Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

A finite hyperreal number $r$ is a number defined as a sum of a real number and an infinitesimal number $\omega$: $$r=a+\omega$$ Do you know if is it possible (and useful) to define a complex number as $$c=a+i\omega\ ?$$ Thanks

share|improve this question
4  
The correct way to define a hypercomplex number would (probably) be $c = r_1 + i r_2$, where $r_1,r_2$ are hyperreal. I don't know whether under this definition, the hypercomplex are algebraically closed. – Yuval Filmus Mar 8 '12 at 13:02
1  
@Yvual: Yes, this is the way to define the hypercomplex numbers, and yes, they are algebraically closed. The easy way to see that they are algebraically closed is that algebraic closure of $\mathbb{C}$ can be expressed in first-order logic over ordered pairs of reals, so by the transfer principle the same applies to the hypercomplexes, $\mathbb{C}^*$. Riccardo's suggestion would not give a $\mathbb{C}^*$ that was algebraically closed, since, e.g., $x^2+1$ does not have any roots whose imaginary part is infinitesimal. – Ben Crowell Mar 8 '12 at 15:58
1  
To define a finite complex number z, just proceed as Yuval suggests and require $|z|$ to be finite. – Ben Crowell Mar 8 '12 at 16:04
Wait, I'm confused. @Ben: Why for the field $\mathbb C^\ast$ to be algebraically closed $x^2+1$ need to have infinitesimal roots? Does $x-1$ need to have infinitesimal roots in $\mathbb R^\ast$ as well? – Asaf Karagila Mar 17 '12 at 8:53
@AsafKaragila: "Why for the field C∗ to be algebraically closed $x^2+1$ need to have infinitesimal roots?" It doesn't, if you define C* correctly. It does, if you define C* as Riccardo suggests (limiting the imaginary part to an infinitesimal value). – Ben Crowell Mar 18 '12 at 19:29

3 Answers

up vote 1 down vote accepted

I am not sure, whether following is helpful or related to your question:

But do you know wick rotation$[(x+t)\to (x+it)]$.. See wikipedia page.

Many system can be change to other system by wick rotation. For example take born infeld equation and minimal surface equation, these are related by wick rotation of one variable.

That is if $\phi(x,t)$ is solution of Born-infeld equation then $\phi(x,it)$ will be solution to minimal surface equation.

So if we define $z= s+t\in\mathbb R$ some variable in solving Born-Infeld equation, then $z= s+it$ will be some point in domain of definition of minimal surface.

I can't explain this transformation completely but you can see wikipedia pages and one article of Rukmini Dey.

This is one particular example. Basically $t\to it$ gives many time very new result in transform equation.

share|improve this answer
up vote 2 down vote

The complex numbers can be defined as numbers of the form $x + i y$ where $x$ and $y$ are real. All constructions of standard analysis work in non-standard analysis, so this statement remains true in the non-standard model: hypercomplex numbers are numbers of the form $x + i y$ for hyperreal $x$ and $y$.

Whether it's useful to single out those complexes with standard real and infinitesimal imaginary part, I think it unlikely. Finite real and infinitesimal imaginary part is more likely to be useful.

For example, in complex analysis, one often constructs contours that include paths of the sort $x + i \epsilon$ to approximate the real line, or circular arcs $a + \epsilon e^{i \theta}$ to approximate a point, where $\epsilon$ is a small positive real number, and then takes the limit as $\epsilon \to 0$.

Replacing $\epsilon$ with a real infinitesimal would probably be useful.

share|improve this answer
up vote 0 down vote

Sorry, this is not an answer, but I cannot figure out how to comment on your question.

Where is this definition from? Is 0 a standard real or an infinitesimal? If the standard reals and the infinitesimals count as finite hyperreals (and they do by the half dozen definitions that I've seen), then you would need to allow both $a$ and $\omega$ to take the value 0 in order to get these numbers as sums of reals and infinitesimals.

share|improve this answer
You need some reputation in order to comment everywhere, see math.stackexchange.com/privileges/comment . You're almost there. – Bruno Stonek Mar 18 '12 at 11:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.