Cauchy derivative formula ($f^{(n)}(w)=\frac{n!}{2\pi i}\oint \frac{f(z)}{(z-w{)}^{n+1}}dz$)

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 satisfiesThe 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. ^[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: $|f^{\prime \prime }(0)|=2$. This is one quantitative reason that holomorphic functions are so rigid.

For a holomorphic function on a neighbourhood of a closed disc, parametrize the Cauchy derivative formula on its boundary. The conjugate Cauchy integral theorem gives $\int \overline{f(r{e}^{i\theta })}{e}^{-i\theta }d\theta =0$. Replacing $f$ in the derivative integral by $f+\bar{f}=2\mathrm{Re}f$ therefore gives the displayed identity. The boundary real part determines every nonconstant analytic coefficient; only an imaginary additive constant is lost.

If $f$ is analytic and bounded by $K$ on $\mathrm{Re}z>0$, then Cauchy's derivative estimate giveson $\mathrm{Re}z>c$. Integrating along line segments gives a Lipschitz bound with the same constant in that smaller half-plane.