If $f(z)=\overline{f(\overline z)}$ for a holomorphic $f$, is the same true for its derivative? Suppose that $f : \mathbb C \to \mathbb C$ is an entire function which satisfies the relation
$$
f(z)=\overline{f(\overline z)}
$$
for every $z \in \mathbb C$. I'm wondering if the same is true for its complex derivative, i.e. so we have $f'(z)=\overline{f'(\overline z)}$?
 A: If you consider "conjugation" for holomorphic functions defined as
$$\bar f (z) \colon = \overline{f(\bar z)}$$
then
$$(\bar f)' = \overline{f'}$$
that is, conjugation and derivative commute.
So the answer to your question is Yes.
Note: the conjugation does the following to the Taylor series at origin: it turns each coefficient into its conjugate.
A: Suppose $f\in \mathcal{H}({\Bbb{C}})$ with the property that $f(z)=\overline{f(\overline z)}$
$\begin{align}f'(z) &=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}\\&=\lim_{h\to 0}\frac{\overline{f(\overline{z+h)}}-\overline{f(\overline{z})}}{h}\\&=\lim_{h\to 0}\frac{\overline{{f(\overline{z+h)}}-{f(\overline{z})}}}{h}\\&=\lim_{h\to 0}\overline{[\frac{f(\overline{z+h}-f(\overline{z})}{\overline{h}}]}\\&=\overline{\{\lim_{\overline{h}\to 0}\frac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}}\}}\\&=\overline{f'(\overline{z})}\end{align}$
Note : $z\to \overline{z}$ is continuous.
A: Yes, you can compute this directly, for example using the quotient definition: if you let $g(z) = f(z^*)^*$ (I use the star as opposed to the bar since it reads better here) then you can check directly that
$$\frac{g(z+h) - g(z)} h \quad\text{ is conjugate to }
\quad \frac{f(z^*+h^*)-f(z^*)}{h^*}$$
Since $h\to 0$ is equivalent to $h^*\to 0$, it follows that $g'(z)= f'(z^*)^*$.
A: Will this work?
The gist of the proof below is that the relation $f(z) = \overline{f(\bar z)}$ forces every coefficient in the power (Taylor) series of $f(z)$ about $0$ to be real, which then carries over to $f'(z)$ forcing $f'(z) = \overline{f'(\bar z)}$.
If
$f(z) = \displaystyle \sum_0^\infty a_n z^n, \tag 1$
then
$f(\bar z) = \displaystyle \sum_0^\infty a_n \bar z^n, \tag 2$
whence
$\overline{f(\bar z)} = \displaystyle \sum_0^\infty \bar a_n \bar {\bar z}^n = \sum_0^\infty \bar a_n z^n; \tag 3$
now given that
$f(z) = \overline{f(\bar z)}, \tag 4$
we have
$\displaystyle \sum_0^\infty a_n z^n = \sum_0^\infty \bar a_n z^n; \tag 5$
with
$z = 0 \tag 6$
this yields
$a_0 = \bar a_0, \tag 7$
and hence
$\displaystyle \sum_1^\infty a_n z^n = \sum_1^\infty \bar a_n z^n; \tag 8$
differentiating this we obtain
$\displaystyle \sum_1^\infty na_n z^{n - 1} = \sum_1^\infty n\bar a_n z^{n - 1}, \tag 9$
or
$a_1 + \displaystyle \sum_2^\infty na_n z^{n - 1} = \bar a_1 + \sum_2^\infty n\bar a_n z^{n - 1}; \tag{10}$
again we invoke (6) and find that
$a_1 = \bar a_1, \tag{11}$
and thus that
$a_0 + a_1z = \bar a_0 + \bar a_1 z, \tag{12}$
from which, in light of (5),
$\displaystyle \sum_2^\infty a_n z^{n - 1} = \sum_2^\infty \bar a_n z^{n - 1}; \tag{13}$
we furthermore see that $a_0$ and $a_1$ are both real. Continuing in this vein, alternating differentiation with the invocation of (6), we are able to conclude that
$a_n = \bar a_n \tag{14}$
for every $n$.  Now
$f'(\bar z) = \displaystyle \sum_1^\infty n a_n \bar z^n, \tag{15}$
and thus
$\overline{f'(\bar z)} = \displaystyle \sum_1^\infty n \bar a_n \bar {\bar z}^n = \sum_1^\infty n a_n z^n = f'(z). \tag{16}$
