3.3 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(z) dz = 0
Cauchy's Theorem for Convex Sets
Let's start with a simple case: f(z)=1.
For any closed curve \gamma in U, we have
\int_\gamma f(z) dz = \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 z_1, z_2 \in U, the line segment [z_1, z_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'(z)=f(z) for all z \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(z) dz = \int_\gamma \left[\frac{d}{dz}g(z)\right] dz = g(z_1)-g(z_2)
Since the curve is closed, z_1=z_2, so \int_\gamma f(z) dz = 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(z)=\int_{[z_0,z]}f(\xi)d\xi.
We claim g\in O(U) and g'=f.
Let z_1 close to z, since f is holomorphic on U, using the Goursat's theorem, we can find a triangle T with \xi\in T and z\in T and T\subset U.
\begin{aligned}
0&=\int_{z_0}^{z}f(\xi)d\xi+\int_{z}^{z_1}f(\xi)d\xi\\
&=g(z)-g(z_1)+\int_{z}^{z_1}f(\xi)d\xi+\int_{z_1}^{z_0}f(\xi)d\xi\\
\frac{g(z)-g(z_1)}{z-z_1}&=-\frac{1}{z-z_1}\left(\int_{z}^{z_1}f(\xi)d\xi\right)\\
\frac{g(z_1)-g(z_0)}{z_1-z_0}-f(z_1)&=-\frac{1}{z_1-z_0}\left(\int_{z}^{z_1}f(\xi)d\xi\right)-f(z_1)\\
&=-\frac{1}{z_1-z_0}\left(\int_{z}^{z_1}f(\xi)-f(z_1)d\xi\right)\\
&=I
\end{aligned}
Use the fact that f is holomorphic on U, then f is continuous on U, so \lim_{z\to z_1}f(z)=f(z_1).
There exists a \delta>0 such that |z-z_1|<\delta implies |f(z)-f(z_1)|<\epsilon.
So
|I|\leq\frac{1}{z_1-z_0}\int_{z}^{z_1}|f(\xi)-f(z_1)|d\xi<\frac{\epsilon}{z_1-z_0}\int_{z}^{z_1}d\xi=\epsilon
So I\to 0 as z_1\to z.
Therefore, g'(z_1)=f(z_1) for all z_1\in U.
QED
Cauchy's Theorem for a disk
Let U be the open set, f\in O(U). Let C be a circle inside U and z be a point inside C.
Then
f(z)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-z} d\xi
Proof:
Let C_\epsilon be a circle with center z and radius \epsilon inside C.
Claim:
\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=\int_{C}\frac{f(\xi)d\xi}{\xi-z}
We divide the integral into four parts:
Notice that \frac{f(\xi)}{\xi-z} is holomorphic whenever f(\xi)\in U and \xi\in \mathbb{C}\setminus\{z\}.
So we can apply Cauchy's theorem to the integral on the inside square.
\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=0
Since \frac{1}{2\pi i}\int_{C_\epsilon}\frac{1}{\xi-z}d\xi=1, \sigma=\epsilon e^{it}+z_0 and \sigma'=\epsilon e^{it}, we have
/* TRACK LOST*/
\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-z}=2\pi i f(z)
QED