Do positive and negative imaginary numbers work exactly the same?

Question

Can you take any true statement involving complex numbers, replace every instance of $i$ with $-i$ and every instance of $-i$ with $i$, and get another true statement? Obviously you can't do this with the real numbers; the positive reals are closed under multiplication and the negative reals aren't. But if you can do it with $i$, how would you prove this very general proposition? Just in case it isn't clear what I mean, here's an example:

$$\text{"}(\alpha+\beta i)^2=\alpha^2+2\alpha\beta i -\beta^2\text{"} \longmapsto \text{"}(\alpha-\beta i)^2=\alpha^2-2\alpha\beta i -\beta^2\text{"}$$ This is probably a naïve question, but I haven't really dealt with complex numbers at all.

Thanks in advance!

Answer 1

It turns out that this actually works, and this operation, a.k.a. complex conjugation, is an "automorphism," which means it's a symmetry under some operations. It is an automorphism of complex numbers under addition, multiplication, limits, differentiation, integration, and more. This is true since $i$ and $-i$ are both square roots of $-1,$ and we can define the complex numbers using either of those as "the real" $i.$ However, arguments and imaginary parts are totally dependent on what is "the real" $i,$ since they are not algebraic.

Answer 2

Can you take any true statement involving complex numbers, replace every instance of $i$ with $−i$ and every instance of $−i$ with $i$, and get another true statement?

Depends on what you mean by any true statement. The following true statement represents a counter example to your assertion.

For all $~x \in \Bbb{R},~$
if $\theta$ represents the argument to $(x + i),$
then $\sin(\theta) > 0.$