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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 $z_0 \in U$.

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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.

$f^\prime(z)=\frac{1}{\pi i}\int_{|\xi-z_0|=r} \frac{f(\xi)}{(\xi -z)^2}d\xi$.

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