Is the complex conjugate of a holomorphic function always the function of the conjugate argument?

Question

Imagine that $f(z)$ is holomorphic in some subsect of the complex plane. Is it true that $$ \bar{f}(z)=f(\bar{z}) $$

Answer 1

It is false. Take for example $f(z)=iz$.

Answer 2

The answer is no.

Consider the function $f(z)=e^{iz}$ which is holomorphic.

$$f(z)=e^{iz}=e^{-y+ix}=e^{-y}(\cos x + i \sin x)$$

$$\bar {f(z)}= e^{-y}(\cos x - i \sin x)$$

$$ f(\bar z ) = e^{i(x-iy)}= e^y(\cos x +i\sin x)$$

As you notice they are different.