Cauchy-Goursat Theorem

This proof is about Cauchy-Goursat Theorem in the context of Complex Analysis. For other uses, see Cauchy's Theorem.

Theorem

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

Let $\partial D$ denote the closed contour bounding $D$.

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

Then:

$\displaystyle \oint_{\partial D} \map f z \rd z = 0$

Proof

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

$f = u + iv$

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

Then we have:

$\displaystyle \oint_{\partial D} \map f z \rd z = \oint_{\partial D} \paren {u + iv} \paren {\d x + i \rd y}$

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

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

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

$\displaystyle \iint_D \paren {-\dfrac {\partial v} {\partial x} - \dfrac {\partial u} {\partial y} } \rd x \rd y + \iint_D \paren {\dfrac {\partial u} {\partial x} - \dfrac {\partial v} {\partial y} } \rd x \rd y$

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.

$\blacksquare$

Example

Let $\map \gamma t = e^{i t}$.

Let the domain of $\gamma$ be $\hointr 0 {2 \pi}$.

Let $\map f z = z^2$.

Then:

\(\displaystyle \oint_\gamma \map f z \rd z\)

\(=\)

\(\displaystyle \int_0^{2 \pi} i e^{i t} e^{2 i t} \rd t\)

\(\displaystyle \)

\(=\)

\(\displaystyle i \int_0^{2 \pi} e^{3 i t} \rd t\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac i {3 i} \paren {e^{6 i \pi} - e^{i 0} }\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac 1 3 \paren {\cos 6 \pi + i \sin 6 \pi - \cos 0 - i \sin 0}\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac 1 3 \paren {1 - 1}\)

\(\displaystyle \)

\(=\)

\(\displaystyle 0\)

$\blacksquare$

Also known as

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

Source of Name

This entry was named for Augustin Louis Cauchy and Édouard Jean-Baptiste Goursat.

Historical Note

The Cauchy-Goursat Theorem was actually first investigated and proved by Carl Friedrich Gauss, but it was just one of the things that he failed to get round to publishing.

He stated it in a letter of $1811$ to his friend Friedrich Wilhelm Bessel, where he says:

This is a very beautiful theorem whose fairly simple proof I will give on a suitable occasion. It is connected with other beautiful truths which are concerned with series expansions.

Augustin Louis Cauchy was, however, the first to publish a proof.

Karl Weierstrass independently discovered this theorem during his exercise to rebuild the theory of complex analysis from first principles.

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Entry: Cauchy's Integral Theorem