Continuity and Differentiability and Analicity Relations between $f(z)$ and $\overline{f(\bar{z})}$ I know that if $f(z)$ is analytic in $D$ then $\overline{f(\bar{z})}$ is analytic in $\overline{D}$.
If $f(z)$ is continuous at $z_{0}$ then $\overline{f(z)}$ is continuos. 
Is it true that  : $f(z)$ is continuous at $z_{0} $$ \Leftrightarrow$ $\overline{f(\bar{z})}$ is continuos ?
and $f(z)$ is differentiable at $z_{0}$ $\Leftrightarrow$ $\overline{f(\bar{z})}$ is differentiable 
This question was in the Final Exam : 
Prove that if $f(z)$ is continuous at a point $z_{0}$ ,then $g(z)=\overline{f(\bar{z})}$ is also continuous .Is it rue that differentiability of $f$ at $z_{0}$ implies differentiability of g ?
Thank You ...
 A: If $f(z)$ is continuous at $z_0$, $\overline{f(\overline z)}$ is not necessarily continuous at $z_0$ :
Let $z_0 = ib \in \Bbb C, b \in \Bbb R \setminus \{0\}$. Consider $f(z) = \frac{1}{z+ib}$ if $z \neq -ib$ and $f(-ib) = \omega \in \Bbb C$. $f(z)$ is continuous at $z = ib = z_0$ but is not continuous at $z = -ib = \overline {z_0}$ because $\lim \limits_{z \rightarrow -ib} \lvert f(z) \rvert = \lim \limits_{z \rightarrow -ib} \lvert\frac{1}{z + ib}\rvert = + \infty$.  Similarly, $\overline{f(z)}$ is not continuous at $\overline {z_0}$ because $\lim \limits_{z \rightarrow -ib} \lvert f(z) \rvert = \lim \limits_{z \rightarrow -ib} \lvert \overline{f(z)} \rvert = + \infty$, i.e. $\overline{f(\overline z)}$ is not continuous at $z_0$ even though $f(z)$ is.
If $f(z)$ is complex-differentiable at $z_0$, then $\overline{f(z)}$ is usually not complex-differentiable at $z_0$ ($\iff$ $\overline{f(\overline z)}$ is usually not complex-differentiable at $\overline{z_0}$). A simple example is $f(z) = z$. It is differentiable but $\overline{f(z)} = \overline z$ is not (look at the Cauchy-Riemann equations to see it).
Obviously, $f(z)$ is continuous at $z_0$ $\iff$ $\overline{f(z)}$ is continuous at $z_0$ ($\iff$ $\overline{f(\overline z)}$ is continuous at $\overline{z_0}$). Indeed, if you write $f(z) = f(x + iy) = u(x, y) + iv(x, y)$, if $f$ is continuous, both $u$ and $v$ are continuous so the function $\overline{f(z)} = u(x, y) - iv(x, y)$ is also continuous and conversely.
Finally, $f(z)$ is complex-differentiable at $z_0$ and on an open neighborhood of $z_0$ $\iff$ $f(z)$ is analytic on a neighborhood of $z_0$ $\iff$ $\overline{f(\overline z)}$ is analytic on a neighborhood of $z_0$ $\iff$ $\overline{f(\overline z)}$ is complex-differentiable at $z_0$ and on an open neighborhood of $z_0$, but you seem to already know that.
A: Take $f(z)=\frac{1}{z+i}$ which is continuous at $z=i$ but $\overline{f(\bar z)}=\frac{1}{\overline{\bar z+i}}=\frac{1}{z-i}$ is not continuous at $z=i$.
Similarly you can check for differentiability of $f(z)$.
However, you may like to know the well-known result $f(z)=\overline{f(\bar z)}$ if $f(x)$ is real. (Reflection principle)
