# Cauchy's Integral formula from Conway's book

Lemma 5.1. Let $$\gamma$$ be a rectifiable curve and suppose $$\varphi$$ is a function defined and continuous on $$\{\gamma\}$$. For each $$m\geq 1$$ let $$F_m(z)=\int \limits_{\gamma}\varphi(w)(w-z)^{-m}dw$$ for $$z\notin \{\gamma\}$$. Then each $$F_m$$ is analytic on $$\mathbb{C}-\{\gamma\}$$ and $$F_m'(z)=mF_{m+1}(z)$$.

Theorem 5.6. Cauchy's Integral Formula. Let $$G$$ be an open subset of the plane and $$f:G\to \mathbb{C}$$ an analytic function. If $$\gamma_1,\dots,\gamma_m$$ are closed rectifiable curves in $$G$$ such that $$n(\gamma_1;w)+\dots+n(\gamma_m;w)=0$$ for all $$w\in \mathbb{C}-G$$, then for $$a\in G-\{\gamma\}$$ $$f(a)\sum \limits_{k=1}^{m}n(\gamma_k;a)=\sum \limits_{k=1}^{m}\dfrac{1}{2\pi i}\int \limits_{\gamma_k}\dfrac{f(z)}{z-a}dz.$$

Theorem 5.8. Let $$G$$ be an open subset of the plane and $$f:G\to \mathbb{C}$$ an analytic function. If $$\gamma_1,\dots,\gamma_m$$ are closed rectifiable curves in $$G$$ such that $$n(\gamma_1;w)+\dots+n(\gamma_m;w)=0$$ for all $$w\in \mathbb{C}-G$$, then for $$a\in G-\{\gamma\}$$ and $$k\geq 1$$ $$f^{(k)}(a)\sum \limits_{j=1}^{m}n(\gamma_j;a)=k!\sum \limits_{j=1}^{m}\dfrac{1}{2\pi i}\int \limits_{\gamma_j}\dfrac{f(z)}{(z-a)^{k+1}}dz.$$

Proof: This follows immediately by differentiating both sides of the formula in Theorem 5.6 and applying Lemma 5.1

I have tried to prove Theorem 5.8 in the following way: let's do it for $$k=1$$. Then:

$$LHS=f'(a)\sum \limits_{k=1}^{m}n(\gamma_k;a)+f(a)\sum \limits_{k=1}^{m}[n(\gamma_k;a)]'$$

$$RHS=\sum \limits_{k=1}^{m}\dfrac{1}{2\pi i} \int \limits_{\gamma_k}\dfrac{f(z)}{(z-a)^2}dz$$

Note that in the RHS I've used Lemma 5.1. Let's use Lemma 5.1 to the second sum in LHS and we get that $$[n(\gamma_k;a)]'=\int \limits_{\gamma_k}\dfrac{dz}{(z-a)^2}$$. Note that $$n(\gamma_k;a)$$ means the winding number of $$\gamma_k$$ around $$a$$.

How to show that the last integral is zero, i.e. $$[n(\gamma_k;a)]'=0$$?

Would be very grateful for any help!

$$n(\gamma_k,a)$$ as a function of $$a$$ is locally constant (since it is a continuous integer valued function) and hence $$\frac{d }{da}n(\gamma_k,a)=0\forall a$$. Alternatively or more analytically u can say $$n'(\gamma_k,a)=\int \limits_{\gamma_k}\dfrac{dz}{(z-a)^2}$$. Now on $$\mathbb C-\{a\}$$ the function $$\frac{1}{(z-a)^2}$$ has a primitive namely $$-\frac{1}{z-a}$$ and hence since $$\gamma_k$$ is a closed curve in $$\mathbb C-\{a\}$$ the fundamental theorem of calculus says $$\int\limits_ {\gamma_k}\frac{1}{(z-a)^2}dz=0$$.
• I know that $n(\gamma_k,a)$ as a function of $a$ is a continuous function and takes integer values. But how it follows that its derivative with respect to $a$ is zero? Could you give more details to this part, please? – ZFR Mar 11 at 21:47
• yes so since it is a continuous function, on an open connected set containing $a$ say a small ball, it is constant (since a cont function from a connected set to $\mathbb Z$ is constant). If a function is constant on a nbhd of $a$ then the derivative at $a$ is $0$ – Ignorant Mathematician Mar 11 at 21:51