4 ms·

Could you elaborate on the "tremendous amount of structure" of complex numbers please?

ssivark 5y ago
Think of the following: there are many many paths to get from one location to another on a plane. If you imagine a function on the plane, it’s behavior as you walk along each part would have a sanity constraint. That (along with the constraints of behavior under complex conjugation, should you choose to impose “analyticity”) adds up to a lot of constraints, which is the “structure” in complex numbers — which makes complex analysis simple.
- btilly 5y ago
  This is the right idea. For example if you draw a loop, specify a continuous function on the loop, and say it has to be differentiable inside of that loop, you can prove from the 2-d structure of differentiable functions that the value of the function at every point must be given by a specific integral around that loop. And from that you can start to recover a lot of the structure. Here is a trivial example. The value of a differentiable function inside of the loop has reach its maximum absolute value at the boundary. Now consider a polynomial p(z) which doesn't have 0 as a root. 1/p(z) is differentiable everywhere that p(z) is not zero. If we draw a large enough circle around the origin, the value of 1/p(z) on that boundary is arbitrarily close to 0. But it isn't arbitrarily close to 0 at the origin, and therefore somewhere in that circle it must not be differentiable. And that point is a root of p(z). This is the first proof that I saw that every polynomial over the complex numbers has a root. And it heavily uses the geometric structure of differentiable functions over the complex numbers. Note, that the reals lack this 2-D structure, and it is likewise easy to come up with polynomials over the reals that do not have real roots.
  - hyperpallium2 5y ago
    Sorry, this trivial eg is still way over my head. Is it related to (high school) polnomials always having complex roots, but not necessarily real roots? But that constraint on reals seems like "structure" to me, and complex numbers lack this structure...? (It may well just require a few more years math on ny part.)
    - btilly 5y ago
      Yes. Polynomials over the complex numbers can always be factored completely based on the roots. Polynomials over the real numbers cannot. The result is easy to prove using complex analysis, which uses the 2-D structure of the complex numbers to prove it.

Think of the following: there are many many paths to get from one location to another on a plane. If you imagine a function on the plane, it’s behavior as you walk along each part would have a sanity constraint. That (along with the constraints of behavior under complex conjugation, should you choose to impose “analyticity”) adds up to a lot of constraints, which is the “structure” in complex numbers — which makes complex analysis simple.

btilly 5y ago
This is the right idea. For example if you draw a loop, specify a continuous function on the loop, and say it has to be differentiable inside of that loop, you can prove from the 2-d structure of differentiable functions that the value of the function at every point must be given by a specific integral around that loop. And from that you can start to recover a lot of the structure. Here is a trivial example. The value of a differentiable function inside of the loop has reach its maximum absolute value at the boundary. Now consider a polynomial p(z) which doesn't have 0 as a root. 1/p(z) is differentiable everywhere that p(z) is not zero. If we draw a large enough circle around the origin, the value of 1/p(z) on that boundary is arbitrarily close to 0. But it isn't arbitrarily close to 0 at the origin, and therefore somewhere in that circle it must not be differentiable. And that point is a root of p(z). This is the first proof that I saw that every polynomial over the complex numbers has a root. And it heavily uses the geometric structure of differentiable functions over the complex numbers. Note, that the reals lack this 2-D structure, and it is likewise easy to come up with polynomials over the reals that do not have real roots.
- hyperpallium2 5y ago
  Sorry, this trivial eg is still way over my head. Is it related to (high school) polnomials always having complex roots, but not necessarily real roots? But that constraint on reals seems like "structure" to me, and complex numbers lack this structure...? (It may well just require a few more years math on ny part.)
  - btilly 5y ago
    Yes. Polynomials over the complex numbers can always be factored completely based on the roots. Polynomials over the real numbers cannot. The result is easy to prove using complex analysis, which uses the 2-D structure of the complex numbers to prove it.