Division by complex infinity, simple question! please help

1. Jul 6, 2011

elegysix

given (-1)! = $\tilde{\infty}$ , which is complex infinity

the real part of $\tilde{\infty}$ is $\overline{?}$ which is "a quantity whose magnitude cannot be determined" as stated in wolfram's site at http://functions.wolfram.com/Constants/ComplexInfinity/introductions/Symbols/ShowAll.html" [Broken]

can we say that taking the real part of x/(-1)!

RE( x / $\tilde{\infty}$ ) = x/c, where c is an arbitrary constant?

Last edited by a moderator: May 5, 2017
2. Jul 6, 2011

pmsrw3

Any finite x divided by infinity gives 0. You can see this in the "Divide" table of the page you reference. So Re(x/inf) = 0, assuming x is finite.

3. Jul 6, 2011

elegysix

there must be some distinction between infinity and complex infinity. I know that what you say is true for division by infinity, but this is not what I am asking. I assume that infinity and complex infinity are not the same thing.

Is c + $\infty$i an example of complex infinity? if it is, then clearly Re( c + $\infty$i ) = c which is not $\infty$ as you suggest.

4. Jul 6, 2011

pmsrw3

"Infinity" is a somewhat vague concept. There are lots of different types of infinity. Complex infinity is one of them. It is a single point, that, adjoined to the complex plane, completes the Riemann sphere and allows you to give sensible answers to many question in complex analysis that, without it, have no answer.

For any finite x, x/$\infty$ is 0 for every type of infinity I know of for which the question makes sense.

When you write "Re( c + $\infty$i )", do you mean that to be complex infinity? If so, it is not correct that Re( c + $\infty$i ) = c. c + $\infty$i is $\infty$, and the question of its real value has no answer. Complex infinity does not have real and imaginary parts. I'm not sure why you think that I suggested that Re( c + $\infty$i ) = $\infty$.

5. Jul 6, 2011

SteveL27

I didn't go through the entire Wolfram page in detail, but there may be some discrepancies in what Wolfram's got, and the way most mathematicians think about the point at infinity.

In fact I believe the Wolfram page describes how Mathematica treats complex infinity. But it doesn't actually tell you what complex infinity is. To understand complex infinity, you should have a look at the Wiki writeup about the Riemann sphere ...

http://en.wikipedia.org/wiki/Riemann_sphere

That article says nothing about the Re and Im of complex infinity. I don't think those are defined. I can't think of a sensible way to define them.

In the complex plane, complex infinity is the (extended) limit of every ray leaving the origin; in the same sense that +/- infinity are the hypothetical "ends" of the extended real line.

In other words if you imagine the complex plane all the way out to infinity (speaking conceptually now), and then think about the "border" all the way around the plane; then if you identify every point of the border with a single point, that's the point at complex infinity.

It's like taking a rectangular piece of cloth and sewing together the entire perimeter, to make a sphere. If you had an infinitely large rectangular cloth and you sewed together its entire boundary to make a finite-sized sphere, that would be the Riemann sphere. And the "point at infinity" is the entire boundary, sewn together into one point.

The best way to think of it is through stereographic projection ... this is the diagram on the Wiki page.

http://upload.wikimedia.org/wikiped...Riemann_sphere1.jpg/250px-Riemann_sphere1.jpg

It's not helpful to regard complex infinity as a number you can just use the same way as any other complex number. It's an ideal point, created through a little mathematical magic. The Wolfram page is telling you how Mathematica treats complex infinity ... it's not telling you what complex infinity really is in terms of mathematics.

I hope this might help put the Wolfram page into context.

By the way, where did (-1)! = $\tilde{\infty}$ come from?

Last edited: Jul 6, 2011
6. Jul 6, 2011

pmsrw3

z! has a first-order pole at every negative integer.

7. Jul 6, 2011

elegysix

plugging (-1)! into wolfram alpha gives (-1)! = $\tilde{\infty}$

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