Can the Complex Plane Extend to Infinity?

[Niles]

Homework Statement

Hi all.

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

Thank you very much in advance.Niles.

 

[HallsofIvy]

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 "i\cdot\infty= \infty". 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".

 

[Niles]

Thanks for replying. Is this single point \infty or i\cdot\infty? Or should I understand it as if they are both this single point?

 

[HallsofIvy]

That single point is called "infinity". You will sometimes see "i\cdot\infty" as shorthand for \lim_{b\to\infty}0+ bi but the limit itself is just the point at infinity.

 

[Niles]

Thanks. It is very kind of you to help me.

 

[Niles]

Hmm, I seem to have stumpled upon something, which is quite odd:

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

<br /> \begin{array}{l}<br /> \mathop {\lim }\limits_{x \to \infty } \left( {x^2 + 4} \right)^{1/2} = \infty \\ <br /> \mathop {\lim }\limits_{y \to \infty } \left( {(iy)^2 + 4} \right)^{1/2} = i \cdot \infty \\ <br /> \end{array}<br />

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?

 

[Count Iblis]

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.

 

[Hurkyl]

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 x \cdot \infty = \infty for all nonzero complex numbers x. (0 \cdot \infty is not in the domain of \cdot -- 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 a + b \mathbf{i} 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 +\infty and -\infty. Topologically, this looks like a square. (In the same sense that the closed interval [-\infty, \infty] has the same shape as the interval [0, 1])