Use Laurent decomposition to prove the next equivalence, assuming that $\gamma_1,\gamma_2$ are closed curves such that they live in the annulus $r<|z|<R$: $$n(\gamma_1,0)=n(\gamma_2,0)\,\text{ iff }\int_{\gamma_1}f(z)dz=\int_{\gamma_2}f(z)dz,\,\forall f\text{ holomorphic in the annulus}$$
$\Rightarrow$)If we suppose that $n(\gamma_1,0)=n(\gamma_2,0)$, then by definition $\frac1{2\pi i}\int_{\gamma_1}\frac{dz}{z}=\frac1{2\pi i}\int_{\gamma_2}\frac{dz}{z}$. I'm considering two cases, if $\gamma_1$ goes around $0$ and if $\gamma_1$ doesn't. In the second case, if $\gamma_1$ doesn't go around $0$ then $n(\gamma_1,0)=0$, hence $\gamma_2$ either goes around $0$, then using Cauchy-Goursat's theorem $\int_{\gamma_1}f(z)dz=\int_{\gamma_2}f(z)dz=0$, and we have what we wanted. Now for the first case, we assume $n(\gamma_1,0)=n(\gamma_2,0)>0$, that means both curves do go around $0$, and moreover they go around $0$ the same amount of times. The problem starts saying that we have to use Laurent decomposition (i.e.,Laurent series development), this is where it gets tricky. We know that: $$f(z)=\sum_{n=-\infty}^\infty a_n z^n=\sum_{n=-\infty}^\infty z^n\frac1{2\pi i}\int_\Gamma \frac{f(z)}{z^{n+1}}dz$$ with $\Gamma$ a circle in the annulus such that goes around $0$, in this case. Then: $$\int_{\gamma_1} f(z)dz=\int_{\gamma_1} \sum_{n=-\infty}^\infty a_n z^n=\sum_{n=-\infty}^\infty \int_{\gamma_1} a_n z^n$$ $$\Rightarrow \int_{\gamma_1} \left( z^n\frac1{2\pi i}\int_\Gamma \frac{f(w)}{w^{n+1}}dw\right) dz$$
And I don't know what to do with this, is it really necessary to use the Laurent decomposition? Or am I using it wrong?
On the other hand I see that $\Gamma$ is not necessarily homotopic to $\gamma_1$, since $\Gamma$ only goes around $0$ one time and we don't know what happens with $\gamma_1$. In the beggining I tried using the fact that $0=\frac1{2\pi i}(\int_{\gamma_1}\frac{dz}{z}-\int_{\gamma_2}\frac{dz}{z})=\frac1{2\pi i}\int_{\gamma_1-\gamma_2}\frac{dz}{z}$ but I didn't got any where.
I made it to the answer, hope someone find this useful.
$\Rightarrow$) Lets assume $n(\gamma_1,0)=n(\gamma_2,0)$. Now, using Laurent's decomposition: $f(z)=\sum_{n=-\infty}^\infty a_n z^n$, where $a_n$ is a constant for each $n\in\mathbb Z$, then $\displaystyle\int_{\gamma_1}f(z)dz=\displaystyle\int_{\gamma_1}\displaystyle\sum_{n=-\infty}^\infty a_n z^n \; dz=\displaystyle\sum_{n=-\infty}^\infty a_n \displaystyle\int_{\gamma_1}z^n dz$. On the other hand, we know that if $n=-1$ then $\displaystyle\int_{\gamma_1}z^n dz=\displaystyle\int_{\gamma_1}\frac{dz}{z}$, and if $n\not=-1$ then $\displaystyle\int_{\gamma_1}z^n dz=0$, hence $\displaystyle\int_{\gamma_1}f(z)dz=a_{-1}\displaystyle\int_{\gamma_1}\frac{dz}{z}=a_{-1}n(\gamma_1,0)$. With a analogous process we have that $\displaystyle\int_{\gamma_2}f(z)dz=a_{-1}\displaystyle\int_{\gamma_12}\frac{dz}{z}=a_{-1}n(\gamma_2,0)$. But by hypothesis we have that $n(\gamma_1,0)=n(\gamma_2,0)\Rightarrow a_{-1}n(\gamma_1,0)=a_{-1}n(\gamma_2,0)$, hence $$\int_{\gamma_1}f(z)dz=\int_{\gamma_2}f(z)dz.$$
$\Leftarrow$) Lets assume $\displaystyle\int_{\gamma_1}f(z)dz=\displaystyle\int_{\gamma_2}f(z)dz$, and using the above, we have that $a_{-1}n(\gamma_1,0)=a_{-1}n(\gamma_2,0)\Rightarrow n(\gamma_1,0)=n(\gamma_2,0)$.