Why does it not matter if we define $i$ as $\sqrt{-1}$ or as $-\sqrt{-1}$?

Question

I'm self learning complex numbers and the definition of complex numbers that I use is as follows:

The set $\mathbb{C}$ of complex numbers is $\mathbb{R}^2$ endowed with some sense of addition and multiplication defined as $(a,b) + (c,d) = (a+c, b+d)$ and $(a,b) \cdot (c,d) = (ac-bd, ad+bc)$ respectively.

Now normally texts like the one I use (Brown Churchill Complex Variables and Applications) just go ahead and define $i$ as $(0,1)$ and thus $i^2 = (-1,0)$ using the definition of multiplication.

While I fully acknowledge that it is a definition and people can define things the way they like, out of curiosity, I'd still like to know what happens if I call $i = (0,-1)$ since $(0,-1)^2 = (-1,0)$ as well.

So far what I've understood by looking up and reading Wikipedia and everything, is that it doesn't matter which square root you choose to label as $i$, which begs another question, how does it not matter? How do I know all the theorems and results would still be equivalently valid? And while looking for an answer, I usually saw people talking about automorphisms and things like that, which, unfortunately, I don't know what they are yet.

And so I was wondering, can anyone just walk me through the "how" part by using simple layman terms, if possible and provide some intuition for it? I'd really appreciate it. Thank you!

Answer 1

Looking at the complex plane $\mathbb{C}$ you see to axes: real and imaginary. Usually the imaginary axis goes upwards. But if you flip it, so that it will go downwards, the complex plane itself will not change. It's like flipping a piece of paper: you go from looking at it from above to looking at it from below. But the piece of paper is the same.

Answer 2

The only fact about $i$ we use to work out the theory of complex numbers is $i^2=-1$. Therefore, no fact in that theory can contradict any choice of $i$ for which $i^2=-1$. Changing the sign of $i$ doesn't change $i^2$, so it can't have detectable consequences.

Answer 3

It wouldn't matter as long as you know which one you're referring to.

In the complex plane $i$ is defined as half the rotation needed to get to $-1$.

The fact though is that you can use two quarter turns or two $3\over4$ turns to get the same result. This is how $x^2=-1$ actually has two solutions according to the fundamental theorem of algebra.

The main takeaway here is that one is the opposite of another. It's trivial knowledge when you consider the negative sign as a flip over. The other thing is that as numbers they're distinct. One causes a $90^\circ$ rotation on $1$ while the other $270^\circ$. In the end, as long as you know that, you can call them whatever you want.

Besides, how would you like a Cartesian plane with the y-axis decreasing upwards?

In your pursuit of self-education:

https://youtu.be/mvmuCPvRoWQ