### Higher derivatives of holomorphic functions

Given a continuous function $\varphi$ on a curve $\gamma\subset\mathbb C$, we may consider the function which is defined for $z\notin\gamma$. Note that this strongly resembles the integral considered in the Cauchy integral formula .

Let's start by showing that $F_1$ is a continuous function. If we fix some $z_0\in\mathbb C\backslash \gamma$, we may choose an open ball of some radius $\delta>0$ about $z_0$ that does not intersect $\gamma$, so that $|\zeta-z_0| > \delta$ for all $\zeta\in\gamma$.

Now, let us consider the difference of the two functions $F_1(z)-F_1(z_0)$ for some $z$ inside of $B_{\delta/2}(z_0)$, the ball centered at $z_0$ with half of the aforementioned radius. We may calculate using linearity of the integral that so we have that Now, because $z\in B_{\delta/2}(z_0)$, we have that $|\zeta-z|>\delta/2$ and $|\zeta-z_0|>\delta/2$ for all $\zeta\in\gamma$. This means that which is a uniform bound in $\zeta$, because we have chosen $\delta$ in terms of $z$ such that this bound holds for all $\zeta\in\gamma$. Hence, we have that or This means that we have and therefore not only is $F_1$ continuous, but it is also differentiable with $F_1'=F_2$.

By induction, we may also establish that $F_n'=nF_{n+1}$ for each $n\in\mathbb N$, making each $F_n$ holomorphic on the regions determined by $\gamma$, or the complement of $\gamma$ in $\mathbb C$. An important consequence of this result is the following:

**Proposition 1.** If $f(z)$ is holomorphic, then it has derivatives of all orders, and further where $\gamma$ is a closed curve such that $\mathrm{ind}(\gamma,z)=1$.

*Proof.* By the Cauchy integral formula , we have that so that if we choose $\varphi=f$, we have $F_1=f$, and all further formulae for the derivatives $f^{(n)}$ follow from the fact that $F_n'=nF_{n+1}$ for each $n\in\mathbb N$. $\blacksquare$

# complex-analysis

# complex-integration

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