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Let $f$ be analytic on the whole all of $\mathbb C$. Assume that $\mathrm{Re}\, f \ge 0$. What can we say about $f$?

I'm thinking $f$ has got to be constant, since otherwise it would map the entire complex plane onto the positive real half-plane. I don't think an analytic function would be capable of this. But I'm not sure how to state it rigorously, if it's even true at all.

I suspect that such a mapping would be 2-1, meaning that $f(z)$ isn't uniquely determined. Is this true/rigorous enough?

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  • $\begingroup$ What you wrote is somewhat confusing. You seem to be saying both that $f$ will be injective, and $2$-$1$. $\endgroup$ – Andrés E. Caicedo Dec 26 '13 at 19:15
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    $\begingroup$ If $\mathrm{Re} f\geq 0$, then the entire function $e^{-f}$ is bounded, hence constant by Liouville. It is not hard to conclude that $f$ is constant after that. But I don't understand your title/question. $\endgroup$ – Julien Dec 26 '13 at 19:16
  • $\begingroup$ Analytic functions can be far from injective, but they are always "almost" surjective, mening they avoid at most one value (i.e. $f(z) = a$ has one or more solutions for all but at most one $a$, for any analytic $f$). $\endgroup$ – Arthur Dec 26 '13 at 19:22
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The question in the title and in the main body are different. An analytic function need not be injective, just look at $z^2$ on the real axis. However as for the question in the main body, the real part of a holomorphic function is harmonic, and harmonic functions that are bounded below are constant. Since the real part is constant, the Cauchy-Riemann equations imply that the function is constant.

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Consider the function

$$g(z) = \frac{1}{f(z) + 1}$$

This is an analytic function (why?) which is bounded (why?).


To answer the question in the title, analytic functions are not always injective: $e^z$ is certainly not, nor is $z^2$.

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