0

Question: Show that multiplication makes $\mathbb{R}^2$ into a field (the field $\mathbb{C}$ of complex numbers)

I know from another forum (Is $\mathbb R^2$ a field?) that $\mathbb{R}^2$ can be made into the field $\mathbb{C}$ by multiplication by computing $(a,b)(x,y) = (ax-by,ay+bx)$

However, can someone explain how $(a,b)(x,y)$ comes out to that answer and why $\mathbb{C}$ is represented by the answer?

I am a little rusty on the topic as I haven't studied this in a while

Community
  • 1
Lindsey G
  • 629

1 Answers1

5

The way one comes up with the definition $(a, b)(x,y) = (ax-by, ay+bx)$ is by identifying $(a, b)$ with the complex number $a + ib$ (and likewise $(x, y) = x+iy$). Thus, we get $$(a,b)(x,y) = (a +ib)(x + iy) = ax + ibx + iay -by = ax-by + i(ay + bx) = (ax-by, ay+bx)$$ By construction, this multiplication structure on $\Bbb R^2$ identifies it with $\Bbb C$ as a field.

Alex G.
  • 9,178