Cauchy-Goursat Theorem

Theorem
Let $U$ be a simply connected open subset of the complex plane $\C$.

Let $\gamma : \left[{a \,.\,.\, b}\right] \to U$ be a closed contour in $U$.

Let $f: U \to \C$ be holomorphic in $U$.

Then:
 * $\displaystyle \oint_\gamma f \left({z}\right) \ \mathrm d z = 0$

Proof
Let $F \left({z}\right)$ be an antiderivative of $f \left({z}\right)$.

Let $z_1 = \gamma \left({a}\right)$.

Let $z_2 = \gamma \left({b}\right)$.

Since $\gamma$ is a closed contour, we have that:


 * $z_1 = z_2$

Thus:


 * $\displaystyle \oint_\gamma f \left({z}\right) \ \mathrm d z = \int_a^b \frac{\ \mathrm d \gamma \left({t}\right)}{\ \mathrm d t} f \left({\gamma\left({t}\right)}\right) \ \mathrm d t = \int_{\gamma \left({a}\right)}^{\gamma \left({b}\right)} f \left({\gamma}\right) \ \mathrm d \gamma = F \left({z_2}\right) - F \left({z_1}\right) = F \left({z_1}\right) - F \left({z_1}\right) = 0$