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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!

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$n(\gamma_k,a)$ 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 $\int\limits_ {\gamma_k}\frac{1}{(z-a)^2}dz=0$.

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  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Mar 11, 2019 at 21:47
  • $\begingroup$ 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 $0$ $\endgroup$
    – user6
    Mar 11, 2019 at 21:51
  • $\begingroup$ image of connected set under continuous mapping is connected but since we are on the integers hence its constant, right? $\endgroup$
    – RFZ
    Mar 11, 2019 at 21:57
  • $\begingroup$ yep.. thats it. $\endgroup$
    – user6
    Mar 11, 2019 at 21:58

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