Fdtxjr

Lemma 5.1. Let $gamma$ be a rectifiable curve and suppose $varphi$ is a function defined and continuous on $gamma$. For
each $mgeq 1$ let $F_m(z)=int
limits_gammavarphi(w)(w-z)^-mdw$ for $znotin gamma$.
Then each $F_m$ is analytic on $mathbbC-gamma$ 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 $f:Gto mathbbC$ 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 $win
mathbbC-G$, then for $ain G-gamma$ $$f(a)sum
limits_k=1^mn(gamma_k;a)=sum limits_k=1^mdfrac12pi
iint limits_gamma_kdfracf(z)z-adz.$$

Theorem 5.8. Let $G$ be an open subset of the plane and $f:Gto mathbbC$ 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 $win mathbbC-G$,
then for $ain G-gamma$ and $kgeq 1$ $$f^(k)(a)sum
limits_j=1^mn(gamma_j;a)=k!sum limits_j=1^mdfrac12pi
iint limits_gamma_jdfracf(z)(z-a)^k+1dz.$$

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:

$$LHS=f'(a)sum
limits_k=1^mn(gamma_k;a)+f(a)sum
limits_k=1^m[n(gamma_k;a)]'$$

$$RHS=sum limits_k=1^mdfrac12pi i int limits_gamma_kdfracf(z)(z-a)^2dz$$

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 $[n(gamma_k;a)]'=int limits_gamma_kdfracdz(z-a)^2$. 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 $fracd dan(gamma_k,a)=0forall a$. Alternatively or more analytically u can say $n'(gamma_k,a)=int limits_gamma_kdfracdz(z-a)^2$. Now on $mathbb C-a$ the function $frac1(z-a)^2$ has a primitive namely $-frac1z-a$ and hence since $gamma_k$ is a closed curve in $mathbb C-a$ the fundamental theorem of calculus says $intlimits_ gamma_kfrac1(z-a)^2dz=0$.