= Cauchy derivative formula
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{title2=$f^{(n)}(w)=\frac{n!}{2\pi i}\oint\frac{f(z)}{(z-w)^{n+1}}\,dz$}
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Suppose $f$ is a <holomorphic function> on an <open set> containing the closed disc $|z-a|\le R$, with $R>0$. For any nonnegative integer $n$ and $|w-a|<R$, the counterclockwise boundary <contour> satisfies
$$
f^{(n)}(w)=\frac{n!}{2\pi i}\oint_{|z-a|=R}\frac{f(z)}{(z-w)^{n+1}}\,dz.
$$
The case $n=0$ is the <Cauchy integral formula>. On a smaller closed disc, the denominator stays uniformly away from zero, allowing repeated differentiation under the <contour integral>. This proves that a <holomorphic function> has complex <derivatives> of every order. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/b1a90df19f643b974555bbbb93138a48_MIT18_04S18_topic4.pdf[Orloff, MIT Topic 4, derivative formula]{ref}
At the centre, if $M=\max_{|z-a|=R}|f(z)|$, taking <absolute values> gives the <Cauchy estimate> $|f^{(n)}(a)|\le n!M/R^n$. Boundary size therefore limits interior variation. For $f(z)=z^2$ on the <unit disc>, $M=1$, and the $n=2$ estimate is exact: $|f''(0)|=2$. This is one quantitative reason that <holomorphic functions> are so rigid.
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