Am I correct in my understanding of the importance of homotopy in my proof? I want to prove that given that $\gamma_0,\gamma_1$ are closed rectifiable curves in a region G with $\gamma_0 \sim \gamma_1\ $(i.e. homotopic) in G and $w \in \Bbb C- \{G\}$.
Then $n(\gamma_0,w)=n(\gamma_1,w).$
Okay so here's my attempt :
First of all we are trying to show that $n(\gamma_0,w)=n(\gamma_1,w). \Rightarrow \tfrac{1}{2\pi i}\int_{\gamma_0}\tfrac{dz}{z-w}=\tfrac{1}{2\pi i}\int_{\gamma_1}\tfrac{dz}{z-w}$
As w lies outside of $\gamma_1,\gamma_0$ then $\tfrac{1}{z-w}$ is holomorphic on a disk which includes $\gamma_1,\gamma_0$ so by Cauchy's theorem 
$$\int_{\gamma_1}\tfrac{dz}{z-w}=0=\int_{\gamma_0}\tfrac{dz}{z-w}$$
This implies that $n(\gamma_0,w)=n(\gamma_1,w).$
Here's my trouble :
I'm not sure I understand why the fact that the two curves are homotopic matters.Here's what I think the reason is but I'm not entirely sure.
As we didn't specify what type of region G was lets say that its an annulus. lets also say that w is in the centre of this annulus. Then the winding number of both curves around the point w is no longer zero infact now they can be different. but then we dont neccarily have it that $\int_{\gamma_0}f(z)dz=\int_{\gamma_1}f(z)dz$ even though f is holomorphic in G. and so we can assume that the two curves are not homotopic.
I'm not sure if this reasoning is correct ,if it is the correct notion I think it's not explained very formally as it were . 
So I have three questions:
i)Is the reasoning in my attempt section correct ?
ii)Is my reasoning as to why homotopy is important here correct ?
iii) Is there any suggestions ( If I am correct in my reasoning behind the importance of homotopy ) as to how I can present it more formally ?
 A: (i) $\ $ Not exactly. As your example in (ii) with $G$ an annulus shows, we don't necessarily have a disk containing $G$ but excluding $w$. 
$\ $ At (ii) you correctly write that $\int_{\gamma_1}\frac{dz}{z-w}$ is not necessarily zero.
(ii) $\ $ Well, if $\gamma_1$ and $\gamma_2$ are not homotopic, then, yes, their winding number might be different. 
$\ $ So, being homotopic is an important condition.
(iii) $\ $ Cut both curves e.g. at their starting points, and connect these two points by a path $p$ within $G$.
$\ $ Consider then the closed curve $\gamma:=\gamma_1p\gamma_2^\smallsmile p^\smallsmile$, where $u^\smallsmile$ denotes the reversed path of the path $u$.
$\ $ Show that $\gamma$ is null homotopic within $G$.
$\ $ Therefore, as $w\notin G$, we have $\int_\gamma\frac{dz}{z-w}=0$.
$\ $ Deduce then $\int_{\gamma_1}\frac{dz}{z-w}=\int_{\gamma_2}\frac{dz}{z-w}$.
A: Let $\gamma_1,\gamma_2$ be two closed rectifiable curves in a region G with $\gamma_0$ homotopic to $\gamma_1$ in G. 
$\Rightarrow \int_{\gamma_0}f(z)dz=\int_{\gamma_1}f(z)dz$. this is from cauchy's theorem for open sets. (It didn't explicitly say G was open but I used it anyway .This may be a mistake )
If f is holomorphic on .
Consider now 
$$n(\gamma_0,w)=\tfrac{1}{2\pi i}\int_{\gamma_0}\tfrac{1}{z-w}dz$$
Well $w\in \Bbb C-G, z\in G$ so therefore $f(z)=\tfrac{1}{z-w}$ is holomorphic in G.
As such $\int_{\gamma_0}f(z)=\int_{\gamma_1}f(z)$
Which means $\int_{\gamma_0}\tfrac{1}{z-w}dz=\int_{\gamma_1}\tfrac{1}{z-w}dz$
In other words
$$\tfrac{1}{2\pi i}\int_{\gamma_0}\tfrac{1}{z-w}=\tfrac{1}{2 \pi i}\int_{\gamma_1}\tfrac{1}{z-w}$$
So therefore
$n(\gamma_0,w)=n(\gamma_1,w)$
