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:


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

, $C_1$ has domain $\left[{a_1, b_1}\right]$, and $C_2$ has 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:


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

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

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

Now, let $f \left({z}\right) = z^2$. Then,

Also known as
This result is also known as Cauchy's Integral Theorem.