# 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

Staff Emeritus
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: 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

Staff Emeritus
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 $[0, 1]$)