by ActorNightly | 2026-02-11 07:34:26+00:00
>The whole point of inventing the complex numbers was to have a set of numbers for which any polynomial has a root

Think about what this implies.

You have an operation, like exponentiation, that has limits. Something squared can never be negative if you are talking about any real number.

In terms of Sets, you essentially have an operation that produces results only in a finite subset of the overall set. And so the inverse of that operation, when applied to the complement of that finite subset, is undefined.

However you can introduce another (ordered) set in complement to your original set and combine them to form a new set, with operations that define how you move around the values of those sets. So in the case of imaginary numbers, you basically redefine all your reals as "real number + 0 i". And now you have a way to apply that inverse operation to the complement of the finite subset, which means you can get answers to the roots of the polynomial.

And in defining the operation of multiplication, you essentially define a way to move around the 2 dimensional set now. And moving around 2 dimensions is exactly the same thing as rotation+scaling.
And note that when you say sqrt(-1) = i, you basically assume that the complex plane is 2d. There is nothing that is stopping you from making a complex plane 3d or 4d or nd. So sqrt(-1) can also be j, or it can be k. To know what it is, you have to specify the axis of the plane when you specify the sqrt operation, which again, brings it back to the concept of rotations.

And thats my whole point, there is nothing special about i, its simply just a construct that bakes in rotations through any way you wanna define it.

>our whole point about Taylor series is also wrong, as Taylor series are not approximations, they are actually equal to the original function if you take their infinite limit

Looking back at what I wrote, I didn't explain it clearly.

I don't have a problem with Eulers identity, which essentially relies on what you are saying - two analytic functions with equal taylor expansions are equal, and analytic functions are by definition holomorphic when complex.

My contention is that when you take taylor expansion of e^x and just plug in ix where you see x, thats doing A LOT Of heavy lifting in assuming a bunch about how derivatives work. If you don't have the Eulers identity and you try to compute the derivative of e^xi through standard means of the limit equation... you can't because you have no idea what it means to take something to the ith power. The general arbitrary power equation is a^b = exp(bln(a)), which for e^ix just leaves you with...e^ix. Thats it. So why is that you can't compute derivative in the traditional way, but Taylor series essentially somehow magically gives you the values of derivatives of e^ix?

And my argument is that because you can't raise something to the ith power means that i is not a number in the traditional sense, so things like the complex field interpretation of imaginary numbers aren't accurate. I is basically like a negative sign, which is an attribute on a number.

Also, if you think about it, the value of e (2.718...) doesn't ever enter anything that you ever do with complex numbers when you work with them in the form of re^ix. It can be any other number, and the only thing you would care about is r and x. You could replace e with 10, when you are actually doing work with the complex numbers, and nothing would change.