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In my book (Serge Lang's Introduction to Complex Analysis at a graduate level) we prove that a function is holomorphic i.f.f it is analytic. My question is, since a holomorphic function is analytic does this imply that because there is a convergent Taylor series expansion at every point on which the function is convergent, that it can be differentiated infinitely many times?

More specifically, say we have $f(z)$ holomorphic on $\mathbb{C}$. This implies that for each point $z_0$ there exists a convergent series such that $f(z)=\sum a_n(z-z_0)^n$. Since $f$ is holomorphic its derivative exists which implies that $f'(z)=\sum n\times a_n (z-z_0)^{n-1}$ converges at any point in $\mathbb{C}$. Now consider $f''(z)=\sum (n)(n-1)a_n(z-z_0)^{n-2}$. Since $f'$ converges this implies that $f'$ has a non zero radius of convergence $K_1$. By the Cauchy-Hamarard formula we have $\lim \frac{1}{(n\times a_n)^\frac{1}{n}}=K_1$ for $f'$. Applying the same formula let $K_2$ be the radius of convergence for $f''$ and we fine that $K_2=\lim \frac{K_1}{(n-1)^\frac{1}{n}}=K_1$. This since $K_1$ is non zero real number it follows that $f''$ is convergent on $\mathbb{C}$. Applying this inductively yields the required result.

Does this work?

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  • $\begingroup$ yes. A holomorphic function is analytic and hence infinitely differentiable (by the way if a power series about some point has a radius of convergence $\rho>0$, then the derived series also has the same radius of convergence). $\endgroup$
    – peek-a-boo
    Jul 30 '20 at 0:05
  • $\begingroup$ @peek-a-boo yes, I think this is what I got as a result when using the cauchy hamarard formula. $\endgroup$ Jul 30 '20 at 0:12
  • $\begingroup$ Oh my bad, I missed the part you showed $K_2=K_1$; I saw you introduced a "different" radius of convergence $K_2$, and just thought I should mention it $\endgroup$
    – peek-a-boo
    Jul 30 '20 at 0:13
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Yes, once you have that a holomorphic function is analytic (the coefficients are given by Cauchy's integral formula), you may proceed by induction to say that a holomorphic function $f(z)$ is infinitely differentiable. You are right that the radius of convergence of $f^{(n)}(z)$ is the same as $f(z)$ for all $n\in \mathbb{N}$. When proving this claim, be sure to note that you can differentiate the power series term by term because the series is uniformly convergent. For the inductive base case, the first derivative's radius of convergence can be computed by the ratio test. Denote the radius of convergence by $ROC$ and let $K$ be the radius of convergence of $f$, then we have the following: $$ROC(f'(z)) = \lim_{n\rightarrow \infty}\frac{(n+1)\cdot a_{n+1}}{n\cdot a_n} = \Big(\lim_{n\rightarrow \infty}\frac{(n+1)}{n}\Big)\Big(\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}\Big)=1\cdot K = K.$$ You should be able to figure out a general formula for the coefficient on the $n$th derivative of the power series, and from there, finish the inductive argument.

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