Cauchy's Integral Formula
 Theorem 1 (Cauchy's Integral Formula): Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ be analytic on $A$. Let $\gamma$ be a simple closed piecewise smooth and positively oriented curve contained in $A$ and the inside of $\gamma$ is contained in $A$. Then for any $z_0$ inside $\gamma$ we have that $\displaystyle{f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} \: dz}$.
• Proof: Let $\epsilon > 0$ be given. Let $r > 0$ be such that $D(z_0, r) \subseteq A$. Let $C_r : z_0 + re^{it}$, $t \in [0, 2\pi]$ denote the positively oriented circle centered at $z_0$ with radius $r$. Since $\gamma$ is a simple closed piecewise smooth curve contained in $A$ we have that $C_r$ is homotopic to $\gamma$. So by the deformation theorem $\displaystyle{\int_{\gamma} \frac{f(z)}{z - z_0} \: dz = \int_{C_r} \frac{f(z)}{z - z_0} \: dz}$ and:
• Since $f$ is analytic on $A$ we have that $f$ is continuous on $A$. So for this given $\epsilon$ here exists a $\delta > 0$ such that if $\mid z - z_0 \mid < \delta$ then $\mid f(z) - f(z_0) \mid < \epsilon$.
• We want $\mid z_0 + re^{it} - z_0 \mid = \mid re^{it} \mid = r < \delta$, so choose $r < \delta$. Then whenever $\mid z_0 + re^{it} - z_0 \mid < \delta$ we have that:
• Since $\epsilon$ is arbitrary we have that $\displaystyle{f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} \: dz}$. $\blacksquare$