# Liberty League International Complex Numbers

1. Jul 17, 2008

### Liberty Leagu

I may be asking a stupid question, but what is the co-relation between the complex plane and the real plane? I know Euler's equation ei\pi+1=0 relates them, but graphically, how are they related?

2. Jul 17, 2008

### tiny-tim

Welcome to PF!

Hi Liberty Leagu ! Welcome to PF!

They're the same!

Except, of course, that you can't multiply points in the real plane.

But addition of comlex numbers is exactly the same as addition of real-plane vectors.

x + iy corresponds to the Cartesian coordinates (x,y) in every way.

And r.e corresponds to the polar coordinates (r,θ) in every way.

3. Jul 17, 2008

### HallsofIvy

As Tiny-tim said, the real plane R2, consists of "points" and there is no arithmetic defined on it- you cannot add or multiply points. You can also think of R2 as the (Euclidean) vector space in which you have addition and scalar multiplication defined but not multiplication of two vectors to give a third vector. In the complex plane C, we have both addition and multiplication of "points" defined and so think of them as numbers.

4. Jul 17, 2008

### chaoseverlasting

5. Jul 17, 2008

### tiny-tim

Hi chaoseverlasting!

Are you and Liberty Leagu the same person … I notice that your original posts were exactly the same?

Have we answered as you wanted?

Your last post (above) is a bit cryptic!

6. Jul 17, 2008

### chaoseverlasting

No. We're not the same person. Which is why I was wondering about this. How are the real plane and complex plane related? Is there any relation?

7. Jul 17, 2008

### qspeechc

BTW, why can we not define inequalities in the complex plane?

8. Jul 17, 2008

### HallsofIvy

You can, it is always possible to assign a linear order to a set, but not in any useful way. The complex numbers cannot be made into an ordered field. An ordered field is a field, together with an order such that if a< b then a+ c< b+ c for any c and, if a< b and 0< c, then ac< bc.

Suppose we were to define an order on the complex numbers. Then, by "trichotomy" we must have exactly one of 0< i or i< 0 or 0= i. Certainly 0 is not equal to i because 02= 0 and i2= -1.

Suppose 0< i. Then 0*i< i*i so 0< -1. That, in itself is not a contradiction, since this is not necessarily our usual order, but from that we must have 0*i< -1*i or 0< -i. If we add i to both sides of that i< 0 which contradicts 0< i.

Suppose i< 0. Then, adding -i to both sides, 0< -i. Now, multiplying both sides of i< 0 by -i, we have -(-1)< 0 or 1< 0. Again, that itself is not a contradiction but multiplying boyh sides by -i gives -i< 0 which contradicts 0< -i.

Since we get a contradiction in every case, such an order is not possible.