Cauchy-Goursat Theorem

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

Let ${\partial D}$ be the closed contour bounding $D$.

Let $f: D \to \C$ be holomorphic everywhere in $D$.

Then:
 * $\displaystyle \oint_{\partial D} f \left({z}\right) \rd z = 0$

Proof
Begin by rewriting the function $f$ and differential $\rd z$ in terms of their real and complex parts


 * $f = u + iv$


 * $\rd z = \rd x + i \rd y$

Then we have


 * $\displaystyle \oint_{\partial D} f \left({z}\right) \rd z = \displaystyle \oint_{\partial D} (u + iv) (\rd x + i \rd y)$

Expanding the result and again separating into real and complex parts yields two integrals of real variables


 * $\displaystyle \oint_{\partial D} (u \rd x - v \rd y) + i \displaystyle \oint_{\partial D} (v \rd x + u \rd y)$

We next apply Green's Theorem to each integral term to convert the contour integrals to surface integrals over $D$


 * $\iint_D \left( -\dfrac{\partial v}{\partial x} - \dfrac{\partial u}{\partial y} \right) \rd x \rd y + \iint_D \left( \dfrac{\partial u}{\partial x} - \dfrac{\partial v}{\partial y} \right) \rd x \rd y$

By the assumption that $f$ is holomorphic, it satisfies the Cauchy-Riemann Equations


 * $\dfrac{\partial v}{\partial x} + \dfrac{\partial u}{\partial y} = 0$
 * $\dfrac{\partial u}{\partial x} - \dfrac{\partial v}{\partial y} = 0$

The integrands are therefore zero and hence the integral is zero.

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