$\newcommand{\compconj}[1]{% \overline{#1}% }$
Let $f$ be holomorphic, then I can expand it to a uniform convergent power series at the origin $$f(z)=\sum_{k=0}^\infty c_k(z-0)^k$$ I know that $$\bar f(z) = \compconj{\sum_{k=0}^\infty c_kz^k}=\sum_{k=0}^\infty \bar c_k \compconj{z^k}$$
Now I wonder if this series does still converge uniformly? Because I really need to use $$\int\sum_{k=0}^\infty \bar c_k \compconj{z^k} dt = \sum_{k=0}^\infty \int\bar c_k \compconj{z^k} dz$$ but I am not sure if this is legit.
The complex conjugate of a uniformly convergent series is a uniformly convergent series. This is obvious because complex conjugation preserves absolute values.