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

Step 1
Let $C_1$ and $C_2$ be two contours such that


 * $\displaystyle \gamma : = C_1 + \left({- C_2}\right)$

, $C_1$ is restricted to domain $\left[{a_1, b_1}\right]$, and $C_2$ is restricted to domain $\left[{a_2 ,  b_2}\right]$.

Then,


 * $\displaystyle C_1 \left({a_1}\right) = C_2 \left({a_2}\right)$

and


 * $\displaystyle C_1 \left({b_1}\right) = C_2 \left({b_2}\right)$

Thus,


 * $\displaystyle \int_{C_1} f \left({z}\right)\ \mathrm d z = \int_{a_1}^{b_1} \dfrac{\ \mathrm d {C_1}}{\ \mathrm d t} f \left({C_1 \left({t}\right)}\right) \ \mathrm d t  = \int{ C_1 \left({a_1}\right)}^{ C_1 \left({b_1}\right)} f \left({C_1}\right) \ \mathrm d {C_1} = \int_{ C_2 \left({a_2}\right)}^{ C_2 \left({b_2}\right)} f \left({C_2}\right) = \int_{a_2}^{b_2} \dfrac{\ \mathrm d {C_2}}{\ \mathrm d t} f \left({C_2 \left({t}\right)}\right) \ \mathrm d t = \int_{C_1} f \left({z}\right)\ \mathrm d z$

Step 2

 * $\displaystyle \oint_\gamma f \left({z}\right) \ \mathrm d z = \oint_{: = C_1 + \left({- C_2}\right)} f \left({z}\right) \ \mathrm d z = \int_{C_1} f \left({z}\right) \ \mathrm d z - \int_{C_2} f \left({z}\right) \ \mathrm d z = \int_{C_1} f \left({z}\right) \ \mathrm d z - \int_{C_1} f \left({z}\right) \ \mathrm d z = 0$

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

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

Now, let $f \left({z}\right) = z^2$. Then,
 * $\displaystyle \oint_\gamma f \left({z}\right) \ \mathrm d z = \int_0^{2 \pi} i e^{i t} e^{2 i t} \ \mathrm d t = i \int_0^{2 \pi} e^{3it} \ \mathrm d t = \frac{i}{3i} \left({e^{6 i \pi} - e^{i 0}}\right) = \frac 1 3 \left({\cos 6 \pi + i \sin 6 \pi - \cos 0 - i \sin 0}\right) = \frac 1 3 \left({1 - 1}\right) = 0$