3.6 KiB
Math 416 Lecture 12
Continue on last class
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.
So
|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
Cauchy's Theorem for a disk
Let U be the open set, f\in O(U). Let C be a circle inside U and \zeta be a point inside C.
Then
f(\zeta)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-\zeta} d\xi
Proof:
Let C_\epsilon be a circle with center \zeta and radius \epsilon inside C.
Claim:
\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_{C}\frac{f(\xi)d\xi}{\xi-\zeta}
We divide the integral into four parts:
Notice that \frac{f(\xi)}{\xi-\zeta} is holomorphic whenever f(\xi)\in U and \xi\in \mathbb{C}\setminus\{\zeta\}.
So we can apply Cauchy's theorem to the integral on the inside square.
\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=0
Since \frac{1}{2\pi i}\int_{C_\epsilon}\frac{1}{\xi-\zeta}d\xi=1, \sigma=\epsilon e^{it}+\zeta_0 and \sigma'=\epsilon e^{it}, we have
/* TRACK LOST*/
\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-\zeta}=2\pi i f(\zeta)
EOP