Let $f: \Omega \rightarrow \mathbb{C}$ be an analytic function on $\Omega$. Show that $f′$ is also analytic on $\Omega$.

Question

I'm trying to understand analytic functions in the correct ways before it's too late so that I don't get lost later.

Problem:

Let $\Omega$ be an open set and let $f: \Omega \rightarrow \mathbb{C}$ be an analytic function on $\Omega$. Show that $f′$ is also analytic on $\Omega$.

My (fixed) approach

Let $z_0$ be an arbitrary point in $\Omega$. Since $f$ is analytic at
$z_0$, $\exists S(z)=\sum a_k(z-z_0)^k$ with radius of convergence
$R>0$ s.t $f(z) = S(z)$, $\forall z\in B(z_0,R)$.

Note that, $S(z)$ is holomorphic on $B(z_0,R)$, and analytic functions are holomorphic.

$\Rightarrow f'(z)=S'(z)$, $\forall z\in B(z_0,R)$ where $S'(z)$ has
radius of convergence $R$ and centered at $z_0$.

$\Rightarrow f'$ is analytic on $B(z_0,R)$ by definition.

But, since $z_0$ is an arbitrary point, $f'$ is analytic on $\Omega$. $\blacksquare $

Any comments or corrections are welcomed. Thank you.

Answer 1

I hesitate to call this proof wrong but it reads a bit weirdly to me. The core idea, which you seem to grasp, is that $f'(z)=\frac{d}{dz}\sum a_k(z-z_0)^k$. However you can use this fact to do much more than merely prove the existence of a power series for $f'(z)$, you can use it to explicitly construct one!

The way I would recommend proving this is showing that $f'(z)=\sum_{k=0}^\infty a_{k+1}(z-z_0)^{k}$, and then computing its radius of convergence. You'll find that it is $R$, and then be done.