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Can we prove that given an entire function $f$ that is also one to one then $f$ must be linear?

Thanks for any help.

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1 (which is overkill if this is homework or something) –  yoyo Mar 29 '11 at 17:06
Maybe you can use the Weierstrass Factorization Theorem? –  Adrián Barquero Mar 29 '11 at 17:06
By 1-1, do you mean injective or bijective? –  lhf Mar 29 '11 at 17:52
@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :) –  Mariano Suárez-Alvarez Mar 29 '11 at 19:56
@Mariano: But it bugs me so... –  Arturo Magidin Mar 29 '11 at 20:40

5 Answers 5

You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.

To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f(\{z:|z|>n\})$ is dense in $\mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=\bigcap_n f(\{z:|z|>n\})$ is dense in $\mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.

I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f(\{z:|z|<1\})$ is open, and $f(\{z:|z|>1\})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.

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Nice application of Baire. Thanks! –  Digital Gal Mar 29 '11 at 19:24

Let $f:\mathbb C\to\mathbb C$ entire and injective. Let $U=f(\mathbb C)$. $U$ is an open subset of the plane.

$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".

Next, if $U\subsetneq\mathbb C$, from Riemann's theorem we know that there is an biholomorphic map $U\to D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $\mathbb C\to D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.

Using this, one can see that the function $1/f(z)$ is bounded at $\infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.

(This avoids Picard but uses Riemann... :( )

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How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez. –  Lebron James Dec 18 '14 at 6:18

By shifting $z$, without loss of generality you can assume $f(z) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 \over f(z)}$ is bounded outside of $U$. Therefore ${z \over f(z)}$ is an entire function that grows no faster than linearly: $|{z \over f(z)}| < A|z| + B$ for some $A$ and $B$.

It's easy from here to show that $g(z) = {z \over f(z)}$ is linear; for any $z_0$ ${g(z) - g(z_0) \over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z \over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {z\over c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 \over c_2} z$.

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I'll give the "usual" proof.

Note that by Little Picard, $f$ misses at most one point; but it is a homeomorphism onto its image, and the plane minus a point is not simply connected. Thus $f$ is onto $\mathbb{C}$, and hence bijective. Then $f$ has a holomorphic inverse, which is enough to imply $f$ is proper, that is, the pre-image of a compact set is compact. This in turn implies $$ \lim_{z\rightarrow\infty} f(z)=\infty,$$ and thus if we define $f(\infty)=\infty$, $f$ becomes a Möbius transformation of the Riemann sphere. So $f$ has the form $f(z) = \frac{az+b}{cz+d},$ and it is easy to see that if $f$ is entire on $\mathbb{C}$, then $c=0$.

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I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $\mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2\times2$ orthogonal. Since $f$ is holomorphic, $A\in SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $\theta$, so $f(z)=ke^{i\theta}z+b$. –  user641 Mar 29 '11 at 20:36

Here is a (longer) proof using very little of complex analysis. Assume that $f:\mathbb{C}\to\mathbb{C}$ is holomorphic and injective. The function $f$ extends to a holomorphic map of the Riemann sphere to itself, $\mathbb{CP}^1\to\mathbb{CP}^1$. Indeed, choose any $z_0$ such that $f'(z_0)\neq 0$ (if it doesn't exist then $f$ is constant (of course from injectivity we know that actually $f'\neq0$ everywhere)). Then a small neighbourhood $U\ni z_0$ is mapped bijectively to a small neighbourhood $V\ni f(z_0)$. The function $1/(f(z)-f(z_0))$ is therefore bounded in $\mathbb{C}-U$, hence by Riemann removable singularity theorem it extends to a holomorphic function on $\mathbb{CP}^1-U$, therefore $f$ extends to a holomorphic map $\mathbb{CP}^1\to\mathbb{CP}^1$.

As an application of Liouville's theorem, any holomorphic map $F:\mathbb{CP}^1\to\mathbb{CP}^1$ such $F(z)\neq\infty$ for $z\neq\infty$ is a polynomial. If we wish, we can also avoid Liuoville theorem and use some topology. If the order of pole of $f$ at $\infty$ is $>1$ then $f$ is not injective in the neighbourhood of $\infty$. Hence there is $a\in\mathbb{C}$ such that $f-az$ is holomorphic $\mathbb{CP}^1\to\mathbb{C}$, hence it's bounded (being a map from a compact space), hence it's a constant: if $f-az$ is not constant then it is a map $\mathbb{CP}^1\to\mathbb{CP}^1$ which is of positive degree but which is not surjective (as it avoids $\infty$).

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