5.4: Proof of Cauchy's integral formula for derivatives

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Recall that Cauchy’s integral formula in Equation 5.3.1 says

\[f^{(n)} (z) = \dfrac{n!}{2 \pi i } \int_C \dfrac{f(w)}{(w - z)^{n + 1}} \ dw, \ \ n = 0, 1, 2, ... \nonumber \]

First we’ll offer a quick proof which captures the reason behind the formula, and then a formal proof.

Quick Proof

We have an integral representation for \(f(z)\), \(z \in A\), we use that to find an integral representation for \(f'(z)\), \(z \in A\).

\[f'(z) = \dfrac{d}{dz} \left[\dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{w - z} \ dw \right] = \dfrac{1}{2\pi i} \int_C \dfrac{d}{dz} \left(\dfrac{f(w)}{w - z}\right)\ dw = \dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{(w - z)^2}\ dw \nonumber \]

(Note, since \(z \in A\) and \(w \in C\), we know that \(w - z \ne 0\)) Thus,

\[f'(z) = \dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{(w - z)^2}\ dw \nonumber \]

Now, by iterating this process, i.e. by mathematical induction, we can show the formula for higher order derivatives.

Formal Proof

We do this by taking the limit of

\[\lim_{\Delta \to 0} \dfrac{f(z + \Delta z) - f(z)}{\Delta z} \nonumber \]

using the integral representation of both terms:

\[f(z + \Delta z) = \dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{w - z - \Delta z}\ dw, \ \ \ \ \ \ f(z) = \dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{w - z}\ dw \nonumber \]

Now, using a little algebraic manipulation we get

\[\begin{align*} \dfrac{f(z + \Delta z) - f(z)}{\Delta z} &= \dfrac{1}{2 \pi i \Delta z} \int_C \dfrac{f(w)}{w - z - \Delta z} - \dfrac{f(w)}{w - z} \ dw \\[4pt] & = \dfrac{1}{2 \pi i \Delta z} \int_C \dfrac{f(w) \Delta z}{(w - z - \Delta z)(w - z)} \ dw \\[4pt] &= \dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{(w - z)^2 - \Delta z (w - z)}\ dw \end{align*} \]

Letting \(\Delta z\) go to 0, we get Cauchy's formula for \(f'(z)\):

\[f' (z) = \dfrac{1}{2\pi i} \int_C \dfrac{f(w)}{(w - z)^2}\ dw \nonumber \]

There is no problem taking the limit under the integral sign because everything is continuous and the denominator is never 0.