Besides the identity map, is there an entire function $f$ that is a bijection from $\mathbb{C}$ to $\mathbb{C}$ and has 2 fixed points? Thank you for the help.

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    $\begingroup$ @Chandru, I'm afraid that is not one to one since -1 and 1 would get mapped to 1, for instance. Thanks though. $\endgroup$ – pel May 16 '11 at 19:51
  • $\begingroup$ Other than $f(x)=-x$, which has the entire imaginary line as fixed points? $\endgroup$ – Thomas Andrews May 16 '11 at 19:52
  • $\begingroup$ @Thomas: Is $f(x+iy)=-x$ entire? $\endgroup$ – pel May 16 '11 at 19:54
  • $\begingroup$ Presumably, you want some sort of continuity/differentiability condition, because set functions can be defined easily with exactly two fixed points. $\endgroup$ – Thomas Andrews May 16 '11 at 19:55
  • $\begingroup$ f(x+iy) = -x - iy is. $\endgroup$ – Thomas Andrews May 16 '11 at 19:56

No. Let $f$ be an entire function, and consider the singularity of $f$ at infinity. If it's removable, then by Liouville's theorem $f$ is constant and so not a bijection. If it's essential, then by Picard's great theorem $f$ is not injective. If it's a pole, then $f$ is a polynomial. If the degree of $f$ is greater than one, then $f$ is not injective. So $f$ is a linear polynomial, and the only linear polynomial that fixes two points is the identity.


Since invoking Picard seems like a bit of an overkill, here's a simple argument (essentially the same as Chris's answer, but replacing Picard by Casorati-Weierstrass and the open mapping theorem):

Claim. If $f: \mathbb{C} \to \mathbb{C}$ is entire and injective then $f$ is linear and non-constant, that is $f(z) = az + b$ for some $a \neq 0$.

It is easy to check that a linear function with two fixed points must be the identity, so we are left with proving the claim.

Proof of the claim. By hypothesis $f: \mathbb{C}^{\times} \to \mathbb{C}$ is injective where $\mathbb{C}^{\times} = \{z \neq 0\}$ is the set of non-zero complex numbers. Now $g(z) = f(1/z)$ is also injective $\mathbb{C}^{\times} \to \mathbb{C}$ and has a singularity at $0$. By injectivity, $g(\{|z| \gt 1\}) \cap g(\{0 \lt |z| \lt 1\}) = \emptyset$. On the other hand, both sets are open by the open mapping theorem, and by the Casorati-Weierstrass theorem $g(\{0 \lt |z| \lt 1\})$ would be dense if $g$ had an essential singularity at $0$. Therefore $g$ has a pole at $0$. But this implies that $f$ is a polynomial and hence it is linear and non-constant by injectivity.

  • $\begingroup$ @Theo: Sorry, but could you explain what the $C^\times$ notation means? Is that the extended complex plane? $\endgroup$ – user1736 May 17 '11 at 4:27
  • $\begingroup$ @user1736: It's the set $\{z \in \mathbb{C}\,:\,z \neq 0\}$. $\endgroup$ – t.b. May 17 '11 at 5:14
  • $\begingroup$ @Theo: Super answer. How did you think about this? $\endgroup$ – user9413 May 17 '11 at 5:24
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    $\begingroup$ @Chandru: Thanks, but the argument is not mine, of course! I probably learned it in my complex analysis class a long time ago but variants of it are quite standard when classifying automorphisms of connected and pointed subsets of $\mathbb{C}$. The point is of course that you want to exclude an essential singularity and the only thing one can say is that around an essential singularity are the results by Casorati-Weierstrass and by Picard. Now Picard is of course much stronger, but also much harder, so it's natural to try to avoid it. $\endgroup$ – t.b. May 17 '11 at 5:34

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