Cauchy's Integral Formula for Derivatives

Recall from the Cauchy's Integral Formula page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ is analytic on $A$, and $\gamma$ is a simple closed piecewise smooth and positively oriented curve in $A$ and whose inside is in $A$ then for any $z_0$ inside $\gamma$ we have that:

(1)

\begin{align} \quad f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} \: dz \end{align}

We will now state a more general form of this formula known as Cauchy's integral formula for derivatives.

Theorem 1 (Cauchy's Integral Formula for Derivatives): Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ be analytic on $A$. Let $\gamma$ be any simple closed piecewise smooth and positively oriented curve contained in $A$ and such that the inside of $\gamma$ is contained in $A$. Then if $z_0$ is inside $\gamma$ we have that for all $k = 0, 1, 2, ...$ that $\displaystyle{f^{(k)}(z_0) = \frac{k!}{2\pi i} \int_{\gamma} \frac{f(z)}{(z - z_0)^{k+1}} \: dz}$.

• Proof: Let $z_0$ be inside of $\gamma$. Since $f$ is analytic on $A$ we have that $f'(z_0)$ exists and:

(2)

\begin{align} \quad f'(z_0) &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \\ &= \lim_{z \to z_0} \frac{1}{z - z_0} [f(z) - f(z_0)] \end{align}

• Using the regular Cauchy's integral formula on $f(z)$ and $f(z_0)$ gives us:

(3)

\begin{align} \quad f'(z_0) &= \lim_{z \to z_0} \frac{1}{z - z_0} \left [ \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{w - z} \: dw - \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{w - z_0} \: dw \right ] \\ &= \lim_{z \to z_0} \frac{1}{z - z_0} \frac{1}{2\pi i} \int_{\gamma} \left [ \frac{f(w)}{w - z} - \frac{f(w)}{w - z_0} \right ] \: dw \\ &= \lim_{z \to z_0} \frac{1}{z - z_0} \frac{1}{2\pi i} \int_{\gamma} \left [ \frac{f(w)(w - z_0) - f(w)(w - z)}{(w - z)(w - z_0)}\right ] \: dw \\ &= \lim_{z \to z_0} \frac{1}{z - z_0} \frac{1}{2\pi i} \int_{\gamma} \left [ \frac{wf(w) - z_0f(w) - wf(w) + zf(w)}{(w - z)(w - z_0)}\right ] \: dw \\ &= \lim_{z \to z_0} \frac{1}{z - z_0} \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)(z - z_0)}{(w - z)(w - z_0)} \: dw \\ &= \lim_{z \to z_0} \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z)(w - z_0)} \: dw \end{align}

• By uniform convergence we can interchange the limit and integral above to get:

(4)

\begin{align} \quad f'(z_0) &= \frac{1}{2\pi i} \int_{\gamma} \lim_{z \to z_0} \frac{f(w)}{(w - z)(w - z_0)} \: dw \\ &= \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z_0)^2} \: dw \\ &= \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{(z - z_0)^2} \: dz \end{align}

• The rest of the proof for $k = 2, 3, ...$ follows by using induction. $\blacksquare$

Cauchy's integral formula for derivatives gives us an outstanding corollary regarding the existence of higher order derivatives for analytic complex functions.

Corollary 1: If $A \subseteq \mathbb{C}$ is an open and if $f : A \to \mathbb{C}$ is analytic on $A$ then all derivatives of $f$ exist on $A$.

• Proof: Since $f$ is analytic on $A$ and $A$ is open for each $z_0 \in A$ there exists an open disk $D(z_0, r_0) \subseteq A$. Let $C_{r_0} : z_0 + r_0e^{it}$, $t \in [0, 2\pi]$ be the circle centered at $z_0$ with radius $r_0$. Then $C_{r_0}$ is a closed, piecewise smooth and positively oriented curve in $A$ and the inside of $C_{r_0}$ is contained in $A$. So for $z_0$ inside $C_{r_0}$ we have by Cauchy's integral formula for derivatives that for each $k = 0, 1, 2, ...$:

(5)

\begin{align} \quad f^{(k)}(z_0) = \frac{1}{2\pi i} \int_{C_{r_0}} \frac{f(z)}{(z - z_0)^{k+1}} \: dz \end{align}

• So the $k^{\mathrm{th}}$ order derivative of $f$ at $z_0$ exists for each $z_0 \in A$. In other words, all derivatives of $f$ exist on $A$. $\blacksquare$