Here is my question: Suppose $f$ is analytic in the region $\{z:|z|>1\}$ and that $\lim_{z\rightarrow\infty}f(z)=0$. Show that if $|z|>2$ then $-f(z)=\frac{1}{2\pi i}\int_{z=2}\frac{f(\zeta}{\zeta-z}d\zeta$.
My thoughts were that if $|z|<2$ then the integral would just equal $f(z)$ by Cauchy's formula since $z$ would lie within the contour. So, since our $z$ lies outside of the contour, then the sign on the integral would be reversed (now that I type this, I am not sure if I can claim this...). On the other hand, I wouldn't be using the given assumptions, so I am thinking that the problem requires a bit more than this. Maybe I could try and use some radius of convergence stuff? Any thoughts, suggestions, ideas, etc. are greatly appreciated! Thank you!
Let $R>|z|>2.$ Then by the residue theorem, $$\frac 1{2\pi i}\int_{|\zeta|=R}\frac{f(\zeta)}{\zeta-z}~d\zeta-\frac 1{2\pi i}\int_{|\zeta|=2}\frac{f(\zeta)}{\zeta-z}~d\zeta=f(z).$$ By the residue theorem again, $$\frac 1{2\pi i}\int_{|\zeta|=R}\frac{f(\zeta)}{\zeta-z}~d\zeta=\frac 1{2\pi i}\int_{|\zeta|=R_1}\frac{f(\zeta)}{\zeta-z}~d\zeta$$ for any $R_1>R.$ Now $$\left|\frac 1{2\pi i}\int_{|\zeta|=R_1}\frac{f(\zeta)}{\zeta-z}~d\zeta\right|\leq \frac{2\pi R_1}{2\pi(R_1-|z|)}\max_{|\zeta|=R_1}|f(\zeta)|$$ tends to $0$ as $R_1\rightarrow \infty.$ It follows that $$-\frac 1{2\pi i}\int_{|\zeta|=2}\frac{f(\zeta)}{\zeta-z}~d\zeta=f(z),$$ hence the result.