# Equation of a line in complex plane

1. May 12, 2013

### 4everphysics

All the derivation of the equation of line on complex plane uses the fact that (x,y) \in R^2 can be identified with x+iy \in C.

Thus, they begin with ax+by+c = 0 then re-write x = (z+\bar{z})/2 and y = (z-\bar{z})/(2i), and substitute it into real plane line equation to get it in complex form.

What I don't quite understand is, since (x,y) is identified with (x,iy),

don't we need to write ax+by+c=0 into ax+biy+c=0 before we proceed with the substitution? Why don't we need to do such thing?

Thank you.

2. May 12, 2013

### tiny-tim

hi 4everphysics!
(x,iy) is not in ℂ

ℂ is a set whose elements are single items (traditionally called "z")

ℂ is not a direct product of two sets, with elements that are ordered pairs

3. May 13, 2013

### HallsofIvy

Staff Emeritus
In other words, we are not "identifying (x, y) with (x, iy)". We are identifying the pair of real numbers, (x, y) with the single complex number x+ iy.

4. May 13, 2013

### Bacle2

Still, one thing that I think is good to take into account is that a complex line is 2-dimensional as a real object. Notice that the 1st complex projective space is a 2-sphere, but 1st real projective space is a circle.