# Cauchy's integral formula, application

Let $$f: U \to \mathbb{C}$$ be a holomorphic function and let $$z_0 \in U$$. How can I use Cauchy's integral formula to express the third derivative of $$f$$ in $$z_0$$. I do not see how to show this, but I need to use the statements in (1) and (2). I appreciate any help!

(1) Cauchy's integral formula:

Let $$f$$ be a holomorphic function on the open disc centered in in $$z_0$$ with radius$$\rho$$. Then

• the number $$a_n = \frac{1}{2 \pi r^n} \int_{0}^{2 \pi} f(r e^{it} + z_0) e^{-int} dt$$ doesn't depend on the choice of $$r < \rho$$

• the power series $$\sum a_n z^n$$ has a convergence radius of at least $$\rho$$

• We have equality: $$f(z) = \sum_{n \geq 0} a_n (z-z_0)^n$$ for $$|z-z_0| < \rho$$

(2) Uniqueness of developement of Taylor series:

Every analytic function $$f: U \to \mathbb{C}$$ has an unique developement in a power series in an environment of its points $$z_0 \in U$$.

The Cauchy integral formula says that if $$f$$ is holomorphic in a neighborhood of a point $$z_0$$ in the complex plane then $$f(z)=\frac{1}{2 \pi i}\int_{|\xi-z_0|=r} \frac{f(\xi)}{\xi -z}d\xi$$ for $$z$$ sufficiently close to $$z_0$$.
Now if you want to obtain a formula for $$f^\prime$$ you just integrate both sides of this formula.
$$f^\prime(z)=\frac{1}{\pi i}\int_{|\xi-z_0|=r} \frac{f(\xi)}{(\xi -z)^2}d\xi$$.