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The question is from the following problem:

If $f(z)$ is an analytic function that maps the entire finite complex plane into the real axis, then the imaginary axis must be mapped onto
A. the entire real axis
B. a point
C. a ray
D. an open finite interval
E. the empty set

A search on Google didn't return any satisfying result of the concept "the entire finite complex plane". The word "entire" and "finite" seem to be a contradiction to me. Here are my questions:

  1. What is "the entire finite complex plane"?
  2. What properties does one need to use about analytic function to solve this problem?
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    @Qiaochu: Heine-Borel says that a sequence of **bounded** real numbers has a converging subsequence, a property that definitely doesn't hold for sequences of **finite** real numbers. (Not saying that your intuition is wrong, because often finiteness does suggest compactness, but one has to be careful in drawing that association.)2011-06-29

4 Answers 4

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The "entire finite complex plane" is just the complex plane. This phrase isn't used very often, but can be used if you want to differentiate between the complex plane $\mathbb{C}$ and the Riemann sphere $\mathbb{C}\cup \{\infty\}$.

The easiest way to answer this is to use Little Picard, but there are certainly easier ways using less powerful machinery. For example, you could consider $\exp(if(z))$ and use Liouville.


Edit: Since you've added the tag reference-request, I will mention that the term "finite complex plane" is used in Silverman's translation of Markushevich's monumental "Theory of Functions of a Complex Variable" which is one of the standard references in complex analysis.

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    @Jack: yes. I mean the Cauchy-Riemann equations. For [Little Picard](http://en.wikipedia.org/wiki/Picard_theorem#Little_Picard), you can see Conway's _Functions of one complex variable I_, Markusevich's text which Corey mentioned in the edit (in which it is called Picard's first theorem), or, IIRC, Ahlfors's _Complex analysis_2011-06-29
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Hint: A non-constant analytic function $f$ is an open mapping, i.e., if $U\subseteq \mathbb{C}$ is an open subset, then the image $f(U)$ is also an open subset of the complex plane.

Complete Solution (using the hint above; please avoid hovering your mouse cursor over the (silver) box below if you do not wish to view the solution):

Since no subset of the real line can be an open subset of the complex plane (except, of course, for the empty subset), $f$ must in fact be constant and the image of the imaginary axis (in particular) under $f$ is a single point. Therefore, $B$ is the correct choice.

Alternate Hint: The maximum modulus principle.

The following exercises are relevant to your question:

Exercise 1: If no non-empty subset of $A$ is open in the complex plane, prove that there is no non-constant entire function $f$ that maps the complex plane into $A$. Can you prove this result using only the Cauchy-Riemann equations?

Exercise 2: If $U$ is an open subset of the complex plane, does there exist a non-constant analytic function $f$ that maps the complex plane into $U$? Does the answer change if you restrict to the unbounded open subsets $U$ of the complex plane?

Exercise 3: Prove that there exists an analytic function $f$ that maps the imaginary axis in the complex plane bijectively onto the real axis.

Exercise 4: Find an invertible holomorphic mapping of the open unit disc (${z:\left|z\right|<1}$) onto the upper half plane ({z:\text{Im}(z)>0}). Hence solve the Dirichlet problem in the upper half plane using the Poisson kernel on the open unit disk.

Challenging Exercise/Important Result (The Riemann Mapping Theorem): If $U$ and $V$ are proper, simply connected (and non-empty), open subsets of the complex plane, prove that there is an invertible holomorphic mapping of $U$ onto $V$. (If you cannot prove this, then you can look up the proof in most texts on complex analysis or online. However, the result is important and therefore you should at least understand the statement.)

Easy Exercise (based on the Riemann Mapping Theorem): Why is the assumption that both open sets be "proper" necessary in the Riemann mapping theorem? More precisely, why does not there exist an invertible holomorphic map from the complex plane onto the unit disk?

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    ... In saying that, of course, I do *hope* that the OP will solve some exercises; if I did not I would not bother giving them! I did not know that the OP did not know about the open mapping theorem when I wrote my answer; my expectation was that anyone who asks this kind of question knows about the open mapping theorem. Although I would give as an exercise the Riemann mapping theorem to someone who has had a first course in complex analysis, I completely agree that in this context the exercise is challenging enough to merit a warning which I have inserted.2011-06-29
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What is "the entire finite complex plane"?

I'm pretty sure "the entire finite complex plane" refers to the plane without the point at infinity.

What properties does one need to use about analytic function to solve this problem?

Knowing what properties one "needs" could be difficult, but I can at least tell you that the open mapping theorem is sufficient.

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Use Cauchy Reimann equations, you will get that function must be a constant