So i've been asked to show that $Res(f,\infty)$ does not depend on the radius R. but first let me give you some back story here.
let $f: \mathbb{C} \longrightarrow \mathbb{C}$ be analytic, except for a finite number of poles $\{z_1,z_2,...,z_m\}$ and define the residue of f at infinity to be $$ Res(f,\infty) = -\frac{1}{2 \pi i}\int_{S_{R}^{+}(0)}f(z)dz$$ Where $R > 0$ is so large that f has no singularities in $A_{R,\infty}(0)$ Show that $Res(f,\infty)$ does not depend on R
Note: Where $S_R^{+}(0)$ is the positively orientated circle around $0$ of radius R (eg, $|z| = R$)
now, i've found this to be a little confusing anyway since we've had to define R large enough to accept all singularities of f. So please by all means extend a helping hand if you see that i'm doing this complete wrong.
The only major thinking i can consider is this. Since $$ Res(f,\infty) = -\frac{1}{2 \pi i}\int_{S_{R}^{+}(0)}f(z)dz$$ we can rewrite this to be the sum of the residues but we could also consider the Cauchy-Goursat Theorem, and essentially the contour $S_{R}^{+}(0)$ will simply be replaced by a sum of smaller contours each containing only a single singularity.
ie. $$Res(f,\infty) = -\frac{1}{2 \pi i}\int_{s_{R}^{+}(0)}f(z)dz = -\frac{1}{2 \pi i}\sum_{i=1}^{m}\int_{S_{\epsilon}^{+}(z_i)}f(z)~dz$$ where $\epsilon>0$ is large enough to capture the single singularity with in question. (for argument sake we could take it to be 1)
Am i on the right lines? is this correct? and if so, how can i make this rigorous (though the last part is not necessary).
Thanks for the help!