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