Why do we negate the imaginary part when conjugating? For $z=x+iy \in \mathbb C$ we all know the definition for the "conjugate" of $z$, $\bar{z}=x-iy$. Geometrically this is the reflection of $z$ across the $y$ axis.
My question is: couldn't we have defined $\underline{z}=-x+iy$ instead? (this is the reflection across the $x$ axis) Is there anything wrong with it?
I agree that the formula $|z|^2=z \bar{z}$ looks better than $|z|^2=-z \underline{z}$, but are there any more serious problems?
 A: No, the nice thing about conjugation is that it is an automorphism: $\overline{zw} = \bar z\bar w$ and $\overline{z+w}=\bar z+\bar w$.
At heart, what conjugacy shows you is that, while $i$ and $-i$ are different complex numbers, they have exactly the same behavior. That's not surprising, because we define $i$ so that $i^2=-1$, but then $(-i)^2=-1$. How do we distinguish these two square roots? What if we defined the complex numbers as $a+bi+cj$ where $i+j=0$ and $i^2=j^2=-1$? Then which would be the "primary" square root of $-1$? There is really no way to tell. With positive real numbers, we can always pick the positive square root, but we can't define "positive" in the complex numbers.  This yields a duality in the complex numbers, represented by conjugation.
A: If we want complex numbers $z$ and their conjugates $\bar{z}$ to have the property that whenever $z$ is a root of a polynomial $f(t)$ with real coefficients, then so
is $\bar{z}$ a root of the same polynomial $f(t)$, then we need to define 
$\bar{z}$ as $x-iy$. This is because $f(t)$ must be a multiple of
$$q(t) = (t - z)(t-\bar{z}) = t^2 - (z+\bar{z})t + |z|^2$$
and the coefficient of $t$ in $q(t)$ is not real if we use your definition of
conjugate, but is indeed real if the standard definition $\bar{z} = x-iy$ is used.
A: The property of a conjugate is that the argument of the conjugate is the additive inverse of the argument of the original.  If $z$ is 30 degrees clockwise from the real axis, then $\bar z$ is 30 degrees counter-clockwise.  The upshot of this is that when you multiply the numbers together, you get a real number, because the arguments cancel out. (As you likely know, multiplication of complex numbers results in addition of arguments.)
Another nice thing about the conjugate is that it basically goes away when we restrict to the real numbers. The conjugate of a real is just that real. So in some cases, $z\bar z$ is an extension of $z^2$ into the complex numbers. For instance distance from the origin is $\sqrt{z^2}$ if $z$ is real, but we generalize it as $\sqrt{z\bar z}$ in the complex numbers. It "specializes back" to $\sqrt{z^2}$.
Nice, simple things like this break if you try to find some other concept for the complex conjugate.
A: A major drawback to this "lower conjugate" is that it ceases to be a ring involution, since it does not send 1 to 1.
The best properties of the conjugate come about because it interacts with the ring properties this way, and with the addition and multiplication operations. Try to find an example of two complex numbers whose product does not obey the multiplicative rule that the ordinary conjugate does.
(Hint: it would pay off to think simply: $i$ hmmmm...)
A: Consider the equation $z^2 + 1$. As you probably know, $i$ is a root of this equation. In fact, this is the most commonly used definition of $i$: a number such that $i^2=-1$.
But $-i$ is also a root of $z^2+1$, so why not choose this one? Ok. Let us write $i'=-i$. Since $i$ and $i'$ share the same definition, "everything" that is satisfied by one of them has to be satisfied by the other one. For example, since $(1+i)^4 = -4 $, we must have $(1+i')^4 = -4$; this is true!
When I say "everything", I actually mean every polynomial equation. The general principle is the following: if $i$ is a root of some polynomial $P(z)$ with real coefficients, then so $i'=-i$. In the previous example, the polynomial would be $P(z) = (1+z)^4 + 4$.
This principle is more powerful than it looks at first sight. For example, if $a+ib$ (with $a,b \in \Bbb R$) is a root of some polynomial $Q(z)$ with real coefficients, then $i$ is a root of $P(z) = Q(a+zb)$. We deduce that $-i$ is a root of $P(z)$, that is $a-ib$ is a root of $Q(z)$. In the same way, you can see that
$$
(a_1+ib_1) + (a_2+ib_2)=(a_3+ib_3) \iff (a_1-ib_1) + (a_2-ib_2)=(a_3-ib_3)
$$
with $P(z) = (a_1+zb_1) + (a_2+zb_2)-(a_3+zb_3)$, and
$$
(a_1+ib_1)(a_2+ib_2)=(a_3+ib_3) \iff (a_1-ib_1)(a_2-ib_2)=(a_3-ib_3)
$$
(which polynomial would you take?)
The notion of conjugate is just a way to formalize all these "symmetries" between $i$ and $-i$. For instance, the two last equivalences write
$$
\overline{z_1}+ \overline{z_2} = \overline{z_1+ z_2}
,\qquad
\overline{z_1}\times \overline{z_2} = \overline{z_1\times z_2}
$$
Remark: the same kind of things could be done with $\sqrt{2}$ and $-\sqrt{2}$ and polynomial equations with rational coefficients. The general idea behind all of this is is that of Galois Theory.
