4.5 KiB
Math416 Lecture 23
Chapter 9: Generalized Cauchy Theorem
Separation lemma
Let \Omega be an open subset in \mathbb{C}, let K\subset \Omega be compact. Then There exists a simple contour \Gamma such that
K\subset \text{int}(\Gamma)\subset \Omega
Corollary 9.9 for separation lemma
Let \Gamma be the contour constructed in the separation lemma. Let f\in O(\Omega) be holomorphic on \Omega. Then \forall z_0\in K such that
f(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{z-z_0}dz
Proof:
Suppose h\in O(G), then \int_{\partial S} h(z)dz=0, by Cauchy's theorem for square, followed from the triangle case.
So \int_{\Gamma} h(z)dz=0=\sum_{j=1}^n \int_{\partial S_j} h(z)dz
Fix z_0\in K,
g(z_0)=\begin{cases}
\frac{f(z)-f(z_0)}{z-z_0} & z\neq z_0 \\
f'(z_0) & z=z_0
\end{cases}
So \int_{\Gamma} g(z)dz=0
Thus
\begin{aligned}
\int_{\Gamma}\frac{f(z)}{z-z_0}dz-\int_{\Gamma}\frac{f(z_0)}{z-z_0}dz&=0 \\
\int_{\Gamma}\frac{f(z)}{z-z_0}dz&=f(z_0)\int_{\Gamma}\frac{1}{z-z_0}dz \\
&=f(z_0)\cdot 2\pi i
\end{aligned}
QED
Theorem 9.10 Cauchy's Theorem
Let \Omega be an open subset in \mathbb{C}, let \Gamma be a contour with int(\Gamma)\subset \Omega. Let f\in O(\Omega) be holomorphic on \Omega. Then
\int_{\Gamma} f(z)dz=0
Proof:
Let K\subset \mathbb{C}\setminus \text{ext}(\Gamma).
By separation lemma, \exists \Gamma_1 s.t. K\subset \text{int}(\Gamma_1)\subset \Omega.
Notice that Separation lemma ensured that w\neq z for all w\in \Gamma_1, z\in \Gamma.
By Corollary 9.9, \forall z\in K, f(z)=\frac{1}{2\pi i}\int_{\Gamma_1}\frac{f(w)}{w-z}dw
\int_{\Gamma} f(z)dz=\frac{1}{2\pi i}\int_{\Gamma}\left[\int_{\Gamma_1}\frac{f(w)}{w-z}dw\right]dz
By Fubini's theorem (In graduate course for analysis),
\begin{aligned}
\int_{\Gamma} f(z)dz&=\frac{1}{2\pi i}\int_{\Gamma_1}\left[\int_{\Gamma}\frac{f(w)}{w-z}dz\right]dw \\
&=\frac{1}{2\pi i}\int_{\Gamma_1}f(w)\left[\int_{\Gamma}\frac{1}{w-z}dz\right]dw \\
&=\frac{1}{2\pi i}\int_{\Gamma_1}f(w)\cdot 2\pi i \ \text{ind}_{\Gamma}(w)dw \\
&=0
\end{aligned}
Since the winding number for \Gamma on w\in \Gamma_1 is 0. (w is outside of \Gamma)
QED
Homotopy
Suppose \gamma_0, \gamma_1 are two curves from
[0,1] to \Omega with same end points P,Q.
A homotopy is a continuous function of curves \gamma_t, 0\leq t\leq 1, deforming \gamma_0 to \gamma_1, keeping the end points fixed.
Formally, if H:[0,1]\times [0,1]\to \Omega is a continuous function satsifying
H(s,0)=\gamma_0(s),\forall s\in [0,1]H(s,1)=\gamma_1(s),\forall s\in [0,1]H(0,t)=P,\forall t\in [0,1]H(1,t)=Q,\forall t\in [0,1]
Then we say H is a homotopy between \gamma_0 and \gamma_1. (If \gamma_0 and \gamma_1 are closed curves, Q=P)
Lemma 9.12 Technical Lemma
Let \phi:[0,1]\times [0,1]\to \mathbb{C}\setminus \{0\} is continuous. Then there exists a continuous map \psi:[0,1]\times [0,1]\to \mathbb{C} such that e^\phi=\psi. Moreover, \psi is unique up to an additive constant in 2\pi i\mathbb{Z}.
Proof:
Let \phi_t(s)=\phi(s,t), 0\leq t\leq 1.
Then \exists \psi_{00} such that e^{\psi_{00}(s)}=\phi(0,t).
\exists \psi_{t}(s) such that e^{\psi_{t}(s)}=\phi_t(s).
We want to show \psi_t(s) is continuous in t.
Since \exists \epsilon>0 such that \phi(s,t) is at least \epsilon away from 0 for all s\in [0,1] and t\in [0,1].
Moreover, \phi(s,t) is uniformly continuous.
So \exists \delta>0 such that |\phi(s,t)-\phi(s,t_0)|<\epsilon if |t-t_0|<\delta.
Therefore,
\begin{aligned}
\left|\frac{\phi(s,t)}{\phi(s,t_0)}-1\right|&<\frac{\epsilon}{\phi(s,t_0)}
&<1
\end{aligned}
So \text {Re} \frac{\phi(s,t)}{\phi(s,t_0)}>0.
Therefore, \text{Log} \frac{\phi(s,t)}{\phi(s,t_0)}=\chi(s,t) is continuous on s\in [0,1], t\in [t_0-\delta, t_0+\delta].
So e^{\chi(s,t)}=\frac{\phi(s,t)}{\phi(s,t_0)}, \chi(s,t_0)=0,\forall s\in [0,1]
Define \tilde{\psi}(s,t)=\chi(s,t)+\chi(s,t_0). So this function is continuous.
And e^{\tilde{\psi}(s,t)}=e^{\chi(s,t)+\chi(s,t_0)}=e^{\chi(s,t)}\cdot e^{\chi(s,t_0)}=\phi(s,t).
\begin{aligned}
\tilde{\psi}(0,t_0)&=\chi(0,t_0)+\psi(0,t_0) \\
&=0+\psi_{00}(t_0) \\
&=\psi_{00}(t_0)
\end{aligned}
\tilde{\psi}(s,0) and \psi(t,0) on t\in[t_0-\delta, t_0+\delta] are both logs of the same function, and agree to each other on t_0.
Therefore, \tilde{\psi}(s,0)=\psi(s,0)+\text{const}
QED