# Cauchy S Integral Theorems for Punctured Domains

Recall from the Cauchy's Integral Theorem for Rectangles page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$, and if $R \subset A$ is a rectangle contained in $A$ and whose interior is contained in $A$, and if $f$ is analytic on $A$ then:

(1)We will now state another version of this theorem which is less restrictive in allowing $f$ to not be analytic at a point $z_0$ (a puncture) in the interior of $R$ while having the conclusion of Cauchy's integral theorem for rectangles hold - provided that $z_0$ satisfies other conditions.

Theorem 1 (Cauchy's Integral Theorem for Rectangles with Punctures 1): Let $A \subseteq \mathbb{C}$ be open, $f : A \to \mathbb{C}$, and let $R \subset A$ be a rectangle in $A$ whose interior is contained in $A$. If $f$ is analytic at on $A$ except at a point $z_0$ which is contained in the interior of $R$ and if $\displaystyle{\lim_{z \to z_0} (z - z_0) \: f(z) = 0}$ exists then $\displaystyle{\int_R f(z) \: dz = 0}$. |

The condition that $\displaystyle{\lim_{z \to z_0} (z - z_0) \: f(z) = 0}$ is equivalent to the following conditions:

- $f$ is continuous (since $\displaystyle{\lim_{z \to z_0} (z - z_0) \: f(z) = 0 \cdot f(z_0) = 0}$).

- $f$ is bounded on some open disk centered at $z_0$.

- $\displaystyle{\lim_{z \to z_0} f(z)}$ exists.

Of course, this result requires that $z_0$ is not on $R$. The following theorem alleviates this necessity.

Theorem 2 (Cauchy's Integral Theorem for Rectangles with Punctures 2): Let $A \subseteq \mathbb{C}$ be open, $f : A \to \mathbb{C}$, and let $R \subset A$ be a rectangle in $A$ whose interior is contained in $A$. If $f$ is analytic at on $A$ except at a point $z_0$ and if $f$ is continuous at $z_0$ then $\displaystyle{\int_{\gamma} f(z) \: dz = 0}$ |

We can also extend the Cauchy-Goursat integral theorem for open disks

Theorem 2 (Cauchy-Goursat Integral Theorem for Open Disks with Punctures): Let $f$ be analytic on the open disk $D(z_0, r) \setminus \{ z_1 \}$ and let $f$ be continuous at $z_1$. Then for any closed, piecewise smooth curve $\gamma$ with $z_1$ contained inside of $\gamma$ we have that $\displaystyle{\int_{\gamma} f(z) \: dz = 0}$. |