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I am having some issues with the following exercise:

Let $\Gamma$ be a chain in $G=\mathbb{C}^*$ , $f$ be a function that is holomorphic in $G$ and bounded on $\mathbb{C} \setminus K_1(0)$. Show that

$n_{\Gamma}(0) f(z) = \frac{1}{2 \pi i} \int_{\Gamma} \frac{zf(w)}{w(z-w)}dw$

for all $z \in \mathbb{C}$ from the unbounded region of $\mathbb{C} \setminus Tr(\Gamma)$ where $n_{\Gamma}$ is the winding number.

To me this seems like a direct application of a generalized version of Cauchy's integral theorem we proved:

Let $U \subset \mathbb{C}$ be open, $f: U \rightarrow \mathbb{C}$ be holomorphic. Then, for a nullhomologous chain $\Gamma$ in $U$:

$n_{\Gamma}(z)f^{(n)}(z)=\frac{n!}{2 \pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta -z)^{n+1}} d\zeta$

The thing that confuses me is the following: Assume that $0$ is in the unbounded region of $\Gamma + z$. Then, $\Gamma'= \Gamma + z$ is nullhomologous in $G$. Hence, by the above theorem we get:

$\frac{1}{2 \pi i} \int_{\Gamma} \frac{zf(w)}{w(z-w)}dw = \frac{1}{2 \pi i} \int_{\Gamma'} \frac{zf(u-z)}{(u-z)(2z-u)} du = n_{\Gamma'}(z)\frac{z f(z-z)}{2z-z}=n_{\Gamma}(0)f(0)$

But this contradicts the exercise. What is going wrong here?

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  • $\begingroup$ I am not sure about notation, but the claim of the exercise seems strange. I cannot see why both points 0 and z should appear on the left hand side. $\endgroup$ Commented Jul 10 at 12:25

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I think the following might work: Since $f$ is bounded on $\mathbb{C}\setminus K_1(0)$ the function $g(z):= f(1/z)$ has a removable singularity in $0$. Hence $g$ can be extended to an entire function (also denoted by $g$). Now $\widetilde{\Gamma}:= 1/\Gamma$ (for each path in the chain $\Gamma$) is a again a chain in $\mathbb{C}^\ast$. If $z \in \mathbb{C}^\ast$ is in the unbounded region of $\mathbb{C}\setminus Tr(\Gamma)$, then $1/z$ and $0$ are in the same region of $\mathbb{C}\setminus Tr(\widetilde{\Gamma})$ and $n_\widetilde{\Gamma}(0)=-n_\Gamma(0)$. Now $$ n_\Gamma(0)f(z)=-n_\widetilde{\Gamma}(0)f(z)=-n_\widetilde{\Gamma}(0)g(1/z)=-n_\widetilde{\Gamma}(1/z)g(1/z) $$ $$ = -\frac{1}{2\pi i} \int_\widetilde{\Gamma} \frac{g(w)}{w-1/z} dz= \frac{1}{2\pi i} \int_\Gamma \frac{zf(w)}{w(z-w)}dw. $$ The last "$=$" (as well as the equation $n_\widetilde{\Gamma}(0)=-n_\Gamma(0)$) can be checked for each path $\gamma:[a,b] \to \mathbb{C}^\ast $ from the chain $\Gamma$. With $\widetilde{\gamma}:=1/\gamma$ we have $$ -\int_\widetilde{\gamma} \frac{g(w)}{w-1/z} dz =\int_a^b \frac{f(\gamma(t))}{1/\gamma(t)-1/z} \frac{\gamma'(t)}{\gamma(t)^2}dt = \int_a^b \frac{zf(\gamma(t))}{\gamma(t)(z-\gamma(t))}\gamma'(t)dt = \int_\gamma \frac{zf(w)}{w(z-w)}dw. $$

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