Prove: if $f(z)$ differentiable in $|z|
Prove: if $f(z)$ differentiable in $|z|<R$  So Does  $\overline {f(\overline{z})}$
I have went over previous similar questions but did not find an answer to the following:
We say that $f(z)$ differentiable in $|z|<R$ therefore
$$u_x=v_y$$ and $$u_y=-v_x$$
Looking at $$\overline {f(\overline{z})}=U(x,-y)-iV(x,-y)$$
We can say that it is defined in $|z|<R$ as $|z|=|\overline{z}|<R$
But looking at its $C-R$ equations we get
$$U_x=--V_y\iff U_x=V_y\iff u_x=v_y$$
$$-U_y=-V_x\neq u_y=-v_x$$
How can we conclude from $C-R$ that $\overline {f(\overline{z})}$ is differentiable?
 A: If $f(z)$ is differentiable in $|z|<R$ (I guess, in the complex sense) then $f(z)$ can be written as a power series
$$ f(z) = a_0 + a_1 z + a_2 z^2 +\ldots $$
uniformly converging to $f(z)$ over any compact subset of $|z|<R$. If you replace $z$ with $\overline{z}$, then conjugate both sides, you get
$$ \overline{f(\overline{z})} = \overline{a_0} + \overline{a_1}z+\overline{a_2} z^2+\ldots $$
and that still is a power series in $z$ that is uniformly converging over any compact subset of $|z|<R$, since the radius of convergence is the same as the previous power series. In particular, $\overline{f(\overline{z})}$ is analytic, hence differentiable, in $|z|<R$.
A: Suppose $f(z)=f(x+yi)=u(x,y)+iv(x,y)$ is differentiable, so that all first-order partial derivatives of $f$ exist and satisfy $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$. Let $g(z) = \overline{f(\overline z)}$. Then $f(\overline z) = f(x-yi) = u(x,-y)+iv(x,-y)$, and so $g(z)=u(x,-y)-iv(x,-y)$ $=U(x,y)+iV(x,y)$. Since the map $z\mapsto \overline z$ is real-differentiable, $g$ is real differentiable, and further
$$U_x(x,y) = \frac{\partial}{\partial x} u(x,-y) = u_x(x,-y) = v_y(x,-y) = \frac{\partial}{\partial y}(-v(x,-y)) = V_y(x,y)$$
$$U_y(x,y) = \frac{\partial}{\partial y}u(x,-y) = -u_y(x,-y) = v_x(x,-y) = -\frac{\partial}{\partial x}(-v(x,-y)) = V_x(x,y)$$
