$ind_{\Gamma}(0)f(z)=\frac{1}{2\pi i}\int_{\Gamma}\frac{zf(w)}{w(z-w)}dw$ Let $\Gamma$ be a cycle in $\mathbb{C}^*$.
f is a holomorphic function on $\mathbb{C}^*$ and bounded on $\mathbb{C}-B_1(0)$.
Show
$$ind_{\Gamma}(0)f(z)=\frac{1}{2\pi i}\int_{\Gamma}\frac{zf(w)}{w(z-w)}dw$$
for every $z\in\mathbb{C}$ and $z$ Element of the unbounded path-connectedness component.
I know that for $\Gamma$ zero-homologous
$$f(z)=\frac{1}{2\pi i }\int_{\Gamma}\frac{f(w)}{w-t}dw$$
$$ind_{\Gamma}(0)=\int_{\Gamma}\frac{1}{v}dv$$
$$\frac{1}{(w(z-w))}=\frac{1}{z}(\frac{1}{w-z}-\frac{1}{w}).$$
I dont know how to put these together...
I also don't know how to proceed since the cycle doesn't have to be zero homologous.
Any help is greatly appreciated!
(I dont know residue thereom yet)
 A: The idea is to consider Cauchy's integral theorem at infinity point.
Let $F(z) = f(\dfrac{1}{z})$. Then $F(z)$ is bounded at a neighborhood of $0$. It implies that $0$ is a removable singularity point of $F(z)$ via Riemann's theorem. Hence $F(z)$ is holormorphic in $\Bbb{C}$.
Suppose the parameterization of $\Gamma$ is $\Gamma: [a, b] \to \Bbb{C}$. Consider the curve $\Gamma^{\prime}: [a, b] \to \Bbb{C}$ defined by $\Gamma^{\prime} = \pi \circ \Gamma$ where $\pi(z) = \dfrac{1}{z}$.
Then $\textrm{ind}_{\Gamma^{\prime}}(0) = -\textrm{ind}_{\Gamma}(0)$ for:
$$
\begin{aligned}
\textrm{ind}_{\Gamma^{\prime}}(0) &=  \dfrac{1}{2\pi i} \int_{\Gamma^{\prime}} \dfrac{1}{w}\textrm{d}w\\
&\stackrel{w \rightarrow \frac{1}{w}}{=} \dfrac{1}{2\pi i} \int_{\Gamma}w \cdot (-\dfrac{1}{w^2})\textrm{d}w \\
 &= - \dfrac{1}{2\pi i} \int_{\Gamma} \dfrac{1}{w}\textrm{d}w\\
&=  -\textrm{ind}_{\Gamma}(0)
\end{aligned}
$$
Using Cauchy's integral theorem for $F(\dfrac{1}{z})$ and curve $\Gamma^{\prime}$:
$$
\begin{aligned}
\textrm{ind}_{\Gamma^{\prime}}(0) F(\dfrac{1}{z}) &= \dfrac{1}{2\pi i}\int_{\Gamma^{\prime}} \dfrac{F(w)}{w - \dfrac{1}{z}} \textrm{d}w \\
& \stackrel{w \rightarrow \frac{1}{w}}{=} \dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{F(\dfrac{1}{w})}{\dfrac{1}{w} - \dfrac{1}{z}} \cdot (-\dfrac{1}{w^2}) \textrm{d}w \\
&=-\dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{zF(\dfrac{1}{w})}{w(z-w)} \textrm{d}w 
\end{aligned}
$$
Since $\textrm{ind}_{\Gamma^{\prime}}(0) = -\textrm{ind}_{\Gamma}(0)$ and $f(z) = F(\dfrac{1}{z})$, we have
$$
\textrm{ind}_{\Gamma}(0) f(z) = \dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{zf(w)}{w(z-w)} \textrm{d}w 
$$
