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:


 * $z_1 = z_2$

Therefore $F\left({ z_2 }\right) = F\left({ z_1 }\right)$.

Thus by the Fundamental Theorem of Contour Integration we have:


 * $\displaystyle \oint_\gamma f \left({z}\right) \ \mathrm d z = F \left({z_2}\right) - F \left({z_1}\right) = 0$

Example
Let $\gamma \left({t}\right) = e^{i t}$.

Restrict $\gamma$ to the domain $\left[{0, 2 \pi}\right)$.

Now, let $f \left({z}\right) = \frac{1}{z^2}$.
 * $\displaystyle \oint_\gamma f \left({z}\right) \ \mathrm d z = \int_0^{2 \pi} i e^{i t} e^{- 2 i t} \ \mathrm d t = i \int_0^{2 \pi} e^{-it} \ \mathrm d t = \frac{i}{-i} \left({e^{- 2 i \pi} - e^{- i 0}}\right) = - \left({\cos 2 \pi + i \sin 2 \pi - \cos 0 - i \sin 0}\right) = - \left({1 - 1}\right) = 0$