Trouble understanding simple Galois Theory example

[PsychonautQQ]

Show that the galois group for (Complex : Reals) is given by {e, y} where y is y: C-->C is the conjugation automorphism defined by y(z) = z~ (Conjugate of z) for all z in C.

if o is an element of gal(C:R) and z = a + bi in C, then o(z) = o(a+bi) = o(a)+o(b)o(i) = a+bo(i)
but o(i)^2 = o(i^2) = o(-1) = -1, so o(i) = i or o(i) = -1.

I am confused on why o(i) can ever equal i... isn't the conjugate of i going to be -1 every time?

 

[jbunniii]

Is there a typo at the end of the second paragarph? Your argument that $o(i)^2 = -1$ is fine, but this implies that $o(i) = i$ or $o(i) = -i$ (not $-1$).

Then:

If $o(i) = i$, then $o$ is the identity, because for an arbitrary element $a+bi \in \mathbb{C}$ we have $o(a+bi) = o(a) + o(b)o(i) = a + bi$.

If $o(i) = -i$, then $o$ is conjugation, because $o(a+bi) = o(a) + o(b)o(i) = a - bi = \overline{a + bi}$.

Since these are the only two possibilities, the Galois group consists of the identity and the conjugation automorphism.