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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,

User001
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  • An argument can also be given for odd-degree polynomials being 1-1 only if the degree is 1. – hardmath Dec 17 '14 at 22:29
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    (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. – 5xum Dec 17 '14 at 22:29
  • This has surely been asked already on the site. – Mariano Suárez-Álvarez Dec 17 '14 at 22:40
  • 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!) – Mariano Suárez-Álvarez Dec 17 '14 at 22:42
  • 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? – User001 Dec 18 '14 at 03:13
  • I think that the substantive changes in the recent edit warrant a different question, rather than simply an edit in this question. – apnorton Dec 18 '14 at 04:28
  • I got it now, @ Mariano Suárez-Alvarez - thanks so much for the warning :). – User001 Dec 18 '14 at 05:40
  • Thanks, @hardmath. I think I got what you are saying now. – User001 Dec 18 '14 at 05:42
  • @5XUM - thanks for the suggestion. Is what your referring to called "analytic continuation"? Extending f to the Riemann sphere, while still assuming that f is one-to-one, so that the only point that maps to infinity...is the point at infinity -- and this makes f a bijection on the Riemann sphere. I have little experience with analytic continuation, so feel free to share any comments with me. Thanks... – User001 Dec 18 '14 at 23:31
  • @LebronJames Actually, analytic continuation is something slightly different than that. Analytic continuation refers to, for example, extending the complex number in such a way that the complex logarithm is well defined on the whole set. What you did was simply expand the function to the Riemann sphere, but you need to be careful, because unless you can prove it, you cannot just assume that any one-to-one entire function on $\mathbb C$ can be continuously extended to the Riemann sphere... – 5xum Dec 19 '14 at 07:10

1 Answers1

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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...

John Hughes
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  • 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. – User001 Dec 18 '14 at 03:02
  • 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... – User001 Dec 18 '14 at 03:05
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    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 $n = \pm 1$. – John Hughes Dec 18 '14 at 03:54
  • 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.. – User001 Dec 18 '14 at 04:03
  • Awesome stuff on the nth root of unity suggestion. Thanks so much, professor :). – User001 Dec 18 '14 at 04:13
  • Actually, I think I see what you are saying now - just because f(0) = 0 does not mean the zero is of order 1...unless f'(0) is not zero. – User001 Dec 18 '14 at 04:18
  • I think I got it now. First using the essential singularity / Picard theorems to show that f is a polynomial, then using your n roots of unity argument, I am able to conclude that we can only have a polynomial of degree 1. Thanks, professor. Have a great night. – User001 Dec 18 '14 at 05:37