Liberty League International Complex Numbers

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
Liberty Leagu
Messages
1
Reaction score
0
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?
 
Physics news on Phys.org
Welcome to PF!

Liberty Leagu said:
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?

Hi Liberty Leagu ! Welcome to PF! :smile:

They're the same!

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

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