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Show that if f is entire and one-to-one, then it must be of the form AZ+B, with A not equal to zero.

I am editing my question, since there are duplicates on this forum to the question of why an entire and one-to-one function must be of the form AZ+B, with A non-zero.

I am currently stuck at f(z)=AZ for A non-zero, from using Liouville's Theorem on g=z/f(z).

I'd like to show that f(z) cannot also take the form AZ^2, AZ^3, and so on...

I think that is done by using the Fundamental Theorem of Algebra and saying that an nth degree polynomial in z (with non-zero coefficient, A) has exactly n roots. But I'm thinking about the situation when all of the roots are at one location, so that we have only one distinct root with multiplicity = n. Then this doesn't rule out the case that f(z) is one-to-one.

What can I do to show the polynomial must only be of degree 1? (I've seen some derivative arguments now, including @JohnHuges ' argument below, where f' is not zero, but I don't understand this argument and why we can conclude from this that f is not one-to-one...)

Thanks in advance,


marked as duplicate by Ayman Hourieh, quid, 5xum, JimmyK4542, user147263 Dec 17 '14 at 23:37

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  • $\begingroup$ An argument can also be given for odd-degree polynomials being 1-1 only if the degree is 1. $\endgroup$ – hardmath Dec 17 '14 at 22:29
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    $\begingroup$ (1) Why can you restrict your discussions to linear fractional transformations? (2) What exactly d you mean when you say $f(\infty)=\infty$? Since $f$ is a function from $\mathbb C$ to $\mathbb C$ and $\infty\notin\mathbb C$, this needs more explanation. $\endgroup$ – 5xum Dec 17 '14 at 22:29
  • $\begingroup$ This has surely been asked already on the site. $\endgroup$ – Mariano Suárez-Álvarez Dec 17 '14 at 22:40
  • $\begingroup$ No even degree polynomial is one to one, in fact. (Odd-degreee polynomials of degree different to one are not much better in that respect!) $\endgroup$ – Mariano Suárez-Álvarez Dec 17 '14 at 22:42
  • $\begingroup$ Hmm...@MarianoSuarez-Alvarez - do you mean polynomials with only even-powered terms that have non-zero coefficients...or a polynomial with even degree? For example, isn't x^4 + x^3 one-to-one? $\endgroup$ – User001 Dec 18 '14 at 3:13

Here's a hack at a proof-sketch.

  1. A bounded entire function is constant -- that's a standard theorem in complex variables.

  2. If your function is unbounded but entire, there must be a sequence $z_i$ with $\lim |f(z_i)| = \infty$, and since the function is bounded on compact sets, we know $\lim |z_i| = \infty$. I'm pretty sure that with a little fiddling, you can conclude that $\lim_{z \to \infty} f(z) = \infty$, so $f$ extends to a function from the Riemann sphere to itself, a function that's 1-1 everywhere (since it sends $\infty$ to $\infty$).

Using this extended function (but still calling it $f$), let $$ g(z) = 1 / f(1/z) $$ and $g$ sends $0$ to $0$, with $g'(0) = a \ne 0$. Then consider

$$ f(z) - \frac{1}{a} z $$

and I'll bet you find that it's a bounded entire function...

  • $\begingroup$ Hi professor: using your suggestions, I have that f is a function from the extended complex plane onto itself, so f is actually a bijection on the Riemann sphere (with the assumption that f is one-to-one). I got a little stuck from studying the last part of your answer, so instead I'd like to now look at g(z) = z/f(z). Since 1/f(z) has a simple pole at zero, g(z) is entire and bounded, hence constant. Then C = z/f(z), which implies f(z)=Az for some complex constant A. I'd like to stay on this path to conclude my proof. The issue is now showing that f(z)=Az^(n) for n>1 is not possible. $\endgroup$ – User001 Dec 18 '14 at 3:02
  • $\begingroup$ Suppose n>1, then z^(n) has exactly n roots, by the Fundamental Theorem of Algebra. I'd like to say that this fact makes f(z) (which of course has n roots) not one-to-one. The problem is...what if those n roots are all at the same location, so that we have one distinct root with multiplicity = n? This doesn't seem to rule out that f(z) is one-to-one... $\endgroup$ – User001 Dec 18 '14 at 3:05
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    $\begingroup$ I don't follow your claim about the simple pole, but for your second question: pick any $z \ne 0$. Then $z^n = (\omega z)^n$ where $\omega = \cos(2\pi/n) + i \sin (2 \pi/n)$ is an $n$th root of unity. So that makes $z^n$ not injective unless $n = \pm 1$. $\endgroup$ – John Hughes Dec 18 '14 at 3:54
  • $\begingroup$ The simple pole comes from the fact that since f(0) = 0 -- a simple zero at 0, then 1/f(z) has a simple pole. We can write f(z) as z^(1) * h(z), where h(z) is analytic and non-zero at zero. So, with g= z/f(z), this is an entire function, since it effectively removes the pole at zero.. $\endgroup$ – User001 Dec 18 '14 at 4:03
  • $\begingroup$ Awesome stuff on the nth root of unity suggestion. Thanks so much, professor :). $\endgroup$ – User001 Dec 18 '14 at 4:13

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