When multiplying by $i$ in inequalities, does the sign flip?

Question

Take $1>-1$, when you multiply by $-1$, to get $-1<1$ you have to flip the sign.

What if you are multiplying by $i$? Would you flip the signs? If you didn't it would net the expression $i>-i$. If you continued that rationale and not flip the signs a second time, you could come to the conclusion that $-1>1$, a clear contradiction. If you were to flip the signs, then from $1>-1$ you would get $i<-i$. Again, if you continue this again, it comes up with the contradiction of $-1>1$. Is it just an undefined inequality? I have no idea.

Answer 1

Well, as far as I understand, you can't really say that one complex number is 'bigger' than another. You could compare the moduli with inequalities, but not the complex numbers themselves.

e.g. $$|1+i|>1$$

Hope I've helped!

Answer 2

Is it just an undefined inequality

Yes, that's it. We do not define inequalities (i.e., order relation) between complex numbers. There is no way to define them in a way consistent with arithmetic operations. For example, if we decided that "$i>0$" or "$i<0$", multiplying both sides by $i$ would give $-1>0$ in either case.

And if we accepted $-1>0$ as true, it would lead deeper down the hole. For one thing, it would imply $(-1)(-1)>0$, hence $1>0$. If both $1>0$ and $-1>0$, we don't have consistency with arithmetics anymore. Adding the inequalities would give $0>0$, and so forth.