# Homework Help: Complex Analysis: Infinity

1. Jun 4, 2009

### Niles

1. The problem statement, all variables and given/known data
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.

Best regards,
Niles.

2. Jun 4, 2009

### 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".

Last edited by a moderator: Jun 7, 2009
3. Jun 7, 2009

### 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?

4. Jun 7, 2009

### 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.

5. Jun 8, 2009

### Niles

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

6. Jun 15, 2009

### Niles

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

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

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

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
7. Jun 15, 2009

### 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.

8. Jun 15, 2009

### Hurkyl

Staff Emeritus
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 $[0, 1]$)