Cauchy’s integral formula for derivatives

It’s soon exam time, so I’m practicing proofs in complex analysis. Right now that means Cauchy’s integral formula for $$n$$’th derivatives.

Let $$G$$ be a domain of the complex numbers and $$f: G\to \mathbb{C}$$ a holomorphic function. We first want to show that $$f$$ can be expressed as a power series, such that $$f(z)=\sum^\infty_{n=0} a_n(z-a).$$ for some $$a\in\mathbb{C}$$, let $$B_\rho(a)$$ the largest open ball at $$a$$ contained in $$G$$. We claim that $$a_n = \frac{1}{2\pi i} \oint\frac{f(z)}{(z-a)^{n+1}}dz.$$ By the Cauchy integral formula we have that, for a fixed $$z_0\in B_\rho(a)$$ we have $$f(z_0)=\frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0}$$ and by elementary calculations we can, for $$z\in \partial B_r(a)$$, write $$\frac{1}{z-z_0} = \frac{1}{z-a} \frac{1}{1-\frac{z_0 -a}{z-a}}=\frac{1}{z-a}\sum^\infty_{n=0} \left(\frac{z_0-a}{z-a}\right)^n,$$ and from above we then have $$\begin{split} f(z_0)& = \frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0} \\ &=\frac{1}{2\pi i}\oint\sum^\infty_{n=0} \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\ &=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\&=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)}{(z-a)^{n+1}}dz(z_0-a)^n\\ &=\sum^\infty_{n=0}a_n (z_0-a)^n,\end{split}$$ as wanted. We see that $$f$$ is a power series and thus infinitely complex differentiable, and the derivatives are $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint\frac{f(z)}{(z-a)^{n+1}}dz,$$ as desired.

The observant reader will have noticed that I didn’t check that the sums were uniformly convergent, which is needed in other to switch the sum and integral signs, but this is an easy application of the Weierstraß $$M$$-test.

References

Berg, C. (2016). Complex analysis. Copenhagen: Department of Mathematical Sciences, University of Copenhagen.