Flipping sign of $i$s Why do we flip the signs of all $i$ s in a complex number when we want to take the conjugate of it?
I mean, conjugating means making $x + iy$ into $x - iy$, but given a number of the form: $$\frac {x+iy}{x-iy}$$ or $$x+iy+e^{iz}$$ or any other form of complex number, why does flipping signs always work?
It works even when taking the complex conjugate of Schrodinger's wave equation. Is there a reason why any complex number, irrespective of their structure can be conjugated by flipping the sign of all the $i$ s (given that the conjugate exists) ?
 A: Conjugation is an automorphism of the field of complex numbers. This just means that $\overline{z+w} = \overline{z} + \overline{w}$ and $\overline{z\cdot w} = \overline{z} \cdot \overline{w}$. In particular, $\overline{a\cdot z^n} = \overline{a} \cdot {\overline{z}}^n$. Any function expressible as a power series with real coefficients will therefore satisfy $\overline{f(z)} = f(\overline{z})$. In particular, this is the case for $f(z) = \exp z$ since
$$ \exp z = \sum_{n=0}^\infty \frac{z^n}{n!}. $$
The coefficients $1/n!$ are all real, and so
$$ \overline{\exp z} = \sum_{n=0}^\infty \overline{\frac{z^n}{n!}} = \sum_{n=0}^\infty \frac{\overline{z^n}}{n!} = \sum_{n=0}^\infty \frac{{\overline{z}}^n}{n!} = \exp \overline{z}. $$
In particular, $\overline{x+iy+e^{iz}} = \overline{x}-i\overline{y}+e^{-i\overline{z}}$.
A: It's because complex conjugation is an automorphism of $\mathbb{C}$. In other words for all $z, w \in \mathbb{C}$ we have $\overline{zw} = \overline{z} \   \overline{w}$ and $\overline{z + w} = \overline{z} + \overline{w}$. In addition, conjugation fixes the reals pointwise: if $r \in \mathbb{R}$ then $\overline{r} = r$. Your observation follows from those properties.
