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本帖最后由 hbghlyj 于 2022-11-2 16:25 编辑 complex.pdf page44
Theorem 7.8. (Cauchy’s Integral Formula) Suppose that $f : U →\Bbb C$ is a holomorphic function on an open set $U$ which contains the disc $\bar B(a, r)$. Then for all $w ∈ B(a, r)$ we have\[
f(w)=\frac{1}{2 \pi i} \int_\gamma \frac{f(z)}{z-w} d z
\]
where $\gamma$ is the path $t \mapsto a+r e^{2 \pi i t}$.
Proof. Fix $w \in B(a, r)$. We use the contours $\Gamma_1$ and $\Gamma_2$ as shown in Figure 3 (where $\Gamma_1$ follows the direction of the blue arrows, and $\Gamma_2$ the directions of the red arrows). These paths join the circular contours $\gamma(a, r)$ and $\gamma(w, \epsilon)^{-}$ where $\epsilon$ is small enough to lie in the interior of $B(a, r)$. By the additivity properties of path integrals, the contributions of the line segments cancel so that
\[
\int_{\Gamma_1} \frac{f(z)}{z-w} d z+\int_{\Gamma_2} \frac{f(z)}{z-w} d z=\int_{\gamma(a, r)} \frac{f(z)}{z-w} d z-\int_{\gamma(w, \epsilon)} \frac{f(z)}{z-w} d z .
\]
On the other hand, each of $\Gamma_1, \Gamma_2$ lies in a primitive domain in which $f /(z-w)$ is holomorphic–indeed by the quotient rule, $f(z) /(z-w)$ is holomorphic on $U \backslash\{w\}$ - so each of the integrals on the left-hand side vanish, and hence
\[
\frac{1}{2 \pi i} \int_{\gamma(a, r)} \frac{f(z)}{z-w} d z=\frac{1}{2 \pi i} \int_{\gamma(w, \epsilon)} \frac{f(z)}{z-w} d z .
\]
Thus we can replace the integral over the circle $\gamma(a, r)$ with an integral over an arbitrary small circle centred at $w$ itself. But for such a small circle,
\[
\begin{aligned}
\frac{1}{2 \pi i} \int_{\gamma(w, \epsilon)} \frac{f(z)}{z-w} d z &=\frac{1}{2 \pi i} \int_{\gamma(w, \epsilon)} \frac{f(z)-f(w)}{z-w} d z+\frac{f(w)}{2 \pi i} \int_{\gamma(w, \epsilon)} \frac{d z}{z-w} . \\
&=\frac{1}{2 \pi i} \int_{\gamma(w, \epsilon)} \frac{f(z)-f(w)}{z-w} d z+f(w) I(\gamma(w, \epsilon), w) \\
&=\frac{1}{2 \pi i} \int_{\gamma(w, \epsilon)} \frac{f(z)-f(w)}{z-w} d z+f(w)
\end{aligned}
\]
But since $f$ is complex differentiable at $z=w$, the term $(f(z)-f(w)) /(z-w)$ is bounded as $\epsilon \rightarrow 0$, so that by the estimation lemma its integral over $\gamma(w, \epsilon)$ tends to zero. Thus as $\epsilon \rightarrow 0$ the path integral around $\gamma(w, \epsilon)$ tends to $f(w)$. But since it is also equal to $(2 \pi i)^{-1} \int_{\gamma(a, r)} f(z) /(z-w) d z$, which is independent of $\epsilon$, we conclude that it must in fact be equal to $f(w)$. The result follows.
Remark 7.9. The same result holds for any oriented curve $\gamma$ once we weight the left-hand side by the winding number ${ }^9$ of a path around the point $w \notin\gamma^*$, provided that $f$ is holomorphic on the inside of $\gamma$.
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${ }^9$ Which, as we used in the proof above, is 1 in the case of a point inside a positively oriented circular path. |
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