I have two queries here,
Can $g(z) = f(\bar{z})$ , and $h(z)$ = $\overline{f(z)}$ be analytic in D?
If $g(z)$ and $h(z)$ both are analytic in D then what can be said about $f(z)$ throughout $D$.
My approach was:
I tried using the equations $$\frac{df}{d\bar{z}} = \frac{d\overline{f}} {d\bar{z}} =0 ---- (1) $$
$\frac{dg(z)}{d\bar{z}} \neq 0 $ as $g(z)$ depends on $\bar{z}$ but $\overline{f(z)}$ can be analytic.
Hence $g(\bar{z}) $ cannot be analytic but $\overline{f(z)} $ is analyic if it is a constant.
But this logic contradicts statement of my query.
Please help
Thanks.
Take $g(z)$ to be any analytic function and define $f(z)=g(\bar{z})\ \ $ ($f$ is anti-holomorphic). Then $g(z)=f(\bar{z})$ is analytic by definition. Now the function $h(z)=\overline{f(z)} = \overline{g(\bar{z})} = \overline{g}(z)$ is again analytic. For the definition of the last function you may e.g. think of $ g(z) = \sum_k a_k z^k$ for which $\bar{g}(z)=\sum_k \bar{a_k} z^k$.
If $g$ is holomorphic in $D$ then $h$ is holomorphic in $\overline{D}$ (which equals $D$ if $D$ means the unit disk). So $f$ is anti-holomorphic in the union of $D$ and $\overline{D}$. (and $h$ and $g$ are analytic there).