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Another number line mind-blown moment is that the complex plane is actually a half-plane since the distinction between i and -i is arbitrary, so any graph in th
by bandrami 2y ago
Another number line mind-blown moment is that the complex plane is actually a half-plane since the distinction between i and -i is arbitrary, so any graph in the complex plane has to be symmetric about the real number line.
- red75prime 2y agoAre you sure? The distinction between i and -i is no more arbitrary than the distinction between 1 and -1. Example of an asymmetric graph: Im(x) = 1
- bandrami 2y agoNope. Take any true mathematical sentence and (consistently) replace i with -i and it remains true; that is not the case for 1 and -1. Im(x) = 1 is meaningless; it would have to be Im(x) = ±1. (In fact you'll only ever see complex numbers in the form of a±bi, never a+bi alone.) It's why you can't say e.g. -i < i; the signs on purely imaginary numbers are not an ordering.
- red75prime 2y agof(x) = |x + i| Non-symmetric real-valued function on C.
- bandrami 2y agoAnd indistinguishable from f(x) = |x - i| The choice of one as +i and the other as -i is arbitrary, which is not true with 1 and -1.
- cgriswald 2y agoSeems pretty true with 1 and -1. Map R with f(x) = -x and f(x) = |x - 1| for x in your new mapping is indistinguishable from f(x) = |x + 1| in R. In any case I’d say this is arbitrary like using + for addition and - for subtraction. It seems like you’re just talking about the symbols themselves. I’m not sure how you get to half plane from there.
- xigoi 2y ago1 and -1 are distinguishable: one of them equals its square, the other does not.
- cgriswald 2y agoSure, but I'm not sure I'm understanding the argument. I don't understand how a function like f(z) = e^z has to be symmetric about the real number line or how i and -i aren't distinguishable with something like Im(z) > 0. Is there a proof somewhere I can read?
- bandrami 2y agoIt falls out of complex numbers satisfying the conditions of a field though I don't know of a specific "proof" of that (you generally don't "prove" definitions). You could equally say "i is indistinguishable from 1/i" or "i's additive inverse is its multiplicative inverse"; in either case it's an arbitrary choice which of the conjugates is positive and which is negative. The key being that you cannot say "i > -i" because of that.
- cgriswald 2y agoAh, gotcha. I think I wasn’t understanding exactly what was being said. Thanks for the explanation.
- red75prime 2y agoI interpreted your words as "the complex plane has topology of a plane where conjugates are glued together".
- bandrami 2y agoAh, sorry, no; it is a half plane in the sense that which way is "up" is completely arbitrary
- xigoi 2y agoExpanding the definition of the imaginary part, this says (x - x∗) / 2i = 1 where x∗ denotes the complement. If you replace i with -i, the graph will be precisely the complement of the original graph.
- bubblyworld 2y agoThey are conjugate elements (both roots of the minimal polynomial of C as an extension of R), so they satisfy all the same algebraic properties over R, but they are certainly distinguishable as elements of C. For example "i" satisfies the polynomial "x-i=0" and "-i" doesn't. It's just that you can't find any such polynomial with real coefficients that differentiates them. Of course there are lots of non-algebraic ways to distinguish them too. Or did you mean something stronger?
- bandrami 2y agoI meant that the distinction between i and -i is entirely arbitrary, because it is
- bubblyworld 2y agoThat's not a mathematically meaningful statement. I'm just trying to understand what you mean in more detail.
- bandrami 2y agoAny true mathematical sentence containing i is still true if you (consistently) replace i with -i. It's why complex numbers are always in the form of a±bi, because there isn't anything that distinguishes a+bi from a-bi other than the conventional sign.
- bubblyworld 2y agoOkay, thanks, I see where you're coming from. That's elementary equivalence, which has some caveats. The truth of sentences that only refer to complex numbers and elementary operations on them is preserved by mapping i to -i. But it's not true if you start involving other structures. Complex numbers are generally only in that form when obtained as roots of a polynomial. There are lots of applications where different signs have different interpretations. You can say it's a convention, which is true, but that's not quite the same as saying the two signs are the same thing.