Is it possible for a holomorphic function on a connected domain to be conformal but not injective? Also, is it possible for a holomorphic function to be injective but not conformal?
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Example. The exponential function is conformal but not injective: its derivative is nowhere vanishing (so conformal) but $\exp 0 = \exp 2 \pi i$ (so not injective).
Proposition. The derivative of an injective holomorphic function $f : D \to \mathbb{C}$, where $D$ is an open subset in $\mathbb{C}$, is nowhere vanishing. In particular, $f$ is conformal.
Proof. We prove the a stronger statement, which will imply the contrapositive of the above proposition. Let $z_0 \in D$. By a change of coordinates if necessary, we may assume that $f(z_0) = 0$ and $z_0 = 0$ without loss of generality. Then, there is a positive integer $n$ and a holomorphic function $h : D \to \mathbb{C}$ such that $f(z) = z^n h(z)$ for all $z$ in $D$ and $h(0) \ne 0$. So in particular there is an open neighbourhood $U$ such that $0 \in U \subseteq D$ and $h$ is nowhere vanishing on $U$. That implies that there is a holomorphic function $g : U \to \mathbb{C}$ such that $g(z)^n = h(z)$ for all $z$ in $U$. (Take a holomorphic branch of the map $w \mapsto w^{1/n}$.) So $f(z) = (z \, g(z))^n$ for all $z$ in $U$.
Note that $z \mapsto z \, g(z)$ has non-zero derivative at $0$ (since $g(0) \ne 0$), so by the inverse function theorem, if $U$ is small enough, $z \mapsto z \, g(z)$ is an invertible holomorphic function on $U$. This implies $f$ is an $n$-to-$1$ function on $U \setminus \{ 0 \}$. But if $f'(0) = 0$, by considering the power series of $f$, we must have $n \ge 2$.
We conclude that a holomorphic function $f : D \to \mathbb{C}$ has derivative vanishing at $z_0$ if and only if it fails to be locally injective at $z_0$.
(This is probably my favourite result in complex analysis.)
Zhen Lin
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1I happen to stumble across this now. Is there a simple way to determine if a given conformal map is injective (aside from trying to find or not find two points, etc)? Perhaps a clever use of the inverse function theorem or something? – davidlowryduda Dec 09 '11 at 07:37
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1@mixedmath: There is no a priori reason to expect anything nice to happen. The inverse function theorem is a result about local behaviour, and conformality is a local property, while injectivity is a global property. – Zhen Lin Dec 09 '11 at 08:12
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10 years later, but, perhaps you could still help me understand this part of your proof: "$z\mapsto zg(z)$ is an invertible holomorphic function in $U$. This implies $f$ is an $n$-to-$1$ function on $U\setminus{0}$". How can we be sure that all $n$ roots of a certain value from $f(U\setminus{0})$ are in the image of $U$ through $z\mapsto zg(z)$? Or perhaps I misunderstood this part. – Stefan Octavian Aug 19 '21 at 15:13
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Write $k (z) = z g (z)$. We have $f (z) = k (z)^n$, and $k$ is invertible, so the solutions of $w = f (z)$ are given by $k^{-1} (v)$ where $v$ are the $n$ different $n$-th roots of $w$. – Zhen Lin Aug 19 '21 at 15:25
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Yes, but how do we know for sure that all the $n$ roots $v$ are in the domain on $k^{-1}$, that is in $k(U)$? – Stefan Octavian Aug 19 '21 at 15:31
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We are assuming $f (0) = 0$ and $w$ is a small non-zero complex number. We already know $k$ is invertible on $U$, so by shrinking $U$ further if necessary, we may assume $k (U)$ is a small disc centred on $0$. This ensures that $w \in f (U)$ has the property that all of its $n$-roots are all in $k (U)$. – Zhen Lin Aug 19 '21 at 15:52
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Oh, I think I got it. It's a little different than you said, though. Since $U$ is open and $k=(k^{-1})^{-1}$, then $k(U)$ is the inverse image of an open set through a continuous function, hence open. It contains $0$, so there is a ball $B$ around $0$ in $k(U)$. Then $f$ is not injective on $k^{-1}(B)$. – Stefan Octavian Aug 19 '21 at 18:04
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It is not possible for a holomorphic function to be injective but not conformal; for a holomorphic f to be injective, its derivative must be non-zero, and conformality of f is precisely equivalent to having $f'(z)\neq 0$.
On the other hand, it is possible for f to be conformal but not injective: $expz:=e^z$ is one example.Conformality follows from the fact that f(z) is analytic , and $\frac{de^z}{dz}=e^z$, and $e^z\neq 0$; (use the identity $e^ze^{-z}=1$ to show $e^z\neq 0$)
EDIT: Conformality follows from the Cauchy-Riemann equations , which show that the effect of f'(z) is that of stretching and rotating. If f'(z) is not zero, then any tangent vector forming an angle $\theta$ will be rotated, under f'(z) by an agle $\theta$, so that two vectors $v,w$ forming an angle $\gamma$ between them, will each be rotated by the same amount, with the effect that the net difference in the angles of the vectors v,w, and their respective images f(v),f(w) is $\theta -\theta=0$
user135041
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gary
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@Braindead: A simpler example of a conformal but not injective function is $z \mapsto z^2$ in a domain that does not include the origin. – lhf Aug 01 '11 at 16:43
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@lhf: but my point in using $e^z$ was that its "non-injectivity" is extreme, in that it is $\infty\rightarrow 1$ – gary Aug 01 '11 at 16:48
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This is strange; my answer had been accepted, and somehow it was unaccepted. I had never seen this before. – gary Aug 01 '11 at 17:56
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@gary - After I accepted your answer, Zhen Lin posted another answer, which I accepted because it completely answered my question by reminding me that all holomorphic functions look like $z\to z^n$ locally. I mean, it feels like you are taking this personally for some reason judging from a series of edits you've made, but I didn't mean any offense. I definitely upvoted your reponse. – Braindead Aug 01 '11 at 18:10
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@Braindead: my apologies, actually; I displayed a lack of class there; my reaction had nothing to see with you. Sorry. – gary Aug 02 '11 at 21:29
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