# If $i^2=-1$, then what about $(-i)^2=-1$? [duplicate]

By definition, $i^2=-1$, right?

But one can then clearly deduce that $(-i)^2=-1$. The only difference I see is that one is $-1$ times the second.

So what allows us to differentiate between $i$ and $-i$? Can they be used synonymous? That is, does nothing happen if all of a sudden we were to switch $i$ and $-i$ in math, so long as we are consistent?

• we can hold the same argument for 2 and -2 right...?
– user394255
Commented Jan 2, 2017 at 13:55
• I think the point here is that there is no way to distinguish in a purely algebraic way the roots of an irreducible polynomial. Here $i$ and $-i$ are the two roots of the polynomial $x^2 + 1$ which is irreducible over the reals. Complex conjugation (an automorphism of the complex field that fixes each real number) exchanges the two, so that they cannot be distinguished algebraically. Commented Jan 2, 2017 at 13:59
• @A.Molendijk No we cannot, for example $1$ and $-1$ can be differentiated intrinsically by the fact that $1$ solves $x^2=x$ while $-1$ does not. I believe the OP is asking whether a similar intrinsic characterization of $i$ vs $-i$ exists -- and the answer is no (which makes the question interesting).
– Did
Commented Jan 2, 2017 at 14:00
• @barakmanos ?? $(-i)=(-1)\times i$ and $i=(-1)\times(-i)$.
– Did
Commented Jan 2, 2017 at 14:04
• If anyone were to think that this question is trivial, it's worth noting that pretty much this single idea can be considered the basis of Galois theory - we can swap $i$ and $-i$ and all of algebra and arithmetic stays the same, so as soon as we ask "what other permutations of a field can do that?" we enter the automorphism theory. Commented Jan 2, 2017 at 14:16

There is some interesting mathematics bound up in your question. There is an automorphism of $\mathbb C$ as a field extension of $\mathbb R$ which is defined by sending $i$ to $-i$. Provided you are consistent (including very great care with signs) everything algebraic works nicely. This reflects the fact that in building $\mathbb C$ from $\mathbb R$ we make an arbitrary choice of a square root of $-1$ to call $i$.
For amusement there is an automorphism of $\mathbb Q(\sqrt 2)$ as an extension of $\mathbb Q$ which sends $\sqrt 2$ to $-\sqrt 2$ - for much the same reason. However, interesting things happen to the metric (distance between points) - so though the algebraic properties are retained, it is not true that "everything" remains the same.
• Indeed, with automorphisms we often have the metric "messing up". This is not the case with complex automorphism, however. And complex conjugation is very special with this regard - if we consider any two embedding of a number field in $\mathbb C$, then they will give rise to very different metrics, unless they are equal or complex conjugates. Commented Jan 2, 2017 at 14:10