How to prove that f(x,y) is complex

1. Aug 12, 2012

friend

Suppose I have a function of two real variables f(x,y). How could I prove that f(x,y) is complex? Thanks.

2. Aug 12, 2012

HallsofIvy

Staff Emeritus
First, what do you mean by a "complex function"? My first thought was "a function that returns complex numbers", but the set of complex numbers includes the real numbers. Do you mean "a function that returns, for some (x, y), non-real values"?

3. Aug 12, 2012

friend

Yes, thank you. I don't know if f(x,y) returns a real or complex value. So I think that means I cannot evaluate its complex conjugate. I think I'm looking for some algebraic properties unique to complex numbers (that include the reals) that I can use to prove my function is complex(including reals). Maybe something like f(x,y)≠f(y,x), but f2(x,y)=f2(y,x), or something like that? Or maybe I can somehow show that there is no sense in saying f(x,y)>0?

Last edited: Aug 12, 2012
4. Aug 12, 2012

DonAntonio

The simplest characteristic of real numbers within the complex ones is that their equal their own conjugate:
$$z\in\Bbb C\,\,,\,\,then\,\,\, \overline z=z\Longleftrightarrow z\in\Bbb R$$

What you talk about "algebraic properties unique to complex numbers" doesn't make much sense to me

in your question's context: what are your function's range and definition domains? Why do you suspect

the function could possible attain some non-complex values? Without this it's hard to know how to help you.

DonAntonio

5. Aug 13, 2012

Bacle2

I would understand your question as saying that you have a function

f: ℝ2→ ℝ2

and you wanted to see if f would be analytic as a complex-valued function, i.e.,

if f(x,y)= (u(x,y), v(x,y)) , then you want to see if:

ux=vy, and then you can express this as an analytic function

f: ℂ→ℂ of a complex variable?

This would be the case with, e.g: f(x,y)= (x2-y2, 2xy),

which can be rewritten as:

f(z)=z2.

Last edited: Aug 13, 2012
6. Aug 17, 2012

friend

Thanks guys for your replies. Sorry it took so long to reply; I've been working.

Suppose I have a set of functions labeled, $\delta (x - {x_0})$, for every value of x0, which have the following property:

$$\delta (x - {x_0}) = \int_{ - \infty }^{ + \infty } {\delta (x - {x_1})\delta ({x_1} - {x_0})} \,d{x_1}$$

I wonder if I can prove that this set of functions can be represented by functions that can take on complex values (which includes the real numbers). For example, if I can prove that these functions are multiplicatively commutative and associative, but lack the property of ordering with respect to ≥, then does that prove that they are complex, or at least can be represented by complex numbers since they share the same algebra? This is opposed to say if it were shown that these functions also lack commutativity, they would be represented by quaternions. Or if they also lacked associativity, they'd be represented by octonions. Your comments are appreciated.

7. Aug 17, 2012

DonAntonio

Once again you don't take the time to tell us what kind of functions those $\,\delta\,$ are: real functions, complex functions...?

You can't take beings out of the blue and decide their functions without first checking some basic stuff: domain, range,...

DonAntonio

8. Aug 17, 2012

friend

Actually, I'm thinking they are tempered distributions that satisfy, among other things, the Chapman-Kolmogorov equation. But please don't ignor the question: if the algebra is the same as the complex numbers: multiplicably commutative and associative but not ordered wrt ≥, then does that prove it IS complex valued? How could it not be? Aren't complex numbers defined by their algebraic properties? Thanks.

Last edited: Aug 17, 2012
9. Aug 17, 2012

DonAntonio

It seems to be we have different requirements: for me, when someone talks of a function and wants to know

something specific about is range or domain, injectivity or whatever, the first thing I require is to specify the

function's domain, range, etc. I don't think one can work mathematically assuming one can find a function (and identifying

it as one), without first knowing or asking for its domain, range and perhaps some other characteristics.

It's like asking "does the word erjfhwd means "dog"? I'd ask: in what language? Where did you hear it? Is it written

correctly or is that a transliteration?

I am addressing your question as mathematically as I can, and I think it is you who's dodging to give answers

DonAntonio

10. Aug 17, 2012

friend

Obviously from the integral, the domain is from plus or minus infinity of the variable x. The range will depend on how you normalize it. But if the output of the function is a complex number with real and imaginary parts, then how can you say which output is more than the other? The real part of one output may be less than the real part of another output, but the absolute value of one may be more than another because the imaginary part is so much greater. So I'm not sure you can specify a range of the output if you can't say which output is larger than another and you cannot even say yet that its output is complex valued.

11. Aug 17, 2012

DonAntonio

$${}$$
Not at all: the function
$$f(a):=\int_{-\infty}^\infty e^{-ax^2} dx\,\,,\,\,a\in\Bbb C$$

has as domain the complex numbers, even that the integration interval is the real line...

$${}$$
You can't if "the output" (or, mathematically, the function's range) is the complex numbers, as there's no order

in the complex as there is in the reals, say.
Well, then my very first answer's relevant here: a complex number is real iff it equals its (complex, of course) conjugate.

If you can tell apart your function's conjugates then you can tell apart when it is real or not, and then from here, knowing

actually the function, we'll be able, perhaps, to say more things about its absolute value, real or imaginary parts, etc.

DonAntonio

12. Aug 17, 2012

friend

For example, in the following paper,

Origin of Complex Quantum Amplitudes and Feynman's Rules,

The authors detail how if a set of two numbers (a vector) is assigned to functions (which they describe as outcomes resulting from setups of experiments), and if those functions obey the Feynman rules, then those two numbers must form a complex number.

Now, it's relatively easy to show that the self-convolution integral of post 6 below forms the same algebra as the Feynman path integral. Just apply the self-convolution integral an infinite number of times and replace the deltas with the complex gaussian. Then the multiplication of the deltas becomes the addition of the exponents in the gaussian, which results in what looks like an action integral of the Feynman path integral of quantum mechanics. See this reference. There's no speculation here; this is easy math.

Anyway, it occurs to me that this self-convolution integral may be justified for other reasons. And this may lead to quantum mechanics derived from those reasons. So now I'm trying to justify the complex nature of the deltas in the self-convolution integral since those deltas obey the Feynman rules (or algebra). Is compliance with the algebra of the Feynman rules enough to justify the deltas being represented with complex numbers? Or do I also have to show that the deltas are a vector? I do have ideas in this regard if I'm not totally misguided here. Any comments are appreciated.

Last edited: Aug 17, 2012