Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?
share|improve this question
2  
Do you have a source for this problem? Offhand my guess is that this term denotes the extended complex plane a.k.a. the Riemann sphere (en.wikipedia.org/wiki/Riemann_sphere). –  Qiaochu Yuan Jun 29 '11 at 0:21
4  
@Qiaochu When I read it, I thought the adjective "finite" meant the opposite as "extended"; i.e. I thought they were referring to the complex plane without the point at infinity. –  Quinn Culver Jun 29 '11 at 0:41
3  
Agreed. "Finite" emphasizes that $\infty$ is not included. Nowadays when we say "complex plane" this is understood. –  GEdgar Jun 29 '11 at 0:59
2  
Curious. To me "finite" suggests "compact," so that's where I guessed the meaning from. –  Qiaochu Yuan Jun 29 '11 at 1:34
1  
@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.) –  Willie Wong Jun 29 '11 at 9:42

3 Answers 3

up vote 8 down vote accepted

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.

share|improve this answer
2  
Another cheap way is to look at Cauchy-Riemann, and $f$ as a function $\mathbb{R}^2\to\mathbb{R}^2$. The dimensional restriction on the image means $df$ is not surjective anywhere, hence by Cauchy-Riemann $df = 0$ everywhere. (Since the Jacobian determinant of $df$ is a sum of squares.) –  Willie Wong Jun 29 '11 at 0:44
    
@Willie Wong: Maybe this is a stupid question: Are you talking about the "Cauchy-Riemann equation" when you say "Cauchy-Riemann"? –  Jack Jun 29 '11 at 1:39
    
@Corey: I am not able to catch up with you in your second paragraph. Could you elaborate it a little more? –  Jack Jun 29 '11 at 1:46
1  
@Jack: The Little Picard theorem says that if the image of an entire function misses 2 points of $\mathbb{C}$, it is constant. The function $e^z$ misses 0, so an analytic function can miss 1 point. But the function in your question misses every point of $\mathbb{C}$ off the real line, so it is constant. As for the second explanation, Liouville's theorem says that a bounded entire function is constant. The function $\exp(if(z))$ is bounded so it is constant. Thus $f$ is constant. –  Corey Jun 29 '11 at 1:53
1  
@Jack: yes. I mean the Cauchy-Riemann equations. For 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 –  Willie Wong Jun 29 '11 at 9:37

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?

share|improve this answer
    
@Amitesh He didn't ask for a full solution. –  Quinn Culver Jun 29 '11 at 1:03
    
@Quinn I have edited my answer and hidden the complete solution. (I apologize for that lapse of concentration in my part; I usually avoid giving complete solutions but I forgot this time.) –  Amitesh Datta Jun 29 '11 at 1:08
    
@Amitesh Datta: Thanks for your answer. Due to my ignorance, I know nothing but the basic calculation with complex variables. Now I see that the "open mapping theorem" in complex analysis is needed. –  Jack Jun 29 '11 at 1:41
    
Dear Jack, you are welcome! Please do not hesitate to ask if you have further questions. (Note that the maximum modulus principle can be used to prove the open mapping theorem; are you familiar with the maximum modulus principle?) –  Amitesh Datta Jun 29 '11 at 1:56
    
Dear Theo, you are absolutely correct, of course (otherwise I would be claiming that a (proper) disconnected open subset of the plane is homeomorphic to a (proper) connected subset of the plane!) I have fixed the exercise; the assumption "simply connected (and non-empty)" should be added (in which case it becomes the Riemann mapping theorem). I apologize for this lapse in concentration on my part (again). –  Amitesh Datta Jun 29 '11 at 9:18

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.