Let $f(z)$ and $f(\bar{z})$ be holomorphic in $|z|\leq 1$. Must $f(z)$ be constant in $|z|\leq 1$?
There is a fact that $f(\bar{z})$ is holomorphic if and only if $\overline{f(z)}$ is holomorphic, for any $z\in\mathbb{C}$.
So we have that $f(z)$ and $\bar{f(z)}$ are both holomorphic in $|z|\leq 1$. How to go from here?
Hint If $f(z)$ and $\overline{f(z)}$ are both holomorphic, so are $f(z)+\overline{f(z)}$ and $\frac{f(z)-\overline{f(z)}}{i}$. But both those functions take only real values. Now prove that if $g$ is holomorphic and only takes real values, it is constant.
Alternate solution Write the Cauchy Riemann equations for $f(z)$ and $f(\bar{z})$, and deduce that $f'=0$.
Just write down the Cauchy-Riemann equations (at some point). If they hold for $z$ and $\overline{z}$ you will see that the derivative has to be zero at that point.
Hint:Use Wirtinger's criterion(which derives from C-R equations) that says that $f$ is holomorphic iff $$\frac {df}{d\overline z}(z)=0$$ for every $z$
$$f(z)-f(\bar{z}) $$ is a holomorphic function in the closed unit disk, which vanishes on its real diameter. By the identity theorem we have $$f(z)-f(\bar{z}) = 0 $$ for all $|z| \leq 1$.
Expand $f(z)$ in its taylor series as $$f(z)= \sum_{n=0}^\infty a_n z^n $$ which is valid in the same disk. Writing $z$ in polar form and using the above we find $$f(z)-f(\bar{z})=\sum_{n=0}^\infty a_n (e^{i n \theta}-e^{-i n \theta}) r^n=2i \sum_{n=1}^\infty a_n \sin n \theta \; r^n \equiv 0 .$$ It follows from the uniqueness of the Fourier series that $a_n =0$ for all $n \geq 1$, which reduces $f(z)$ to the constant $a_0$.
