Is a complex continous function of $f$ on a disk around the origen also continous as a function of $\bar{z}$?

Question

I am trying to use the Leibnitz theorem on a continous funktion $f: B \rightarrow \mathds{C}$, with $B$ being an open disk around the origen of Radius $R$, on $f$ as a function $f(z, \bar{z})$. It is given that $f$ is continous on $B$. According to my textbook I can use the Leibnitz theorem on this function. For this to be true, that would mean that $f$ also has to be continous as a function of $\bar{z}$. Why would that be the case?

Answer 1

Think of $z\longmapsto f\big(\overline{z}\big)$ it as a composition of $$ z\longmapsto \overline{z} $$ and $$ z \longmapsto f(z) $$ And use the fact that a composition of continuous functions is a continuous function.