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)
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
$$ \textrm{ind}_{\Gamma}(0) f(z) = \dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{zf(w)}{w(z-w)} \textrm{d}w $$