$ to $\mathbb{R}$ as you might want (e.g. it is real analytic) but $f'(0) = 0$. Of course the $'$ in $f'$ means something different here, but it's certainly analogous. So any proof of your statement will really have to do something. Just my two cents. $\endgroup$
    – anon
    Apr 27, 2011 at 4:51

3 Answers 3

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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 kth 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 kth 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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  • $\begingroup$ 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. $\endgroup$ Apr 28, 2011 at 1:04
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    $\begingroup$ I don't understand the last step. How do we know there exists $z_1,z_2$ such that $\phi(z_2)=\delta exp(\frac{2i\pi}{k})$? $\endgroup$ Feb 9, 2015 at 7:51
  • $\begingroup$ What ensures $z_{1}\neq z_{2}$? $\endgroup$ Apr 5, 2016 at 22:32
  • $\begingroup$ Very nice and neat solution. May I ask you a question; does the converse of this result true? $\endgroup$
    – Bumblebee
    Aug 4, 2017 at 3:23
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    $\begingroup$ @DiegoFonseca well they map to different elements, hence they must be different $\endgroup$
    – Luigi M
    Jan 5, 2018 at 18:17
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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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    $\begingroup$ Thanks, I think that this is the argument principle. $\endgroup$ Jan 18, 2014 at 14:30
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    $\begingroup$ In fact, Rouche's theorem is proven along similar lines. $\endgroup$ Dec 7, 2018 at 17:19
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    $\begingroup$ How did you get that the number of zeros is equal to k for p near $f(z_0)$ ? $\endgroup$ Jan 28, 2019 at 6:36
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    $\begingroup$ It's a continuous function of $p$ near $f(z_0)$, but it's $k$ at $f(z_0)$ and its values are integers. $\endgroup$ Jan 28, 2019 at 13:33
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    $\begingroup$ @RobertIsrael What is your $p$? And what is $r$? $\endgroup$
    – Bach
    Jun 12, 2019 at 11:55
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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 $B_1$.

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  • $\begingroup$ Excuse me, you writed “let δ>0 be so that |g(z)|>δ for all z on the boundary of B” I want to ask how we know there is a δ like that? Why is it bounded? $\endgroup$
    – Remas
    Jan 13 at 10:44

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