If $f:\mathbb{C}\to\mathbb{C}$ is holomorphic, is $f':\mathbb{C}\to\mathbb{C}$ holomorphic? Is it true that if $f:\mathbb{C}\to\mathbb{C}$ is holomorphic, then its derivative function is also holomorphic?
How can it be proved? In case this is true.
Thanks.
 A: Since you can use Cauchy's integral formula, consider a $z_0$ in the domain of definition of $f$, and a disk $D_{3r}(z_0) = \{ z : \lvert z - z_0\rvert < 3r\}$ that is contained in the domain. Then, for $z \in D_r(z_0)$, we have the representation
$$f(z) = \frac{1}{2\pi i}\int_{\partial D_{2r}(z_0)} \frac{f(\zeta)}{\zeta - z}\,d\zeta.\tag{1}$$
For $z \in D_r(z_0)$ (actually, $z\in D_{2r}(z_0)$, but let's stay far enough away from the contour to have a uniform bound for the integrand), you can differentiate under the integral by the dominated convergence theorem, and differentiating yields
$$f'(z) = \frac{1}{2\pi i} \int_{\partial D_{2r}(z_0)} \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta,$$
which can again be differentiated under the integral. Etc. ad infinitum.
Alternatively, we can develop the integrand in $(1)$ into a geometric series,
$$\begin{align}
\frac{1}{\zeta - z} &= \frac{1}{\zeta-z_0}\cdot \frac{1}{1- \frac{z-z_0}{\zeta-z_0}}\\
&= \sum_{n=0}^\infty \frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}}.
\end{align}$$
Since the series converges uniformly on the contour, and $f$ is bounded on it, we can interchange integration and summation, so
$$f(z) = \sum_{n=0}^\infty \left(\frac{1}{2\pi i}\int_{\partial D_{2r}(z_0)} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta \right)(z-z_0)^n\tag{2}$$
and the power series $(2)$ can be differentiated arbitrarily often by general considerations about power series.
