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> The complex numbers come about by simply adding one dimension Complex numbers are also best thought of (in my opinion) as an abstract completion of the reals

by j2kun 10y ago

> The complex numbers come about by simply adding one dimension Complex numbers are also best thought of (in my opinion) as an abstract completion of the reals under the operation of taking roots of polynomials. Because otherwise, what would be the difference between the real plane and the complex numbers? They are topologically identical, after all. There has to be something more substantial than simply adding a dimension.

nilkn 10y ago
If by real plane you just mean the standard vector space R^2, the answer is algebraic rather than topological. R^2 as a vector space has an addition operation, but no multiplication operation. The complex plane is obtained by simply picking the appropriate multiplication operator. In general, completing the rationals into the reals is more complex than constructing the complex plane from the real numbers. For the latter, you just need to adjoin a single element (sqrt(-1)), enforce existing arithmetic rules, and the rest falls into place. For the former, you can't just adjoin a single new element like sqrt(2). Doing so will get you the ring (actually field) Q[sqrt(2)], but not R. If you take R and adjoin two special new elements (sqrt(-1) and the point at infinity), you do obtain a topologically different result: the Riemann sphere. This sphere is in many ways the more natural domain for complex analysis than the complex plane.
bzbarsky 10y ago
They're also isomorphic as abelian groups under addition. At least if you assume the axiom of choice. See eg. http://math.stackexchange.com/questions/925706/is-it-true-that-group-of-real-numbers-under-addition-isomorphic-to-group-of-comp/925723#925723 http://math.stackexchange.com/questions/925706/is-it-true-th...
jacobolus 10y ago
Complex numbers are (isomorphic to) a certain subset of linear operators on a 2-dimensional Euclidean vector space which rotate and/or scale the vectors in the plane without skewing or anisotropically stretching them. (We usually also include a zero operator here; depending on use case we sometimes omit it (the “punctured plane”), or sometimes add a point at infinity (the “Riemann sphere”).) If you want you can write them down as matrices acting on vectors in an orthonormal basis by matrix multiplication: [a -b] [b a] Or if you prefer you can consider i to be a unit bivector in the plane, with a complex a + bi acting on vectors by Clifford’s geometric product. Or if you want you can write them using a length and an angle measure, and use high school trigonometry to figure out how they apply to vectors. The difference between the real plane and the complex numbers is that for two vectors in the plane, the product is not a vector. (Indeed, if you use Clifford’s geometric product, then the product of two vectors is a complex number (scalar + bivector).)

by j2kun 10y ago

nilkn 10y ago

If by real plane you just mean the standard vector space R^2, the answer is algebraic rather than topological. R^2 as a vector space has an addition operation, but no multiplication operation. The complex plane is obtained by simply picking the appropriate multiplication operator. In general, completing the rationals into the reals is more complex than constructing the complex plane from the real numbers. For the latter, you just need to adjoin a single element (sqrt(-1)), enforce existing arithmetic rules, and the rest falls into place. For the former, you can't just adjoin a single new element like sqrt(2). Doing so will get you the ring (actually field) Q[sqrt(2)], but not R. If you take R and adjoin two special new elements (sqrt(-1) and the point at infinity), you do obtain a topologically different result: the Riemann sphere. This sphere is in many ways the more natural domain for complex analysis than the complex plane.

bzbarsky 10y ago

They're also isomorphic as abelian groups under addition. At least if you assume the axiom of choice. See eg. http://math.stackexchange.com/questions/925706/is-it-true-that-group-of-real-numbers-under-addition-isomorphic-to-group-of-comp/925723#925723 http://math.stackexchange.com/questions/925706/is-it-true-th...

jacobolus 10y ago

Complex numbers are (isomorphic to) a certain subset of linear operators on a 2-dimensional Euclidean vector space which rotate and/or scale the vectors in the plane without skewing or anisotropically stretching them. (We usually also include a zero operator here; depending on use case we sometimes omit it (the “punctured plane”), or sometimes add a point at infinity (the “Riemann sphere”).) If you want you can write them down as matrices acting on vectors in an orthonormal basis by matrix multiplication: [a -b] [b a] Or if you prefer you can consider i to be a unit bivector in the plane, with a complex a + bi acting on vectors by Clifford’s geometric product. Or if you want you can write them using a length and an angle measure, and use high school trigonometry to figure out how they apply to vectors. The difference between the real plane and the complex numbers is that for two vectors in the plane, the product is not a vector. (Indeed, if you use Clifford’s geometric product, then the product of two vectors is a complex number (scalar + bivector).)