Suppose that \(f: U \to \mathbb{C}\) is a holomorphic function in an open set \(U \subseteq \mathbb{C}\) which is of class \(C^1(U)\) as a function defined on \(U \subseteq \mathbb{R}^2\), without loss of generality we can assume that the ball of radius \(R\) and center \(0\) is contained in \(U\), we can consider for any \(r < R\) the function

\[F(r,\theta) = f(r e^{i \theta}), \qquad r > 0,\ \theta \in [0, 2\pi).\]

The idea is to study the behavior of \(f\) in terms of the Fourier transform of \(F\) in its second variable and use this to prove Cauchy’s theorem. This is a well-understood fact that is not usually noted in a complex analysis course.

First let us note that by Cauchy’s identities \(f\) is holomorphic if and only if \(\overline{\partial} f = 0\) where, \(\overline{\partial} = \partial_x + i \partial_y\) and its derivative is given by \(\partial f = \partial_x f -i \partial_y f\).

Proposition: If \(f\) is infinitely differentiable we have that \(f^{(k)}\) is also an holomorphic function for all \(k\geq 0\).

Proof: Note that \(\overline{\partial} \partial = \partial\overline{\partial}\) so in particular \(\overline{\partial} \partial^k f = \partial^k \overline{\partial}f = 0\).\(\square\)

Using the chain rule

\[\begin{aligned} \partial_x &= \cos(\theta)\partial_r - \frac{1}{r}\sin(\theta) \partial_\theta\\ \partial_y &= \sin(\theta)\partial_r + \frac{1}{r}\cos(\theta) \partial_\theta \end{aligned}\]


\[\overline{\partial} = e^{i\theta}\left( \partial_r + \frac{i}{r} \partial_\theta\right).\]

So in particular for a holomorphic function \(F\), using the identities \(\widehat f' (k) = i\,k\, \widehat f(k)\) and \(\widehat{e^{i\theta} f(\theta)} (k) = \widehat f(k-1)\), we have

\[0 = \widehat {\overline{\partial}F}(r,k+1) = \partial_r \widehat{F} (r,k) - \frac{k}{r} \widehat{F}(r,k).\]

Note that here we need to use the fact that \(f\) is of class \(C^1\) to differentiate under the integral sign. If such \(F\) comes from an holomorphic function then, solving the differential equation given above, we get

\[\widehat{F}(r,k) = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi} F(r,\theta) e^{-i k\theta}\,d\theta= C_k r^k,\]

for some constants \(C_k \in \mathbb{C}\), which are given by

\[C_k = r^{-k} \widehat{F}(r,k) = \frac{1}{\sqrt{2\pi}} r^{-k} \int_0^{2 \pi} f(re^{i k \theta}) e^{- i k \theta} d\theta,\]

Let us compute the Fourier inversion formula for \(F\),

\[F(r,\theta) = \frac{1}{\sqrt{2\pi}} \sum_{k=-\infty}^{\infty} \widehat{F}(r,k) e^{i k \theta} = \frac{1}{\sqrt{2\pi}} \sum_{k=-\infty}^{\infty} C_k r^k e^{i k \theta},\]

and let \(r_0 < R\). If we do the substitution \(\gamma(\theta) = r_0 e^{i\theta}\), we obtain

\[a_k := \frac{1}{\sqrt{2\pi}} C_k = \frac{1}{2 \pi i}\int_{\gamma} \frac{f(w)}{w^{k+1}} dw.\]

Consider the function \(g(z) = \sum_k a_k z^k\) for any \(z\) with \(\vert z\rvert = r < r_0\) then, since

\[\lvert a_k\rvert \leq \frac{C(r)}{r_0^k},\qquad C(r) = \sup_{\lvert z \rvert = r}\lvert f(z)\rvert\]

we have that

\[\sum_k \lvert a_k z^k \rvert < C(r) \sum_k \left(\frac{r}{r_0} \right)^k < \infty,\]

the series for \(g\) converges absolutely, so it is an analytic function and therefore holomorphic. A small computation shows that if \(G(r,\theta) = g(re^{i\theta})\) then for all \(k,r\) we have \(\widehat{F}(r,k) = \widehat{G}(r,k)\), so by uniqueness of the Fourier transform \(f=g\) for all \(r < r_0\), since this analysis was done for arbitrary \(r_0 < R\) we have that

\[f(z) = \sum_k a_k z^k, \qquad \lvert z \rvert < R,\]

where by definition \(a_k = \frac{f^{(k)}(0)}{k!}\) and by our previous analysis we get

Theorem: Let \(f\) be a holomorphic function of class \(C^1\) on a disk of radius \(R > 0\) around \(0\). Then \(f\) is infinitely differentiable and for all \(k \geq 0\) we have for all \(r < R\)

\[f^{(k)}(0) = \frac{k!}{2 \pi i}\int_{\gamma} \frac{f(w)}{w^{k+1}}\,dw,\]

where \(\gamma\) is the closed loop given by \(\gamma(\theta) = re^{i\theta}\).

This analysis shows that in fact holomorphic functions are the same as analytic functions under the mild hypothesis that their first derivative is continuous.

Corollary: Let \(f:U \to \mathbb{C}\) be a holomorphic function of class \(C^1(U)\), then \(f\) is analytic on any ball contained in \(U\).