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Complex Analysis problem

  1. Jan 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Let a continuous function ##f:\mathbb{C}\rightarrow\mathbb{C}## satisfy ##|f(\mathbb{C})|\rightarrow\infty## as ##|z|\rightarrow\infty## and let ##f(\mathbb{C})## be an open set. Then ##f(\mathbb{C})=\mathbb{C}##.

    3. The attempt at a solution
    Suppose for contradiction that ##G=f(\mathbb{C})\not=\mathbb{C}##. Then ##G\subset\mathbb{C}## so ##\partial G\cap\mathbb{C}\not=\emptyset##. (This fact was proved in class.)

    The professor gave a hint that I should be using Bolzano-Weierstrauss and the fact that if ##f## is a continuous function and ##z_{n}\rightarrow z## then ##f(z_{n})\rightarrow f(z)##. I started by contradiction because that was the hint in the book but I can't see how to incorporate the professor's hint.
     
  2. jcsd
  3. Jan 12, 2014 #2

    Dick

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    Ok, suppose the boundary of ##f(\mathbb{C})## is not empty. Then pick a point ##z_0## in the boundary. Then there is a sequence ##z_n## such that ##f(z_n) \rightarrow z_0##, right? Now can you see that ##|f(z)|\rightarrow\infty## as ##|z|\rightarrow\infty## means ##z_n## must be a bounded sequence? Apply Bolzano-Weierstrass from there.
     
  4. Jan 13, 2014 #3
    By Bolzano-Weierstrauss, there exists a convergent subsequence of ##z_{n}##,##z_{n_{k}}## which converges to some ##z\in\mathbb{C}##. Then since ##f## is continuous, ##f(z_{n_{k}})\rightarrow f(z)=z_{0}##.

    I am not sure how to proceed. I keep thinking that the goal is to derive a contradiction by showing that ##\partial G\cup\mathbb{C}=\emptyset##. I think the problem is that I can't picture what I am trying to show in my head. How would you recommend going about picturing this?
     
  5. Jan 13, 2014 #4
    Suppose x is a complex number, and not in f(C). Let M > |x|. Then there exists N such that if |z| > N, then |f(z)| > M. The set {y: |y| ≤ N} is compact.

    Edit: x is either a limit point of f(C), or it isn't. If it's not, choose M so that the open ball with radius M centered at (0,0) contains the closure of the open ball with radius r centered at x, such that every open ball centered at x with radius greater than r contains a point in f(C), and the open ball with radius r centered at x contains no points in f(C).
     
    Last edited: Jan 13, 2014
  6. Jan 13, 2014 #5

    Dick

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    ##f(z)=z_0## mean ##z_0## is in ##f(\mathbb{C})##. Can an element of an open set also belong to the boundary? Think about definitions.
     
  7. Jan 13, 2014 #6
    Ah! Then since ##z_{0}## is contained in ##f(\mathbb{C})##, this is a contradiction because ##G## does not contain its boundary. If it did then ##G## would also have to be closed and the only sets which are both open and closed in ##\mathbb{C}## are ##\emptyset## and ##\mathbb{C}## but since ##G## contains ##z_{0}##, ##G## must be the entirety of ##\mathbb{C}##.
     
  8. Jan 13, 2014 #7

    Dick

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    Yes, that's it. The real trick is the argument that the ##z_n## must be bounded. That's how you use the limit property.
     
    Last edited: Jan 13, 2014
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