Equation of a line in complex plane

[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.

 

[tiny-tim]

hi 4everphysics!

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

(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

 

[HallsofIvy]

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.

 

[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.

 

[blue_raver22]

I can provide an explanation for why we don't need to write ax+biy+c=0 before proceeding with the substitution. In the complex plane, the coordinates are represented by a complex number z=x+iy, where x and y are real numbers. This means that when we substitute the values of x and y into the equation ax+by+c=0, we are essentially substituting the complex number z into the equation. Since z is already in the form of x+iy, there is no need to rewrite the equation in the form of ax+biy+c=0. By substituting z into the equation, we are essentially taking into account both the real and imaginary components of the complex number. This is why we can proceed with the substitution without needing to rewrite the equation in the form of ax+biy+c=0.