Math 416 Lecture 12

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Cauchy's Theorem on triangles

Let T be a triangle in \mathbb{C} and f be holomorphic on T. Then


\int_T f(\zeta) d\zeta = 0

Cauchy's Theorem for Convex Sets

Let's start with a simple case: f(\zeta)=1.

For any closed curve \gamma in U, we have


\int_\gamma f(\zeta) d\zeta = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i

Definition of a convex set

A set U is convex if for any two points \zeta_1, \zeta_2 \in U, the line segment [\zeta_1, \zeta_2] \subset U.

Let O(U) be the set of all holomorphic functions on U.

Definition of primitive

Say f has a primitive on U. If there exists a holomorphic function g on U such that g'(\zeta)=f(\zeta) for all \zeta \in U, then g is called a primitive of f on U.

Cauchy's Theorem for a Convex region

Cauchy's Theorem holds if f has a primitive on a convex region U.


\int_\gamma f(\zeta) d\zeta = \int_\gamma \left[\frac{d}{d\zeta}g(\zeta)\right] d\zeta = g(\zeta_1)-g(\zeta_2)

Since the curve is closed, \zeta_1=\zeta_2, so \int_\gamma f(\zeta) d\zeta = 0.

Proof:

It is sufficient to prove that if U is convex, f is holomorphic on U, then f=g' for some g holomorphic on U.

We pick a point z_0\in U and define g(\zeta)=\int_{[\zeta_0,\zeta]}f(\xi)d\xi.

We claim g\in O(U) and g'=f.

Let \zeta_1 close to \zeta, since f is holomorphic on U, using the Goursat's theorem, we can find a triangle T with \xi\in T and \zeta\in T and T\subset U.


\begin{aligned}
0&=\int_{\zeta_0}^{\zeta}f(\xi)d\xi+\int_{\zeta}^{\zeta_1}f(\xi)d\xi\\
&=g(\zeta)-g(\zeta_1)+\int_{\zeta}^{\zeta_1}f(\xi)d\xi+\int_{\zeta_1}^{\zeta_0}f(\xi)d\xi\\
\frac{g(\zeta)-g(\zeta_1)}{\zeta-\zeta_1}&=-\frac{1}{\zeta-\zeta_1}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)\\
\frac{g(\zeta_1)-g(\zeta_0)}{\zeta_1-\zeta_0}-f(\zeta_1)&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)-f(\zeta_1)\\
&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)-f(\zeta_1)d\xi\right)\\
&=I
\end{aligned}

Use the fact that f is holomorphic on U, then f is continuous on U, so \lim_{\zeta\to\zeta_1}f(\zeta)=f(\zeta_1).

There exists a \delta>0 such that |\zeta-\zeta_1|<\delta implies |f(\zeta)-f(\zeta_1)|<\epsilon.


|I|\leq\frac{1}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}|f(\xi)-f(\zeta_1)|d\xi<\frac{\epsilon}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}d\xi=\epsilon

So I\to 0 as \zeta_1\to\zeta.

Therefore, g'(\zeta_1)=f(\zeta_1) for all \zeta_1\in U.

EOP

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