# Applying the generalized version of Cauchy's Integral Formula

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?

• 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. Commented Jul 10 at 12:25

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 $$-\int_\widetilde{\gamma} \frac{g(w)}{w-1/z} dz =\int_a^b \frac{f(\gamma(t))}{1/\gamma(t)-1/z} \frac{\gamma'(t)}{\gamma(t)^2}dt = \int_a^b \frac{zf(\gamma(t))}{\gamma(t)(z-\gamma(t))}\gamma'(t)dt = \int_\gamma \frac{zf(w)}{w(z-w)}dw.$$