How do I rigorously show $f(z)$ is analytic if and only if $\overline{f(\bar{z})}$ is? I'm doing a bit of self study, but I'm uncomfortable with a certain idea. I want to show that $f(z)$ is analytic if and only if $\overline{f(\bar{z})}$ is analytic, and by analytic I mean differentiable at each point. Here $f$ is a complex valued function.
What I do is write $f(z)=u(x,y)+iv(x,y)$, where $u$ and $v$ are real functions of two variables. Then $\overline{f(\bar{z})}=u(x,-y)-iv(x,-y)$.
These two forms look very similar, in the sense the one function being differentiable should immediately imply the other is differentiable, since the only thing really changing might be a $-$ sign popping out due to the chain rule. 
How can I more rigorously express that $f(z)$ is analytic iff $\overline{f(\bar{z})}$ using this? Many thanks.
 A: Define $g(z)=\overline{f(\overline z)}$.  Then note that $f(z)=\overline{g(\overline z)}$, so there is symmetry allowing only one implication to be shown directly.  Suppose that $f$ is analytic.  Then for all $z$, 
$$\begin{align*}g'(z)&=\lim\limits_{h\to 0}\frac{g(z+h)-g(z)}{h}\\
&=\lim\limits_{h\to 0}\frac{\overline{f(\overline{z+h})}-\overline{f(\overline z)}}{h}\\
&=\lim\limits_{h\to 0}\overline{\left( \frac{f(\overline z+\overline h)-f(\overline z)}{\overline h} \right)    }\\
&=\overline{\left( \lim\limits_{h\to 0}\frac{f(\overline z+\overline h)-f(\overline z)}{\overline h} \right)    }\\
&=\overline{f'(\overline z)}.
\end{align*}$$
That is, $g$ is differentiable, with $g'(z)=\overline{f'(\overline z)}$.  

Alternatively, as other answers have indicated, you could check that the Cauchy-Riemann equations hold for $g$ if they hold for $f$, with $f(x+iy)=u(x,y)+iv(x,y)$ and $g(x+iy)=u(x,-y)+i(-v(x,-y))$ as you indicated.

Another perhaps more conceptual way to think of this is that complex analytic maps are conformal (where their derivatives are nonzero), preserving orientation and angles.  Complex conjugation preserves angles but reverses orientation.  Reversing orientation twice gets you back where you started, so the result is that $g$ is conformal.  (I have given an idea here rather than anything close to a rigorous proof.)

Another approach is to look at power series expansions.  If $f$ has power series expansion in a neighborhood of $\overline{c}$, $\displaystyle{f(z)=\sum\limits_{k=0}^\infty a_k(z-\overline c)^k}$, then in a neighborhood of $c$, $g$ has the power series expansion $\displaystyle{g(z)=\sum\limits_{k=0}^\infty\overline{a_k}(z-c)^k}$.  That is, you just conjugate the coefficients and conjugate the base point for the expansion.  This shows that $g$ is analytic if $f$ is.
A: Hint: Cauchy-Riemann equations.
A: Let $K$ be a Hausdorff topological field, let $f$ be a $K$-valued function defined on an open subset $U$ of $K$, and let $a$ be a point of $U$. 
Say that $f$ is differentiable at $a$ if there is a function $g$ from $U$ to $K$ which is continuous at $a$ and satisfies 
$$
f(z)=f(a)+(z-a)\ g(z)
$$
for all $z$ in $U$. In this case, we write $f'(a):=g(a)$. (One easily checks that this makes sense.)
Assume that this condition holds, and that $\phi$ is an automorphism of $K$ viewed as a topological field. 
Then $z\mapsto\phi(f(\phi^{-1}(z))$ is differentiable at $\phi(a)$, and we have 
$$
(\phi\circ f\circ\phi^{-1})'(\phi(a))=\phi(f'(a)).
$$
The proof is straightforward.
EDIT. Here is the proof. By assumption we have 
$$
f(z)=f(a)+(z-a)\ g(z)\ \ \forall\ z\in U,\quad f'(a)=g(a).
$$
Put 
$$
\widetilde f:=\phi\circ f\circ\phi^{-1},\quad \widetilde g:=\phi\circ g\circ\phi^{-1},\quad 
\widetilde a:=\phi(a),\quad \widetilde z:=\phi(z),\quad \widetilde U:=\phi(U).
$$ 
This yields
$$
\widetilde f(\widetilde z\,)=\widetilde f(\widetilde a\,)+(\widetilde z-\widetilde a\,)\ \widetilde g(\widetilde z\,)\ \ \forall\ \widetilde z\in\widetilde U,\quad\left(\widetilde f\ \right)'(\widetilde a\,)=\widetilde g(\widetilde a\,)=\phi(f'(a)).
$$
This fashion of stating the definition of differentiability is due Carathéodory. See this answer.
A: Use the Cauchy-Riemann equations. Say $\overline{f(\bar{z})}=\alpha(x,y)+i\beta(x,y)$, so that you have
$$\alpha(x,y)=u(x,-y);$$
$$\beta(x,y)=-v(x,-y).$$
Thus
$$\frac{\partial\alpha}{\partial x}=u_x(x,-y),\qquad \frac{\partial\beta}{\partial y}=(-1)\cdot(-v_y(x,-y))=v_y(x,-y);$$
$$\frac{\partial\alpha}{\partial y}=(-1)\cdot u_y(x,-y),\qquad\frac{\partial\beta}{\partial x}= -v_x(x,-y).$$
Show that if one or the other of $\alpha,\beta$ and $u,v$ satisfies CR, so does the other pair.
A: Another way: Morera's Theorem.  Of course you have to figure our how two convert between integral $dz$ and integral $d\overline{z}$.
