Little Problem with Complex Mathematics on Calculators

[Walter]

I came across the following quite simple calculation:

0 [1/x] [+/-] returns -∞.

We all know that SQRT(-∞) = i ∞.
The calculator returns 0. + i ∞ which is perfectly equivalent.

Now let's revert this last operation by squaring: (i ∞)^2 = -∞. So far so good.
But (0 + i ∞)^2 = 0 + (-∞) + 2 i 0 ∞ = -∞ + 2 i 0 ∞.
So we must conclude that 0 times ∞ equals 0 which is in contradiction to other people stating 0 times ∞ is undefined or even non-numeric (the calculator returns -∞ + i NaN).

What am I missing? Competent help will be appreciated.

[rprosperi]

I came across the following quite simple calculation:

0 [1/x] [+/-] returns -∞.

We all know that SQRT(-∞) = i ∞.
The calculator returns 0. + i ∞ which is perfectly equivalent.

Now let's revert this last operation by squaring: (i ∞)^2 = -∞. So far so good.
But (0 + i ∞)^2 = 0 + (-∞) + 2 i 0 ∞ = -∞ + 2 i 0 ∞.
So we must conclude that 0 times ∞ equals 0 which is in contradiction to other people stating 0 times ∞ is undefined or even non-numeric (the calculator returns -∞ + i NaN).

What am I missing? Competent help will be appreciated.

Probably not competent, but anyway...

What machine are you using here Walter?

I can't find any (I did not try all of them!) machine that yields SQRT(-∞) => 0 + i*∞ (thus I believe implying that SQRT(∞) = ∞, which although possibly sound mathematically (I don't know for sure) it also doesn't seem to be handled my many (any?) machine).

[H2X]

But (0 + i ∞)^2 = 0 + (-∞) + 2 i 0 ∞ = -∞ + 2 i 0 ∞.

Aren't you assuming that 0 times ∞ is defined yourself? If 0 times ∞ is undefined, 2 i 0 ∞ is not defined either?

[dlachieze]

We all know that SQRT(-∞) = i ∞.
The calculator returns 0. + i ∞ which is perfectly equivalent.

Probably not competent either, but I know that infinity is tricky...

For example you can also say that -i + i ∞ is perfectly equivalent to i ∞.

But (-i + i ∞)*(-i + i ∞) = -1 -∞ +2∞ which is undefined

[H2X]

We all know that SQRT(-∞) = i ∞.
The calculator returns 0. + i ∞ which is perfectly equivalent.

Probably not competent either, but I know that infinity is tricky...

For example you can also say that -i + i ∞ is perfectly equivalent to i ∞.

But (-i + i ∞)*(-i + i ∞) = -1 -∞ +2∞ which is undefined

Complex multiplication has a rotational element which is tricky too. Infinite number of rotations? I get dizzy all the time...

[dlachieze]

Doing some research I’ve found that for both Wolfram Alpha and the HP Prime: SQRT(-1/0) returns the complex infinity (which is not i∞).

As you’re taking the square root of a negative value you're moving to the complex plane where there is only one infinity, the complex infinity which is linked to the extended complex plane and the Riemann sphere.

So when you're working with infinity you need to know if your are in the real or complex domain.

[Walter]

Thanks to Bob, Didier, and Haakon for responding so far.

My basic question was: why is i ∞ = 0 + i ∞ but it looks that (i ∞)^2 ≠ (0 + i ∞)^2 ?

Reasoning for the second equation:
(i ∞)^2 = -∞ while (0 + i ∞)^2 = 0 - ∞ + 2 i 0 ∞ (this expansion of the squared parenthesis doesn't assume anything IMO).
For making both sides equal, 0 × ∞ must be zero.
OTOH, WolframAlpha claims 0 × ∞ being undefined, and the 34S returns 0 × ∞ being 'not numeric'.
At the same time, WolframAlpha claims (0 + i ∞)^2 = -∞ while the 34S claims (0 + i ∞)^2 = -∞ + i × NaN; and I don't see that NaN = 0.
There must be a fundamental error in this reasoning but I can't find it.

[dlachieze]

Well, I think that the real question is how you add infinity to ℂ. As far as I know this is called the compactification of ℂ.
With the Riemann sphere and the associated extended complex plane there is only one complex infinity, so if I understand it correctly there is nothing like i ∞, juste one complex infinity which means that 0 + i ∞ doesn't make sense in this extended complex plane.

I don't know if you can define an extended complex plane with multiple infinity, but if so you need also to define the appropriate arithmetic operations to handle these infinity.

Anyway I'm not an expert in these matters so I would like to ear from more competent people.

[H2X]

Anyway I'm not an expert in these matters so I would like to ear from more competent people.

I am not an expert either, but clearly there are kinds of infinities which cannot be easily compared. While there is an infinite number of integer values, it seems logical that the also infinite number of rational numbers is larger, and real numbers larger still.

This does not extend further into complex numbers, because where - along which line - would this infinity be? Apparently it needs to be along some line, which is the numerical axis of the absolute value or magnitude of the complex number, not some coordinate in the complex plane.

Interesting stuff, though. I'd welcome an expert statement about this!

[H2X]

... while we're waiting, here's an interesting take on imaginary numbers: