What is the connection between complex numbers and vectors in R2?

[Ratzinger]

What is the relationship between complex numbers and vectors in a plane?

I read they have the same mathematical structure. What does that mean and how far does that sameness go?

If the complex numbers are all ordered pair that obey (a,b)+(c,d)=(a+c,b+d) and (a,b)(c,d)=(ac-bd,ad-bc), can we then equate these ordered pairs with 2-dim vectors? I believe not. What does then (a,b)(c,d)=(ac-bd,ad-bc) mean?

Could someone help?

thanks

 

[matt grime]

As a set, the complex numbers are in bijection with R^2. They also carry the same additive structure as R^2, and they behave properly with respect to multiplication by elements of R, so the complex numbers are a 2 dimensional real vector space. They also carry an extra structure, multiplication of complex numbers, that the plane does not naturally carry. What this means is that the complex numbers are an algebra over the reals, 2-dimensional. More than that they are a division algebra, meaning that you can divide elements, ie form inverses under multiplication, and even more than that they are a commutative division algebra, i.e. a field too.

So, think of the complex numbers as a real vector space with extra structure, that of an algebra.

Really, this is the best way to think about complex numbers, if you ask me. No one can deny they exist...

 

[HallsofIvy]

No, they don't have the same "mathematical structure". If we associate the vector $x\vec{i}+ y\vec{j}$ with the complex number x+ iy, then addition and scalar multiplication are the same (so they are "isomorphic" as vector spaces) but the complex numbers have a multiplication that is not defined for vectors in R².

 

[Swapnil]

No, they don't have the same "mathematical structure". If we associate the vector $x\vec{i}+ y\vec{j}$ with the complex number x+ iy, then addition and scalar multiplication are the same (so they are "isomorphic" as vector spaces) but the complex numbers have a multiplication that is not defined for vectors in R².

So is the property of multiplication the *only* thing that separates complex numbers from vectors in R^2?

 

[Swapnil]

So is the property of multiplication the *only* thing that separates complex numbers from vectors in R^2?

Anyone...?

 

[Hurkyl]

So is the property of multiplication the *only* thing that separates complex numbers from vectors in R^2?

Essentially. (The exact answer depends on the fine print of what you mean by R² and C)

 

[HallsofIvy]

In technical terms, there is an "isomorphism" between the vector space, R², and the set of complex numbers with the operations of addition and multiplication by a real number (NOT multiplication by a complex number).