Notice that since the curve $\gamma$ is a compact set, for any point $z \in I(\gamma)$ the expression $w-z$ found in the denominator in Cauchy's formula is bounded away from zero, suggesting that we can differentiate the formula with respect to $z$ to obtain $ f^{\prime}(z)=\frac{1}{2 \pi \mathrm{i}} \int_\gamma \frac{f(w)}{(w-z)^2} \mathrm{~d} w $ Of course we need to justify moving the derivative inside the integral sign. We assumed that $f$ was analytic, which means that $f^{\prime}(z)$ exists. The expression above would produce a formula for it, a way to compute it. The key observation is that without assuming that $f$ has more derivatives it seems that the right hand side can be differentiated arbitrarily many times, which would suggest that $f$ has infinitely many derivatives. This is indeed the case as we will show in the next Theorem. > [!NOTE] Theorem > Let $\gamma:[a, b] \rightarrow \mathbb{C}$ be a positively oriented simple closed $C^1$ curve. Assume that $f$ is analytic in $\gamma$ and on the interior of $\gamma, I(\gamma)$. Then $f^{(n)}(z)$ exists for all $n \in \mathbb{N}$ and the derivative is given by $f^{(n)}(z)=\frac{n!}{2 \pi \mathrm{i}} \int_\gamma \frac{f(w)}{(w-z)^{(n+1)}} \mathrm{d} w \quad \text { for all } z \in I(\gamma).$ ###### Proof Notice that [[Cauchy's Integral Formula]] would correspond to the case $n=0$ in the current Theorem. In order to prove the result for $n=1$ we consider the incremental quotient, and use (7.21) to obtain $ \frac{f(z+h)-f(z)}{h}=\frac{1}{h}\left[\frac{1}{2 \pi \mathrm{i}} \int_\gamma \frac{f(w)}{w-z-h} \mathrm{~d} w-\frac{1}{2 \pi \mathrm{i}} \int_\gamma \frac{f(w)}{w-z} \mathrm{~d} w\right] $ By the deformation of contours Theorem we can choose $\gamma$ as $\partial B_{2 r}(z)$, with $B_{2 r}(z) \subset I(\gamma)$. We have, operating on the right-hand side $ \begin{gathered} \frac{f(z+h)-f(z)}{h}=\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} \frac{f(w)}{(w-z-h)(w-z)} \mathrm{d} w \\ =\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} \frac{f(w)}{(w-z)^2} \mathrm{~d} w+\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} f(w)\left[\frac{1}{(w-z-h)(w-z)}-\frac{1}{(w-z)^2}\right] \mathrm{d} w \\ =\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} \frac{f(w)}{(w-z)^2} \mathrm{~d} w+\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)}\left[\frac{h f(w)}{(w-z-h)(w-z)^2}\right] \mathrm{d} w . \end{gathered} $ To conclude the proof all that we need to do is show that the limit of the last integral as $h$ tends to zero is zero, that is (ignoring factors of $2 \pi \mathrm{i}$ ) $ \lim _{h \rightarrow 0} \int_{\partial B_{2 r}(z)}\left[\frac{h f(w)}{(w-z-h)(w-z)^2}\right] \mathrm{d} w=0 $ and recall that we are able to choose $r$ arbitrarily small without affecting the value of the integrals above. First we choose $|h|<r$ so that for all $w \in \partial B_{2 r}(z)$ we have $ |w-z-h| \geqslant w-z|-|h|>r . $ Here we have used the reverse triangle inequality in the first case, and the fact that $\mid w-z=2 r$ for points $w \in \partial B_{2 r}(z)$. Choosing $\gamma(t)=z+2 r \mathrm{e}^{\mathrm{it}}$ for $t \in[0,2 \pi)$, we have $\gamma^{\prime}(t)=2 r \mathrm{ie}^{\mathrm{i} t}$, and therefore $\mid \gamma^{\prime}(t) \leqslant 2 r$. Since $f$ is analytic, in particular it is continuous and therefore there exists $M>0$ such that $f(w) \leqslant M$ for all $w \in \partial B_{2 r}(z)$. Using these facts we have $ \left|\int_{\partial B_{2 r}(z)}\left[\frac{h f(w)}{(w-z-h)(w-z)^2}\right] \mathrm{d} w\right| \leqslant \int_0^{2 \pi} \frac{h M}{r(2 r)^2} 2 r \mathrm{~d} t=\frac{\pi M}{r^2} h $ which goes to zero as $h$ goes to zero, proving the result for $n=1$. The general case is proven by induction. If we assume the result for $n=1,2, \cdots, k-1$ we want to prove it for $n=k$. That is, in particular we assume $ f^{(k-1)}(z)=\frac{(k-1)!}{2 \pi \mathrm{i}} \int_\gamma \frac{f(w)}{(w-z)^{(k)}} \mathrm{d} w \quad \text { for all } z \in I(\gamma) $ We write the corresponding incremental quotient, just as before $ \frac{f^{(k-1)}(z+h)-f^{(k-1)}(z)}{h}=\frac{1}{h}\left[\frac{(k-1)!}{2 \pi \mathrm{i}} \int_\gamma \frac{f(w)}{(w-z-h)^k} \mathrm{~d} w-\frac{(k-1)!}{2 \pi \mathrm{i}} \int_\gamma \frac{f(w)}{(w-z)^k} \mathrm{~d} w\right] $ By the deformation of contours Theorem we can choose $\gamma$ as $\partial B_{2 r}(z)$, with $B_{2 r}(z) \subset I(\gamma)$. We have, operating on the right-hand side $ \begin{gathered} \frac{f^{(k-1)}(z+h)-f^{(k-1)}(z)}{h}=\frac{(k-1)!}{2 \pi \mathrm{i} h} \int_{\partial B_{2 r}(z)} \frac{f(w)\left[(w-z)^k-(w-z-h)^k\right]}{(w-z-h)^k(w-z)^k} \mathrm{~d} w \\ =\frac{k!}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} \frac{f(w)}{(w-z)^{(k+1)} \mathrm{d} w} \\ +\frac{(k-1)!}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} f(w)\left[\frac{\left[(w-z)^k-(w-z-h)^k\right]}{h(w-z-h)^k(w-z)^k}-\frac{k}{(w-z)^{(k+1)}}\right] \mathrm{d} w \\ =\frac{k!}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} \frac{f(w)}{(w-z)^{k+1}} \mathrm{~d} w+\frac{(k-1)!}{2 \pi \mathrm{i}} \int_{\partial B_{2 r}(z)} f(w)\left[\frac{(w-z)^{k+1}-(w-z-h)^k(w-z)-k h(w-z-h)^k}{h(w-z-h)^k(w-z)^{k+1}}\right] \mathrm{d} w . \end{gathered} $ As before, all that remains is to show that the last integral tends to zero as $h$ tends to zero. We choose $h$ and the parametrisation as above. The result will follow if we show that $ \left|\frac{(w-z)^{k+1}-(w-z-h)^k(w-z)-k h(w-z-h)^k}{h}\right| \leqslant C|h| $ where the constant might depend on $r$. This is the case because, as before $|f| \leqslant M$ and $|w-z-h| \geqslant$ $|w-z|-|h|>r$ implies $ \left|\frac{1}{(w-z-h)^k(w-z)^k}\right| \leqslant \frac{1}{(2 r)^k r^k} $ In order to prove (7.23), notice that the binomial formula implies $ (w-z-h)^k=\sum_{j=0}^k\binom{k}{j}(w-z)^{k-j}(-h)^j $ and therefore $\begin{aligned} &\begin{gathered} (w-z)^{k+1}-(w-z-h)^k(w-z)-k h(w-z-h)^k \\ =-\sum_{j=2}^k\binom{k}{j}(w-z)^{k+1-j}(-h)^j-k h \sum_{j=1}^k\binom{k}{j}(w-z)^{k-j}(-h)^j \end{gathered}\\ &\text { which is of order } h^2 \text {, proving the result. } \end{aligned}$