5 ms·

Honestly, after spending months studying the subject, I don't think it's really possible to "get" complex numbers. I just view them as affine transformations wr

by E6300 10y ago

Honestly, after spending months studying the subject, I don't think it's really possible to "get" complex numbers. I just view them as affine transformations written in an unusual notation. I don't think they make sense as anything but a recontextualization of R^2.

akvadrako 10y ago
So what don't you get? Are you implying there is something more to complex numbers?
- E6300 10y ago
  What I don't get is why someone would bother with complex numbers and their silly notation when linear algebra would work perfectly fine. I know that in certain contexts they are useful. E.g. to avoid losing information when solving polynomials. But that's a minuscule fraction of their range of application. Nearly every practical use I've seen of complex numbers just use them as a vector representation.
  - akvadrako 10y ago
    Well the two uses I'm most familiar with are AC circuit analysis and quantum mechanics. They can both be reformulated without complex numbers of course, since nothing is special about i. Yet the complex versions are a lot easier to work with, because even in manifestly real formulations, the complex structure is still there, but in disguise: - http://www.scottaaronson.com/democritus/lec9.html http://www.scottaaronson.com/democritus/lec9.html - http://physics.stackexchange.com/questions/32422/qm-without-complex-numbers http://physics.stackexchange.com/questions/32422/qm-without-...
jameshart 10y ago
The problem with that is that complex numbers initially emerge as roots of polynomials with real coefficients. Getting to affine transformations from there seems a much bigger leap than asserting there is a square root of -1.
- E6300 10y ago
  The video linked in this post will only make sense if you accept that x e^(theta i) and x k are respectively the rotation part and the scaling part of a linear transformation of x. I'm not aware of any other way to intuitively grasp an expression like e^(pi i).
  - ouid 10y ago
    this is an old comment, but e^z (over C) can be defined as the analytic continuation of e^x (over R). This is a much more interesting definition, since it depends on the fact that that function is unique.

by E6300 10y ago

akvadrako 10y ago

So what don't you get? Are you implying there is something more to complex numbers?

E6300 10y ago
What I don't get is why someone would bother with complex numbers and their silly notation when linear algebra would work perfectly fine. I know that in certain contexts they are useful. E.g. to avoid losing information when solving polynomials. But that's a minuscule fraction of their range of application. Nearly every practical use I've seen of complex numbers just use them as a vector representation.
- akvadrako 10y ago
  Well the two uses I'm most familiar with are AC circuit analysis and quantum mechanics. They can both be reformulated without complex numbers of course, since nothing is special about i. Yet the complex versions are a lot easier to work with, because even in manifestly real formulations, the complex structure is still there, but in disguise: - http://www.scottaaronson.com/democritus/lec9.html http://www.scottaaronson.com/democritus/lec9.html - http://physics.stackexchange.com/questions/32422/qm-without-complex-numbers http://physics.stackexchange.com/questions/32422/qm-without-...

jameshart 10y ago

The problem with that is that complex numbers initially emerge as roots of polynomials with real coefficients. Getting to affine transformations from there seems a much bigger leap than asserting there is a square root of -1.

E6300 10y ago
The video linked in this post will only make sense if you accept that x e^(theta i) and x k are respectively the rotation part and the scaling part of a linear transformation of x. I'm not aware of any other way to intuitively grasp an expression like e^(pi i).
- ouid 10y ago
  this is an old comment, but e^z (over C) can be defined as the analytic continuation of e^x (over R). This is a much more interesting definition, since it depends on the fact that that function is unique.