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Complex Analysis: Infinity

  1. Jun 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    We we look at [itex]z\rightarrow \infty[/itex], does this include both [itex]z=x[/itex] for [itex]x \rightarrow \infty[/itex] AND [itex]z=iy[/itex] for [itex]y\rightarrow \infty[/itex]? So, I guess what I am asking is, when [itex]z\rightarrow \infty[/itex], am I allowed to go to infinity from both the real and imaginary axis? If yes, then this implies that [itex]i\cdot \infty = \infty[/itex]?

    Thank you very much in advance.

    Best regards,
  2. jcsd
  3. Jun 4, 2009 #2


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    Yes, the complex plane has a single "point at infinity". Topologically, every set of the form {z | |z|> R} is a neighborhood of the "point at infinity". That does NOT mean that "[itex]i\cdot\infty= \infty[/itex]". This is purely "topological" or "geometric". Infinity is NOT a complex number and we do not multiply it.

    Added: that is the most common way to extend the complex numbers (the "one-point compactification") and makes the complex plane topologically equivalent to a sphere, but it is also possible to add a "point at infinity" at the end of every ray from 0 (The "Stone-Cech" compactication). That makes the complex plane topologically equivalent to a closed disk. However, both of those are topological (geometric) concepts and do not change the algebra of the complex numbers. We still do not define arithmetic involving "infinity".
    Last edited: Jun 7, 2009
  4. Jun 7, 2009 #3
    Thanks for replying. Is this single point [itex]\infty[/itex] or [itex]i\cdot\infty[/itex]? Or should I understand it as if they are both this single point?
  5. Jun 7, 2009 #4


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    That single point is called "infinity". You will sometimes see "[itex]i\cdot\infty[/itex]" as shorthand for [itex]\lim_{b\to\infty}0+ bi[/itex] but the limit itself is just the point at infinity.
  6. Jun 8, 2009 #5
    Thanks. It is very kind of you to help me.
  7. Jun 15, 2009 #6
    Hmm, I seem to have stumpled upon something, which is quite odd:

    Lets look at the function [itex]
    f(z) = \sqrt{z^2+4}
    [/itex]. This function has a simple pole at [itex]z_0=\infty[/itex], but we also have the limits

    \mathop {\lim }\limits_{x \to \infty } \left( {x^2 + 4} \right)^{1/2} = \infty \\
    \mathop {\lim }\limits_{y \to \infty } \left( {(iy)^2 + 4} \right)^{1/2} = i \cdot \infty \\

    and hence the limits are both infinite, and thus there is no singularity according to the limits - but we just found a pole at infinity. What part of my reasoning is wrong here?
    Last edited: Jun 15, 2009
  8. Jun 15, 2009 #7
    The fact that the limits are infinite is consistent with there being a pole. You can compute the residue at the pole, e.g. by first using the transformation z ---> 1/z to move the (alleged) pole to the origin and then compute the residue at zero.
  9. Jun 15, 2009 #8


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    There are many different compactifications of the complex numbers -- essentially, ways to add points "at infinity" to make calculus behave nicely.

    The most common is the projective complex numbers. In that number system, there is only one infinite number, and [itex]x \cdot \infty = \infty[/itex] for all nonzero complex numbers x. ([itex]0 \cdot \infty[/itex] is not in the domain of [itex]\cdot[/itex] -- i.e. it's undefined) This can be pictured as the Riemann sphere.

    Another one that crops up sometimes is to consider the set of all [itex]a + b \mathbf{i}[/itex] where a and b are extended real numbers. The extended real numbers are the ones you're probably familiar with from calculus (although you weren't taught to use them as numbers) -- it has two infinite numbers called [itex]+\infty[/itex] and [itex]-\infty[/itex]. Topologically, this looks like a square. (In the same sense that the closed interval [itex][-\infty, \infty][/itex] has the same shape as the interval [itex][0, 1][/itex])
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