Asymmetry in the complex plane. When $\mathbb{C}$ is viewed as a topological field, it has precisely two automorphisms, namely the identity automorphism, and complex conjugation. Thus we can flip the complex plane upside down without truly changing anything. So, given functions $f,g : \mathbb{C} \rightarrow \mathbb{C}$ such that $f(\bar{z})=g(z),$ I would've thought that $f$ and $g$ were "essentially the same."
However, just because $f$ is complex differentiable, does not mean $g$ will be complex differentiable, and vice versa. So my intuition is telling me that $f$ and $g$ should be essentially the same, when really they're very different.
What's going on?
 A: Note that $h(z)=\bar{z}$ is not complex differentiable. Complex differentiability, geometrically, preserves very small angles when the derivative isn't zero. 
You cam see this by taking some small $\Delta z$ and approximating:
$$f(z+\Delta z)-f(z)\approx\Delta z f'(z)$$
and
$$f(z+e^{i\theta}\Delta z)-f(z) \approx e^{i\theta}{\Delta z}f'(z)$$
$h(z)=\bar{z}$ sends angle $\theta$ to angle $-\theta$ so it cannot be complex-differentiable.
Thus you have to do the above trick several answerers have noted, defining:
$$g(z)=\overline{f(\bar{z})}$$
This does preserve orientation of angles.
Another way to look at it is that knowing that:
$$\lim_{z\to z_0} \frac{f(z_0)-f(z)}{z_0-z}$$
means by symmetry that:
$$\lim_{z\to z_0} \frac{f(\bar{z})-f(\bar z_0)}{\bar z-\bar z_0}$$
exists. If $g(z)=f(\bar z)$, then $g$ holomorphic would mean:
$$\lim_{z\to z_0} \frac{f(\bar{z})-f(\bar z_0)}{ z-z_0}$$
Note the difference in the denominator.
A: I'd like to propose an answer that follows the way you started thinking about it. You noted that there are two automorphisms of $\Bbb{C}$ as a topological field - identity and complex conjugation. Well, great, but if you're going to apply the second automorphism, you need to apply it to the definition of "complex differentiable" as well.
In the definition of the complex derivative, we're dividing by $z - z_0$, which is an operation in the "original" copy of $\Bbb{C}$. In the "conjugated" copy we ought to be dividing by the conjugate of $z - z_0$.
Given a complex function $f$, define the complex conjugate-derivative $f^*$ (note: this notation is nonstandard) by:
$$f^*(z_0) = \lim_{z\to z_0} \frac {f(z) - f(z_0)} {\bar z - \bar z_0}$$
Now observe: Given $g$ analytic and $f(\bar z) = g(z)$, we have:
$$f^*(z_0) = \lim_{z\to z_0} \frac {f(z) - f(z_0)} {\bar z - \bar z_0} = \lim_{z\to z_0} \frac {g(\bar z) - g(\bar z_0)} {\bar z - \bar z_0} = g'(\bar z_0)$$
So indeed, $f$ is the "same" as $g$ and $f^*$ is the "same" as $g'$. (This means $f$ has a derivative $\frac d {d\bar z}$, instead of $\frac d {dz}$.)
A: They should be "the same" if $g(z) = \overline{f(\bar{z})}$, and indeed if this is the case, then $f$ is also differentiable when $g$ is differentiable.
A: If $f$ is differentiable and $f(\mathbb{R})\subseteq\mathbb{R}$ then $f(z)=\sum a_n z^n$ with $a_n\in \mathbb{R}$. Then $\overline{f(\bar{z})}=\overline{\sum a_n\bar{z}^n}=\overline{\sum\overline{a_nz^n}}=\sum a_nz^n=f(z)$.
Without the assumption $f(\mathbb{R})\subseteq \mathbb{R}$ I do not know how much you can say. For example $f(z)=z+i$ then $\overline{f(\bar{z})}=z-i$.
