Computing with Infinity and complex numbers

Question

Complex analysis text books extend the complex numbers to the Riemann sphere. Is there anything wrong with my use of Infinity (from the extended real line) below?

z/Infinity = 0  for any complex number z.

z^Infinity = 0  when Abs(z) < 1

z^(-Infinity) = 0  when 1 < Abs(z) < Infinity 

***** CLARIFICATION ******************************************************************** In the Wolfram Lauguage the point at infinity on a Riemann sphere is called ComplexInfinity. So if you use it to compute 1/0 it returns ComplexInfinity. The Wolfram Launguage also knows about the affine infinity, and it knows how to do things like the following. Does the mathematical community agree with the next three results that I get using the Wolfram language? The infinity used below is the affine infinity.

Answer 1

Yes, there is something wrong:

First, you start talking about infinity on the Riemann sphere, but then talk about infinity on the extended real line; given that you use $-\infty$ in (3), I assume you're talking about the affine infinity (that is, extending the real line on both sides, with $\infty$ on the positive side and $-\infty\ne\infty$ on the negative side).

Those two concepts are not compatible; in particular, in the Riemann sphere there is no separate $-\infty$; negation on the Riemann sphere means rotation by $180^\circ$ around its vertical axis; besides $0$, also $\infty$ is a fixed point of that rotation, which means that on Riemann's sphere, $-\infty=\infty$.

Your case (1) makes sense on the Riemann sphere, as long as you don't expect this to be a full-fledged algebraic inverse, that is, the equation $a\cdot\frac{1}{a}=1$ does not hold for $a=\infty$, rather the product of $0$ and $\infty$ is not defined (that product cannot be defined without contradiction, unless you are willing to sacrifice the normal laws of arithmetics). On the Riemann sphere, $x\mapsto 1/x$ is just mirroring on the equatorial plane, and that indeed exchanges $0$ and $\infty$.

This BTW reinforces the fact that $\infty=-\infty$, since you get $$\infty = \frac{1}{0} = \frac{1}{-0} = -\frac{1}{0} = -\infty$$

However (2) and (3) don't really make sense on the Riemann sphere, which is already shown by the fact that you did differentiate between $\infty$ and $-\infty$.

Indeed, exponentials of complex numbers are a complicated topic, and if you want to keep it simple, you should restrict the exponent to integers. You might extend the integers with $\pm\infty$ by declaring $f(\pm\infty)=\lim_{n\to\pm\infty} f(n)$ whenever that limit exists, as you apparently did in (2) and (3), since with $f(n)=z^n$ you'll get exactly the conditions you wrote. However note that the $\infty$ you get that way in the exponent is very different from the $\infty$ in the Riemann sphere. It can, however, be identified with $\pm\infty$ of the usual extended real line.

Note that while the real line is embedded in the complex plane, the extended real line (with $\infty\ne-\infty$) is not embedded in the Riemann sphere.

Answer 2

In your question you used the term "Infinity" in three distinct senses that should not be confused.

(1) In the equation $\frac{z}{\infty}$ you are using $\infty$ as the point $\infty\in\hat {\mathbb C}$ added to the complex plane to form the Riemann sphere.

(2) In the second equation $z^\infty=0$ you are using the symbol as shorthand for the limit of $z^n$ as $n$ increases without bound.

(3) In the third equation when you write $1<|z|<\infty$ you simply mean to say that $z$ is an ordinary complex number whose absolute value $|z|$ is an ordinary real number greater than $1$, so the use of infinity here is merely poetic.

It is important to realize that the symbol $\infty$ is not being used in the same senses here.